point-set-topology

[gritty widget]

if you had instead {0} u [1/70, 2], 0 would still be isolated there, but you'd need to use a different delta to prove it

[chilly lynx]

but any real could be isolated? if I take a delta large enough so that the intersection is different from empty

in you example if a take 20 and delta = 19

so 20 is a isolated?

[gritty widget]

20 is not an element of that set

also 1 is not isolated here

[chilly lynx]

ahh

[sterile fossil]

That's a topological concept so it makes sense being in here

[gritty widget]

the way its stated makes it clear that it comes from some calculus course rather than topology, even if technically it is a topology concept

[worthy olive]

Is there an inverse to homology and cohomology groups?

Like a way to go from a group to an object?

[unreal stratus]

There are a few ways to look at this I guess. But if you mean in the sense of getting a space X from a collection of groups in a manner inverse to forming homology groups, then this won't really work in general since different spaces can have isomorphic homology groups

But given some sequence of (abelian) groups A_0, A_1,... we can construct spaces X with H_i(X) = A_i. Probably the most important example are the so called Moore spaces M(A,n) which satisfy H_n = A and H_i = 0 for i not 0 or n (and H_0 = Z)

[vestal herald]

how can i show that the punctured unit circle is homeomorphic to (0,1)?

[gentle girder]

well it depends on how much rigor you have to have, you could actually just write down an explicit homeomorphism (and then show that the map you wrote down is actually a homeomorphism)

if you wanted to

[rancid umbra]

think complex exponential

[merry geode]

In Morse theory, given smooth M -> R with 2 critical points one can homeomorphically recover M as sphere.
But Morse theory is homotopic, so how do you go figure out that n-cell is involved?

[limpid fern]

this isnt true? punctured unit circle is homeomorphic to S^1 which is compact, but (0,1) is not

did you mean [0,1]/{0,1}

where the points 0,1 are identified

[vestal herald]

maybe i used the wrong terminology i meant S^1 with a hole in it

[umbral panther]

Puncturing usually makes things not compact

[gentle girder]

punctured unit circle isn’t compact of course, it’s not closed

it’s also not homeomorphic to S^1

[limpid fern]

oh i thought they meant closed unit circle, whoops.

[unreal stratus]

Wdym by closed unit circle?

I guess you mean closed unit disk. But the punctured (closed) unit disk is not homeomorphic to S^1 anyway, but rather homotopy equivalent

[empty grove]

You're good, derpz was just derping. You can explicitly write down a homeomorphism (0,1) → S¹ - point, in terms of the exponential function e^(ix). Visually it would look like wrapping the interval in a circle once, and the puncture point will correspond to the fact that 0 and 1 are missing from (0,1)

[unreal stratus]

in other words this is that you can write everthing on the unit circle as e^it and you usually have issues with making this well-defined etc but now you've removed a point it's fine

[stuck geyser]

So, is it relatively easy to show that R^n under the supremum norm and standard norm generate the same topology?

[red yoke]

Yes

[stuck geyser]

More specifically, that every L^\infty ball on R^n is open

actually if y is in that hypercube of side length d then there is a ball contained in that hypercube that is centered at y

[haughty yew]

I have just learnt that the multiplication of a topological group need not be a closed map, for example, [0,sqrt(2)) + [0,2-sqrt(2)) = [0,2) in Q.

What are some sufficient conditions for the multiplication map to be closed?

[plain raven]

You may very well be already familiar with this fact, but if G is compact and Hausdorff, then the multiplication is closed.

More generally, a continuous map from a compact space to a Hausdorff one is closed. If G is compact and Hausdorff, then so is G x G.

[gritty widget]

Whats the intuition for open and closed maps?

[haughty yew]

Ohhh

Thanks

[plain raven]

I don't know if they have many interesting properties on their own but covering space maps are closed, and this is useful in proving theorems about them.

Here's a cool proof that there are no non-constant holomorphic functions on a compact Riemann surface (such as the projective complex plane)

[rich trail]

when we talk about graphs as topological spaces, how should one think of them

as CW complexes?

[plain raven]

there are a few definitions of graph, but all of them are CW complexes

more restrictive definitions of graph will give abstract simplicial complexes or delta complexes depending on how restrictive you get

but probably the most general notion of directed graph most people care about is simplicial sets which are degenerate above degree 1

for undirected, replace simplicial set with symmetric simplicial set

[iron lynx]

hi idk if this is the right channel but how is an open set in the product metric X x Y defined?
not sure on how to use latex on discord but i’ll try. is a set U open in $X \times Y$ if $\forall x \in X, \exists \epsilon > 0$ such that $B_{\epsilon}(x) \subseteq U$ and if $\forall y \in Y, \exists \epsilon > 0$ such that $B_{\epsilon}(Y) \subseteq U$

[gentle osprey]

Percy Jackson

[iron lynx]

U is a set in X x Y tho so idk how that works out

[iron lynx]

ah nvm i got it

[red yoke]

Finite degree coverings?

[plain raven]

Yes, this is what I had in mind.

[rich trail]

the topology on graphs is the weak topology?

[plain raven]

if G is a graph, a function G -> X should be continuous if and only if it is continuous when restricted to each edge.

[earnest ibex]

Could somebody help me understand the equality at the end?

[plain raven]

Ah, I see. When they say "W intersects the closures of only finitely many cells", you should keep in mind that the closures of the cells form a cover of the space, and this more strongly means that W is contained in the union of those finitely many cells. So you can read everything after that as happening in the union $\overline{e}_1\cup\dots\cup \overline{e}_m$. In particular, $W\setminus A \subset \overline{e}_1\cup\dots\cup \overline{e}_m$, so of course $(W\setminus A) = (W\setminus A)\cap \overline{e}_1\cup\dots\cup \overline{e}_m$, and then the rest is clear I think

[gentle osprey]

diligentClerk

[tidal lynx]

I know that a square is homeomorphic to a circle, so a square is then a 1-manifold. But I don’t see what the charts could possibly look like at the corners. I know I could just compose with the homeomorphism to get something but that doesn’t really give intuition

e.g if I keep zooming into the corner the area around it doesn’t look any straighter

[gentle girder]

you can make the chart piecewise, the charts don't have to be smooth as a map from the subspace to the line

in fact, they will not be smooth charts, and you are right that if you keep zooming in, the corner will remain a corner, but if you are thinking about it that way you implicitly are assuming there is extra structure in the square that is not actually there just as a topological manifold

[tidal lynx]

Oh I see what you mean

like V is homeo to —

[gentle girder]

yeah that’s right

[rich trail]

is that what uses the lifts and stuff

[rich trail]

I dont understand what people mean when they say $[\gamma]=0$

[gentle osprey]

kevinhardy2

[unreal stratus]

Presumably these are elements of the fundamental group. since this is a group the 0 should be referring to the identity i.e. class of the constant loopat basepoint

though imo usually you should say 1, since the fundamental group is generically non-abelian and multiplicative notation is preferred

[umbral panther]

Probably homology, not fundamental group

[unreal stratus]

Maybe yeah

[shut vessel]

Hey! Topology is it relates to application might be fundamentally flawed!

It places the focus on a voided position relative to the observer and axis notation.

This voided position of focus creates a flawed picture.

It proposes the center of focus as the center of mass. Sentient beings we are, can't help but place our consciousness in this focal point.

To address this, you assert that 0 is 1, through means of higher dimensional rotation and arrangement of axis to a new focal point.

If 0 is 1 in topology, doesn't that mean 0 is 1 in every field related to topology, through manner of reproduction and refraction?

This creates the presumption that 0 is solid, and tangible. However, 0 is only solid and tangible, relative to logical arrangement with surrounding points, as arranged and interpreted by a neutral, extradimensional 0

[topaz plover]

Can this please be on the topology qual
I think I might be able to pass it in that case

[swift fjord]

Are you actually about to understand any of this?

[topaz plover]

No but it's probably way easier to BS

[empty grove]

@true robin is the last line true?

[true robin]

The last line is true, but completely vacuous since zero is the most illogical of all mathematical concepts. When we talk about the solidity of zero relative to the logical arrangement, we implicitly assume we are talking about Galilean relativity, but since we now know about Einstein’s general relativity, the arrow of time becomes meaningless and we can only talk about logical arrangements with respect to lorentzian space time

[gritty widget]

beautiful

[empty grove]

Exactly the kind of answer I wanted

[true robin]

It is really sad that the average man will never get to see the depth of understanding we now have about the natural and platonic world.

[gritty widget]

🤧

[unreal stratus]

"Hey! This entire field I probably know little about is possibly completely wrong"

[ebon galleon]

Hello GRAPE APE,

What the fuck are you talking about.

Best,
cosimplicial ryxolution of Δ[1]

[ivory dragon]

this is a field of study FOUNDED on being visited by the spirit of your late best friend in your dreams

lets not judge nonconventional sources here

[gentle girder]

lol what is happening

[rich trail]

re: proof of fundamental thm of algebra using fundamental group

why is the polynomial that is used monic?

[gentle girder]

you’re a crank lol

[rich trail]

any1 know

[empty grove]

You got an issue with what he said? Then rebut the content, don't call the guy a crank

[rich trail]

first line is always “We may assume the polynomial is of the form p(z) = zn + a1zn−1 + ··· + an .

[true robin]

This guy gets it

[ivory dragon]

divide by the leading coefficient

[gentle girder]

yeah because you can consider p(z)/a_n

[ivory dragon]

if ax^n + whatever = 0, then dividing by a gives x^n + whatever = 0

[gentle girder]

and if p(z) has

okay

thanks nami

[ivory dragon]

owned

[ebon galleon]

pwned

[gentle girder]

i’m on my phone lol

[rich trail]

dont you need an equation to do this

[ivory dragon]

not sure what you mean

[rich trail]

why is the lhs not p(z)/a_n

[ivory dragon]

the point is that it suffices to prove it for monic polynomials

since we can just multiply back by the leading coefficient

and the roots are the same

[rich trail]

oh right

the roots are the same

sorry

thanks

[true robin]

No problem

[ebon galleon]

Okay now back to this, got any rebuttal of what they said, smartie?

[rich trail]

Shush

[gentle girder]

they didn’t say anything

oh my god

sorry i didn’t realize that the thing that you were saying was a response to an actual nonsense statement, and the things that you’re saying actually make sense. just reading it by itself made me think that you weren’t saying anything at all @true robin

[ivory dragon]

i thought you were in on the bit

[gentle girder]

i literally don’t know any of these people i just got here

oh it’s still nonsense

lol

[empty grove]

As long as you have repented

[true robin]

That’s ok, general relativity is hard even for gifted individuals such as myself

[gentle girder]

i love you

[true robin]

I know

[gentle girder]

i know you know

it was easy for a gifted individual such as myself to figure out that you knew i loved you

[true robin]

Now you’re getting it

[gentle girder]

nope, they’re completely right, thanks for pointing that out ryx

[lusty trench]

Where can I find motivation for doing homotopy theory with simplicial sets instead of CW complexes?

(At least in my case, the motivation for CW complexes comes from fiber bundles, characteristic classes and obstruction theory.)

[umbral panther]

Simplicial sets form a category. CW complexes with cellular maps are lousy. They or spaces homotopy equivalent to them usually make sense as a full subcategory of spaces. Whereas simplicial sets are even simpler by simplifying the morphisms

Nerves of categories require you to consider simplicial sets. Homotopy limits and colimits require you to consider simplicial spaces. Why not consider spaces as formal homotopy colimits?

[lusty trench]

Yeah, it's clear to me that simplicial sets contain the combinatorial information necessary to do homotopy theory, without any homotopically superfluous nonsense. But somehow that doesn't seem like geometric motivation. It's just “what we need to make the algebra / category theory nicer”.

[umbral panther]

There are two aspects. One is that you are making it discrete. There are many ways you could do this. But the other aspect is that lots of things work out correctly. The most basic is that the product in simplicial sets yields the product in spaces. I don’t think this is a necessary detail, but it is very convenient and special to simplicial sets

[lusty trench]

Oh, okay, that's nice. Yeah, products in the category of CW complexes don't always match the products of the same spaces in the category of topological spaces.

[umbral panther]

Compare: chain complexes and simplicial abelian groups are equivalent, but the tensor product in sAB is correct

[umbral panther]

What do you mean? What category of CW complexes? Isn’t the finite product of CW complexes a CW complex? Just that there is no canonical CW structure?

[lusty trench]

Mmm, category of spaces homotopically equivalent to CW complexes, then?

Also, a category doesn't have to have products, me thinks.

I was originally thinking just the full subcategory of Top whose objects are CW complexes.

[swift fjord]

No but categories with products are much better

[zinc siren]

should get the Klein bottle shaped one

[lusty trench]

In what way is the tensor product of chain complexes $(A \otimes B)k = \bigoplus{i+j=k} A_i \otimes B_j$ not “correct”?

[gentle osprey]

Eduardo León

[umbral panther]

It derives badly.
There is no model structure on commutative dgas in positive characteristic. You can try to avoid model categories and just localize, but the forgetful functor from commutative dgas to chain complexes does not have a left adjoint

[formal tide]

What's the name of the operator B in expression such as "B Z = S1"?

seen it a couple of times but noone ever names it so can't search it

[cedar pebble]

delooping

[formal tide]

thank you

[lusty trench]

I guess what I really want is an example of a pedestrian geometric problem that has a nice solution if we use simplicial stuff.

"Such and such categorical construction works nicely" is, well, nice, but not really geometric.

[unreal stratus]

You'll also want to look at classifying spaces ig

[umbral panther]

Since the geometric realization preserves products, the classifying space of a topological abelian group is again a topological abelian group

[lusty trench]

What's the Abelian group structure on BC^* = CP^oo?

[umbral panther]

There are ad hoc answers, like the projectivization of the vector space of rational functions on S^2 is a group under multiplication, but this gives a general answer

The Dold-Thom theorem that the homotopy groups of the infinite symmetric product are the homology is easy in the simplicial setting. It’s the Dold-Kan identification of simplicial abelian groups with chain complexes plus the definition of homology

[lusty trench]

I can see why CP^oo could be an Abelian group in the homotopy category. But I don't see any on-the-nose Abelian group structure on the point-set space.

[umbral panther]

If to take the projectivization of polynomials, it’s a topological monoid

If you take rational functions it has inverses, but people get nervous about the topology

[lusty trench]

Oh, so we think of C^oo as the underlying vector space of the polynomial ring C[x]. I see, thanks.

[unreal stratus]

One nice thing is that you can show the H-space structure on CP^oo is unique up to homotopy

[lusty trench]

Do these homotopies come from C^*-equivariant self-homotopy equivalences of C^oo?

(Where C^* acts linearly.)

[plain raven]

there is a theory of fiber bundles of simplicial sets, idk if you're familiar with this.

see May, "Simplicial Objects in Algebraic Topology"

obstruction theory also works, at least postnikov towers

[nova fjord]

Question: I'm trying to show that no other closed surface except S^2 is homeo to a suspension over any other space.

My rough argument is suppose a surface S were iso to a suspension Sigma X over some space X. Since S is a manifold, small neighborhoods are homeo to disks. Neighborhoods of the vertices of the suspension though are iso to the cone CX over X.

I want to argue that this means X = S^1 since CX must be iso to a disk, but I don't think disks arise just from cones over circles. Why can I exclude the other possibilities in this case?

[umbral panther]

I think that’s true, but it’s easier to prove the stronger statement that the other surfaces aren’t homotopy equivalent to a suspension

[nova fjord]

Just from classification theorem for compact surfaces?

[plain raven]

I kind of don't agree with your viewpoint, algebraic topology has always considered combinatorial structures such as simplicial complexes. Simplicial sets are just one particular formalism for simplicial complexes. CW complexes are kind of a blend between combinatorial and topological, they're combinatorially generated by families of disks/balls but "topological" in the sense that the attaching maps can be arbitrary continuous maps. the combinatorial structure is exploited to define CW homology. A map between CW complexes which sends n-cells to n-cells obviously induces a map between CW homology groups; then I would hazard a guess there's a known theorem that states every map between CW complexes can be homotoped to one which sends n-cells to n-cells, and that lets you explain what the induced homology map is of an arbitrary continuous map. So you're exploiting the combinatorial structure, trying to reduce general concepts of topology such as continuous maps to ones that respect the combinatorial structure (cellular maps)

[unreal stratus]

Algebraic topology was originally known as combinatorial topology anyway ig

[plain raven]

Right, yeah.

[unreal stratus]

Or at least that was a precursor to much modern stuff

[plain raven]

I mean you can always take the geometric realization of a simplicial set, and you can think that the geometric realization "is" the simplicial set if you want, and then simplicial maps are just continuous maps respecting the simplicial structure and which are given by extending in the obvious affine-linear way, using barycentric coordinates on the interior of cells

[unreal stratus]

Silly q, how useful are simplicial complexes as simplicial complexes (as opposed to simplicial sets or their minor variants)

Like one of my advisors said you "should" be using simplicial sets instead at least for stuff like singular homology and personally I've not used them at all

[umbral panther]

You have to subdivide the source before you can homotope to preserve cells

[unreal stratus]

But I imagine they are useful for, say, applications to computations w computers

[unreal stratus]

Ye

Simplicial objects feel like a kind of amazing discovery tbh

[plain raven]

@unreal stratus I might have said this before but there's a pretty clear nerve-realization adjunction between simplicial complexes and presheaves on FinCard and i'm pretty sure (did the proof in my head and handwaved some stuff) that via the nerve, simplicial complexes embed as a fully faithful reflective subcategory of presheaves on FinCard (call them "symmetric simplicial sets" if you like, idk what the prevailing term is)

it's fully faithful iff the counit is an isomorphism and my gut strongly says the counit is an isomorphism here, mentally visualizing the colimit lol

[plain raven]

I would think that abstract simplicial complexes are just as adequate for computers? They're all discrete combinatorial structures

[umbral panther]

Simplicial complexes don’t have an obvious definition of morphism. Simplicial sets being a category is a big advantage

[plain raven]

You can define the "singular simplicial complex" as a delta set (semisimplicial set), simplicial set, or symmetric simplicial set and you get the same homology theory either way.

so your advisor would have to provide some other reason why simplicial sets are "Right" for this

[lusty trench]

Yes, I agree that the combinatorial structure is what makes CW complex nice. However, my advisor wouldn't allow me to take it as a given that the combinatorial structure is interesting on its own. I have to start with a problem with manifolds, and only then would he allow me to progressively transform it into a more algebraic or combinatorial one.

[plain raven]

I would accept that different parties could come to different conclusions about what the definition of a morphism is/should be.
To me, there is a clear notion of morphism. A simplicial complex is a pair (K, V) with K a set and V a set of finite subsets of K, closed downward. A morphism (K, V) -> (K', V') is a function f : K -> K' such that for each simplex sigma in V, the image f(sigma) is a simplex in V'.
It's not clear to me whether this definition of morphism has close cousins elsewhere

[umbral panther]

That doesn’t admit a functor to topological spaces, does it?

You said they embed fully faithfully in simplicial sets. I really doubt it. I think that defines a different notion of morphism

[plain raven]

Just to clarify, I said presheaves on FinCard. Here the distinction matters because in simplicial sets the vertices are inherently ordered, but in presheaves on FinCard the vertices are not ordered. If (K, V) is a simplicial complex and sigma is an n-simplex, then in the presheaf on fincard associated to (K,V) there would be n! nondegenerate simplices of degree n associated to sigma, each one coding one possible permutation of the vertices; because in FinCard all automorphisms of [n] are in Hom([n],[n]), in the geometric realization all of these will be identified as a simple simplex.

But if you're interested i'll try and write up a careful proof later and we'll see if there are significant problems that arise

[plain raven]

If f : K -> K' is a map on vertices, and sigma is a vertex in K, with interior point (t0 * v0 + t1 * v1 + ... + tn * vn), then we send this to (t0 * f(v0) + t1 * f(v1) + ... + tn * f(vn)). The assumption that (f(v0), .... f(vn)) is a simplex of (K', V') means this point is at least well defined; to me it's clear that it's continuous on individual simplices and thus continuous overall by definition of the colimit topology

Ok, cool. I found a reference. This is in Algebraic Topology by Edwin Spanier, chapter 3, Polyhedra.

And it is a functor.

[solemn oar]

I knew I'd seen it somewhere

[knotty vine]

Depends on the application: for TDA simplicial complexes work well, but for effective algebraic topology (see Kenzo) you really need simplicial sets

[plain raven]

I do concede your advisor's point, I mean, certainly manifolds are a rich and interesting field. However manifolds are such rich structures, having so many interesting properties that for many of the beautiful theorems you might naturally think: Do I really need the full strength of smooth manifolds for this theorem to be true? Is all this essential to the proof?
And that leads us to try and abstract away and strip things down in some quest for the "deeper reason" these things are true, trying to elucidate the core logical structure that makes it work or create some more parsimonious mathematical objects with which we can prove the same theorems, thus highlighting the essential aspects of the proof.
Maybe not all mathematicians share this POV, but for me at least, a proof may leave me unsatisfied in the sense that I still haven't appreciated the deeper reason it is true and it might spur me to go look for that deeper reason, see if there's some structure that can be elucidated to tease out the really fundamental ideas.

I think there's something to be said for the fact that the existing literature on simplicial homotopy theory has been able to reproduce many known theorems about fiber bundles, that they really have succeeded in this mission of stripping down to the essentials the bare minimum that is really needed for the proof to work. It's maybe a shame that the smooth manifolds are gone, but honestly I think that just means it wasn't really a theorem about manifolds to begin with, it was a more general theorem about the structure of these objects and maps which can be specialized to manifolds and smooth maps

[plain raven]

Thank you for making that statement. Can you justify it?
"Kenzo uses simplicial sets" is of course not a justification for "For effective algebraic topology you really need simplicial sets."

I'm not denying the utility of simplicial sets. I think simplicial sets are great. They are obviously necessary for some things. For example, the category of delta sets does not possess an identity homotopy id_f from a map f : X -> Y to itself. That, obviously, is fatal to some applications.

[knotty vine]

For one it's much easier to construct, say, and n-torus as a simplicial set than a complex

[plain raven]

Ok, I see, different interpretations of "need".

[knotty vine]

Well it becomes more of a need when the simplicial sets aren't compact anymore

There's a bunch of theory of how Kenzo is able to compute homology groups of such non-effective spaces which I don't understand

[plain raven]

Interesting, I'll have to look into that.

[knotty vine]

Would it be possible to make something like the infinite dimensional complex projective space as a simplicial complex?

[lusty trench]

I'm actually sympathetic to your POV, and I think I even agree to some point. However, most professors at my university have trouble caring about abstract nonsense. (There's just one, and he's often busy traveling abroad.) In fact, so much that they would rather do laborious computations in local coordinates of manifolds, just to avoid having to work with fancier abstractions.

For my master's thesis, I had to count singular points of 1-dimensional holomorphic foliations on compact complex manifolds. Of course, the neat conceptual way to do this is to notice that

• 1-dimensional holomorphic foliations on a manifold M are given by holomorphic sections of E = TM (x) L, for some holomorphic line bundle L on M.
• Assuming we have a section s of E with isolated zeros, the number of zeros of s is the cap product of E's top Chern class with M's fundamental class (compatible with the canonical orientation of M as a complex manifold).

And this is how I came to care about CW complexes and obstruction theory. It actually saved me the effort of doing tedious computations with differential forms in local coordinates.

Maybe simplicial sets are more natural if I start with problems in combinatorics rather than differential geometry? I'm not opposed to combinatorics myself, but again, I'd rather start with a concrete pedestrian problem, and only then use it to motivate fancier abstractions.

[unreal stratus]

sure i mean i am conflating ASC and SC

[wicked ledge]

What do you all find interesting in alg-top particularly

[rancid umbra]

the first homotopy group uses loops to detect properties about a space. is it incorrect to think that the second homotopy group uses surfaces to detect properties about a space?

[lusty trench]

"Detect properties" sounds rather vague.

[visual rock]

you are specifically using the two sphere

[kindred cairn]

the fundamental group looks at (homotopy classes of) maps from S^1 to a space. higher homotopy groups look at (homotopy classes of) maps from S^n to your space

[rancid umbra]

so how do people typically think about the image of what a homotopy class representative function looks like?

in the first homotopy group, we think of them as loops based at a point. what is the generalization from here (if any)?

[visual rock]

spheres based at a point

[lusty trench]

I can't speak for others, but for me, algebraic topology solves many differential geometry problems much more elegantly than if you only have differential geometry tools. Algebraic topology has just the right tools to discuss the global features of a space directly and efficiently (e.g., "this manifold contains a noncontractible loop"), rather than indirectly through the effect of coordinate changes ("you can construct a nontrivial line bundle if you use these transition functions").

Also, in some sense, algebraic topology deals with very fundamental and unavoidable objects. My advisor couldn't care less about Eilenberg-MacLane spaces or Postnikov towers, and yet they appeared in a solution to a problem he gave me.

[rancid umbra]

thanks 🙂

[lusty trench]

The full answer is in a Postnikov tower, of course.

[cedar pebble]

this is a reasonable way to think about it yes

in some ways you can think of higher homotopy groups as detecting "higher loops"

like \pi_2 of a space is \pi_1 of its loop space

there is also a way you can think of higher homotopy groups as classifying "higher covering spaces" in the same way that fundamental groups classify covering spaces, it's just hard to write this down explicitly

[rancid umbra]

sweet

thank you

[cedar pebble]

the other nice interpretation is like

\pi_1 sees locally constant sheaves on your space

you can generalize sheaves to things like stacks or even infinity-stacks (by looking at sheaves with values in groupoids or even infinity groupoids, with some much more complicated sheaf gluing condition needed to deal with higher coherences)

\pi_2 sees locally constant stacks on your space

the whole homotopy type sees all locally constant infinity stacks

[umbral panther]

Does it make any more sense to say that pi_2 sees locally constant stacks than to say that H_2 does?

[cedar pebble]

yes

I mean one should really say "truncated homotopy type" here but whatever

[west brook]

Is there some sort of structure that models p-adic manifolds similarly to how simplicial complexes model real manifolds?

[toxic garnet]

true or false, any vector bundle of dim n over B is sub bundle of the product bundle B x R^n. If true, can you formulate bunldes in terms of (co)lim?

[coral pawn]

By P(Sm_S), do homotopy theorists usually mean presheaves valued in sets or simplicial sets?

[umbral panther]

Ultimately they must mean sheaves of spaces. But maybe this object is a sheaf of sets, which can then be thought of as a sheaf of spaces

[unreal stratus]

Well there aren't many dimension n subbundles

[prime elbow]

I am not sure about C intersect Bd A, what is meant by Bd ?

[alpine nest]

Boundary, usually

[prime elbow]

Okay, thank you

[heady skiff]

not every connected set is convex right?

wtf is emeritus

[limber wren]

nope

[quiet thorn]

Very active but not anymore

[heady skiff]

:(

[limber wren]

you've retired, but maintain your titles and glories

[quiet thorn]

You keep your channel privileges without the color

[kindred cairn]

of course not. what about the converse ?

[umbral panther]

The empty set is convex but not connected

[knotty vine]

thats mean

[kindred cairn]

the empty set is a trouble maker

[slender glen]

is the empty set convex?

[knotty vine]

Yeah, it's the convex hull of the empty set

[slender glen]

I vaguely remember a text I read explicitly requiring convex sets to be nonempty

[knotty vine]

You can define it as whatever you want of course. If you take that approach you might sometimes have to say: "Let S be either empty or a convex set". If you take the other approach you might sometimes have to say: "Let S be a non-empty convex set". Doesn't really matter.
But it does seem to me that the empty set being convex is fairly natural, for the convex hull reason. Then again I don't know much about this field.

[cosmic oracle]

I have a question.
How are these two definitions equivalent?
(1) Let X be a topological space. A point p in X is called a limit of a subset of a X if and only if every open set G containing p contains point a point of A different from p.
(2) A point p in X is a limit point of A - a subset of X, if and only if every open neighborhood contains a point of A other than p.

Note : The definition of Neighbourhood in the book is this : Let p be a point of a topological space X. Then a subset N of X is called neighborhood of p if and only if N is a superset of an open set G containing p.

[iron kite]

Did you mean “… of a subset A of X if and only if every open set G…” in (1)?

[cosmic oracle]

Yes yes.

Sorry for the typo.

[iron kite]

What have you tried? One of these definitions implies the other, can you figure out which one? What happens in situations where the converse implication is falsified?

[uneven bronze]

I have a basic question (from my measure theory notes). Let F be a compact subset of R^n that contains A (here A is a bounded Lebesgue measurable set, but I don't think this fact is important). Let G be an open set such that it contains F \ A. Then the claim is that K=F \ G is compact. I can not work this out. Is it true, and if yes, how can I convince myself of this? My drawings haven't helped that much.

[winged viper]

K is a closed subset of F right? Or am I missing something

[uneven bronze]

I think K will simply be a subset of A.

[winged viper]

But G is open, so F - G should be closed in F

[unreal stratus]

Well it's closed by virtue of being the complement

ye

[winged viper]

So being a closed subset of a compact set implies it is compact

[uneven bronze]

ok, I have to think this through 👍

[sterile fossil]

You don't need most of the claim for the latter result actually, only the fact that G is open

and that F is compact

[uneven bronze]

yeah, I think the other claims were distracting me 😄 now it's a lot clearer, thank you

[prime elbow]

Every open interval in R is connected?

[toxic garnet]

does anybody know what is the definition of collapsed vector bundle (in Husemoller) in standard literature? Can't find anything about it online

[kindred cairn]

every interval is connected

[prime elbow]

So also closed intervals?

Is R/ {a } disconnected space ? If yes, how can I show that?

[kindred cairn]

every open interval is connected, so is every closed interval

R minus one point is disconnected. what is the definition of connected ? from there you can show it

[prime elbow]

There is no other subset which is both open and closed

[kindred cairn]

i suggest working with the definition: a set is connected if it cannot be written as the disjoint union of two open subsets

[prime elbow]

Because every closed interval can be written as union of open intervals ?

[kindred cairn]

no. because any union of open intervals will give you an open set. and a closed interval is not an open set (unless its the whole line)

[prime elbow]

Yes, got it.

But I took a set (a+1/n , b-1/n ) and took Union

[prime elbow]

It can be generalized to arbitrary X connected space?

[kindred cairn]

yes

not only in R

[kindred cairn]

the union over N is open and not closed

[prime elbow]

Okay, thank you

[prime elbow]

If I let R\ {a } be disconnected, there exists A and B which are open and closed in R{a}. I have no idea after that

[kindred cairn]

just show that R \ {a} is the union of two disjoint open sets

[prime elbow]

A = (-∞, a) and B = (a, ∞) ?

If I take an intersection with R\ {a} then new sets A_0 and B_0 are both disjoint and open in R\ {a} whose union is R \ {a}, right ?

[kindred cairn]

yes

[prime elbow]

How can I generalize it to arbitrary X connected space?

And is this only valid for finite points, X is connected then X \ {a_1,a_2,...,a_n } is connected?

[prime elbow]

For open intervals in R, intermediate value theorem works?

[foggy cairn]

Hate this solution. Can someone please help me understand why p* sends generators of pi1(T) to a^2 and b in pi1(K)? Actually what even is p*?? Is it the induced homomorphism from p that sends T to T/sigma?

[limpid fern]

Yes perfectly well

Any interval whatsoever

[prime elbow]

Okay, thank you

[hidden crag]

p* is the induced hom from the map you're supposed to construct

the generator part will depend on the concrete descriptions of the maps you end up with

[prime elbow]

I want an example such that the closure of intersection A and B is not equal to the intersection of the closure of A and B.

So if I let A = Q and B = R\ Q then its intersection is empty and the closure of the empty set is empty, right?

[prime elbow]

A topological space (X, τ) is said to be separable if it has a dense subset which
is countable.

So if X is countable with any Topology, X has dense subset X itself which is dense in X, right?

[rancid umbra]

correct

[rancid umbra]

correct

[prime elbow]

If (X, T ) is topological space and T is finite then T is Separable space depends on X.

If I take X = R and T be indiscrete topology then it is not Separable space.
And if I take X = Q and T be indiscrete topology then it is.

Is it correct?

[alpine nest]

If T is finite or countable, then the space is separable whatever X is

You just take one element from each open set, and that's your countable dense subset of X

[prime elbow]

Got it, thank you

[alpine nest]

In general the weaker the topology (the fewer open sets) the easier it is for a subset to be dense

[prime elbow]

If X is an uncountable set and T is discrete topology then T is not Separable because if any subset A, A is closed in T.

Therefore, closure of A is exactly A and thus for dense , A must be X but X is uncountable.

Is it correct?

[alpine nest]

Yes

[prime elbow]

Any hint if T is finite closed topology then every infinite subset of X is dense in X.

[alpine nest]

The closure of every set is the intersection of all closed sets that contain it.

[prime elbow]

Oh so there is no closed set which contains A except X, right ?

[alpine nest]

Indeed

[prime elbow]

Thank you

[prime elbow]

Let M =Q, the set of rational numbers with the Euclidean metric of R. Let S consist of all rational numbers in the open interval (a,b), where both a and b are irrational.

Then how S is closed subset of Q. Is this a and b are fixed?

[alpine nest]

Yes, a and b are fixed

So for example S is the set of all rational numbers between -sqrt(2) and sqrt(2)

And indeed that's a closed subset of the rationals

(it's also an open subset of the rationals)

[prime elbow]

Because it is the intersection of the closed set [a , b ] and Q ?

[alpine nest]

Yes, although you can also do it directly from definition, which is a nice exercise

[prime elbow]

Definition?

[alpine nest]

Of being a closed/open set

[prime elbow]

Okay, thank you

[umbral panther]

@lusty trench I gave you another application of simplicial sets here, in case you mixed it

The infinite symmetric product is the union of the finite symmetric products S^nX. Riemann Roch commutes this for surfaces. Dold-Thom generalizes it to say that the homotopy groups of the infinite symmetric product are the homology of the original space. Simplicial sets and the Dold-Kan correspondence make this easier to see

[lusty trench]

Wait, I just had to Google what a symmetric product is. I guess that's nice.

Mmm, how do we construct the n-th symmetric product of a simplicial set, then?

[umbral panther]

The symmetric product is the product followed by the quotient by a finite group. You have to check that geometric realization preserves both operations

[lusty trench]

Dumb question, never mind.

Yeah

I had a bruh moment. Sorry.

[umbral panther]

This shuffled around the difficulty. But under the Dold-Kan correspondence this is exactly the definition of homology

[lusty trench]

Oh, that seems cool. It gives homology a nice interpretation, similar to how Eilenberg-MacLane spaces classify cohomology groups.

[umbral panther]

Leaving side simplicial sets, an application of Dold-Thom is that the infinite symmetric product (or free abelian monoid) on S^n is a model for K(Z,n)

[lusty trench]

Oooh.

[umbral panther]

(Dold-Thom requires a base point to define the maps S^nX->S^n+1X and that acts as the unit in the monoid)

[lusty trench]

Is it possible to do homotopy theory with schemes or analytic spaces, rather than topological spaces?

[iron kite]

Yes, but from what little I understand, it’s a lot more complicated than the case of topological spaces.

[lusty trench]

Is it because of the lack of a concrete combinatorial model as in the case of CW complexes or simplicial sets?

[stuck geyser]

If C(X,Y) is the space of continuous functions from X to Y, is there any neat topology you can put on it

[alpine nest]

If Y is metric, then the topology of uniform convergence is a popular choice

You've also got the compact-open topology

Which even works without metricness

[lusty trench]

I think it suffices X being locally compact and Y Hausdorff...

[stuck geyser]

Never heard of that?

[alpine nest]

You can look it up!

[gentle osprey]

Leno

[rotund halo]

Did you try to figure out what they meant by preimages of open rays in the first link?

Idk about the mse links so maybe they have a better solution

But that first link, they are saying the two unions on the rhs are open as a union of open sets because you can write each set in either union in terms of preimages of open rays

For ex ${x : g(x) < y < f(x)}=g^{-1}[(-\infty,y)]\cap f^{-1}[(y,\infty)]$ I think

[gentle osprey]

DootDooter

[rotund halo]

I could be mixing something silly up

But I'm pretty sure the sets in the first union work out to something like that

Pretty sure the other union can be handled along similar lines.

[rotund halo]

Try and prove this to yourself first

[lusty trench]

Is it obvious that the infinite symmetric product of a connected pointed CW complex is again a CW complex?

[lusty trench]

There has to be something like a notion of "cellular group action", so that the orbit space is again a CW complex for free. But I can't find such a definition online.

[round oyster]

is anyone good with combinatorics?

ive got a simple problem

and idont havesoltuion for it i wanna confirm

[umbral panther]

I am very good at unstated combinatorics questions
The answer is 17

[mortal briar]

Assuming the Continuum Hypothesis, the answer is 6

[alpine nest]

Assuming the continuum hypothesis, the answer is strictly greater than aleph 0, but strictly less than continuum.

[red yoke]

According to all known laws of ZFC, there is no reason that CH should be false. The cofinality of א₁ is too large to rule out by König's theorem. CH, of course, is false anyway because sets don't care what youtube math cranks yap about א₁

[tender halo]

wtf is an א₁

[rugged canyon]

||no, it's just "scissors" ||

[alpine nest]

No, that's the answer to homotopy questions.

[fading vale]

this is called A^1 homotopy theory

[lusty trench]

I was under the impression that A^1 homotopy theory deals with spaces that are actually much more general than ordinary schemes. More like "what would schemes look like if their coordinate rings were ring objects in groupoids?" or something like that.

(Not that I actually know anything about it.)

[umbral panther]

It embeds then in a category that includes things that are not schemes, but the point is to learn about schemes

There are schemes and already existing cohomology theories and the main point was to define steenrod operations on these cohomology theories

[placid cave]

Hello! I'm scratching my head around the argument that intM U intN is an open saturated subset of M U_h N, does anyone know why? I've been trying to write it in the from U=q^(-1)(V) for some subset V but can't figure out who it should be 😦

[candid wyvern]

could someone explain

[hidden crag]

what's K

[candid wyvern]

a knot embeded in a torus S1 x S1

[paper wedge]

sadly no

you have to visualize this hsit

shit*

like look for it on youtube

[mortal briar]

@placid cave well Int M U Int N is saturated because points in Int M or Int N are only equivalent to themselves (the adjunction space identifies points on the boundaries only)

and it's open because of the definition of the topology on the disjoint union (and Int M, Int N are open in M, N respectively)

[placid cave]

So pretty much int M U intN=q^(-1)(q(int M U int N))?

[mortal briar]

yeah

[placid cave]

Okok, thanks!

[red yoke]

There are diagrams in the book

The torus cuts S³ into 2 solid tori

So we can look at each of those separately

If the knot has n loops longitudinalally and m loops meridionally, then the solid torus - knot deformation retracts onto a screw-like shape

Imagine placing this star with n edges in a meridional disk

And sliding the disk longitudinally to trace a shape in the solid torus

Applying a rotation so that the disk rotates m times when it goes around longitudinally once

[red yoke]

https://www.geogebra.org/m/pg5eqruw

I made a thing

[candid wyvern]

[candid wyvern]

ill check it out later thanks

[quartz edge]

thats cool

looks like rotini

i think u might have just invented campbell rotini-os

[candid wyvern]

im still lost

[foggy cairn]

Hello, why is P^2(C) minus a line the same as C^2?

[red yoke]

Screwish spaghetti be pog

[red yoke]

There is a deformation retract (grey solid torus - red torus link → green rotini)

[cedar pebble]

in general if you fix an open A^n in a P^n the complement is a P^{n-1}

[red yoke]

That restricts to a deformation retract (grey torus → green rotini ∩ torus)

[foggy cairn]

Umm not sure I follow

What even is a line here

[cedar pebble]

yeah this is my next question, by line do you mean a CP^1?

[candid wyvern]

oh yes i get this

this sounds better

but how does the green and torus manage to deformation retract into a cylinder

[foggy cairn]

The problem asks to find the fundamental group of P^2(C) minus a line and a point not on the line, but doesn't specify what line means

[foggy cairn]

The solution begins by saying "note that..this^"

[red yoke]

The green rotini is actually (S¹ × [0, 1]) / (z, 1) ~ (ze^(2πi/n), 1)

[foggy cairn]

Btw is there some sort of way I'm supposed to think of these spaces over C? For like R^n, S^n there's always some idea but like P^2(C) or C^2 just isn't really intuitive

[cedar pebble]

there is a nice way to write down CP^n in terms of homogeneous coordinates

CP^n is the quotient of C^{n+1} by the action of C* by scaling

[candid wyvern]

how do i see get from one to the other? the green to that. i can get why we have S^1 x I but i dont get the quotient part

[cedar pebble]

so you can represent points of CP^n as homogeneous coordinates [z_0:...:z_n]

where [z_0:...:z_n]=[cz_0:...:cz_n] for any nonzero complex number c

you can cover this by open subsets of the form C^n in the following way

choose one of the coordinates z_i to be nonzero

then you can rescale everything so that this z_i is equal to 1, and then the remaining coordiantes describe a point of C^n

[red yoke]

The green rotini can be parametrized by a strip S¹ × [0, 1] so that S¹ goes around the hole n times, S¹ × 0 is on the torus, and S¹ × 1 is in the black circle

[unreal stratus]

rotini

what does that mean lol

[red yoke]

^

[red yoke]

Since S¹ × 1 maps onto the black circle n times, it is quotiented

[unreal stratus]

makes sense lol

[candid wyvern]

[cedar pebble]

for example CP^1 is described by homogeneous coordinates [z_0:z_1], you have one patch where z_0 is nonzero which gives you homogeneous coordinates of the form [1:z] where z is in C, and you have another patch where z_1 is nonzero which gives you homogeneous coordinates of the form [z:1] where z is in C

[foggy cairn]

Oh. Thanks. Hold on just trying to wrap my head around this, sorry - not heard of homogeneous coords before..

[cedar pebble]

so this gives you the usual open cover of the sphere by two complex planes

[foggy cairn]

Uhm hmm

[cedar pebble]

if you think about the complement of the patch where z_i is nonzero, it's exactly where z_i=0, so these are homogeneous coordinates where the i-th coordinate is 0, which describes a CP^{n-1}

so topologicaly you get a nice stratification like

CP^n=C^n \cup CP^{n-1}

so inductively you have

CP^n=C^n \cup C^{n-1} \cup ... \cup C^0

[candid wyvern]

sorry could u mention it more explicitly or in another way, maybe i would understand that better

[foggy cairn]

Thanks a lot for your explanation, though I think I need a little more background first. 😰 What type of course are these topics usually covered in? So I can go read up a bit

[cedar pebble]

I think this perspective on projective space usually gets covered in very basic algebraic geometry (classical AG, not fancy modern AG)

[red yoke]

Set this to n = 3, m = 2

[cedar pebble]

sometimes it's covered in topology courses

[foggy cairn]

We've been following hatcher's AT chapters 1, 2

[foggy cairn]

Ah I see

[red yoke]

The rotini locally looks like three edges glued to a black circle

[cedar pebble]

algebraic geometers use this a lot because they are typically describing projective varieties as the zero sets of homogeneous polynomials in projective space

[red yoke]

It is actually a single looped strip S¹ × [0, 1] that travels three cycles around the solid torus

Thus gluing to the black circle thrice

[cedar pebble]

like if you're looking at {y^2=x^3+1} in C^2 you can look at this instead as {y^2z=x^3+z^3} in CP^2 with homogeneous coordinates [x:y:z]

[warm quiver]

I literally never would’ve guessed that rotini is an american term, that’s so interesting

[unreal stratus]

Oh lol fair

Yeah I'd never heard rotini (as a Brit), only fusilli

[foggy cairn]

Right, thanks. I've not done any algebraic geometry, so I'll dig around and see

[candid wyvern]

it looks so weird now D:

[cedar pebble]

you probably don't need much, at least the basic description of projective space is mostly just playing around with quotient spaces and homogeneous coordinates

although in general it's very interesting to study the topology of complex algebraic varieties like this, it's a very rich source of examples for which you can say a lot about

[red yoke]

But extruded short helicoidal pasta

[foggy cairn]

Cool cool, good to know👍if you happen to know of any good introductory sources I'd really appreciate it

[winged viper]

Lol someone called this pasta fusilli yesterday (the Italian name) and I was confused cuz I was like I’ve never heard of this pasta

Now I know why

[candid wyvern]

im dizzy, ive been watching it spin awhile

[cedar pebble]

poking around wikipedia might be a good start, also Harris' First Course has some basic stuff around this at the beginning

[candid wyvern]

but yes i get that now

[winged viper]

I have a question, is there a way to show pi_1(SO(n)) -> pi_1(SO(n + 1)) is surjective (for n large) without appealing to the fibration SO(n) -> SO(n + 1) -> S^n

[cedar pebble]

deep question: if you have the zero set of some system of complex polynomial equations, how is its fundamental group constrained (compared to more general topological spaces that might not arise algebraically like this)

there are all sorts of very subtle constraints on the topology that come from algebra

very basic question that motivates a ton of modern research

[foggy cairn]

Hmm..sounds interesting and also very difficult 🥲

[red yoke]

Then combining the constructions on the inner and outer solid tori you get the result

[cedar pebble]

in general it's very difficult yeah

there are coarser questions you can ask about Betti numbers and cohomology (although I guess you haven't got to cohomology yet)

[foggy cairn]

Not yet, but I plan to read up on it over the summer

[candid wyvern]

im still trying to understand them being mapped to the hole by spinning twice TnT

[cedar pebble]

one basic constraint is that if you have the zero set of some homogeneous polynomials in CP^n, the first Betti number (equivalently the rank of the fundamental group) has to be even

[unreal stratus]

how do you see this btw? oop

i've seen smth like that for kaehler manifolds

[cedar pebble]

the Betti number thing?

[unreal stratus]

ye

[cedar pebble]

yeah it comes from Hodge theory

[unreal stratus]

yeah cool assumed so

[cedar pebble]

H^1 decomposes into H^{1,0} and H^{0,1}

Hodge symmetry says these both have the same dimension

[unreal stratus]

and symmetry

gotem

i only learnt a lil bit because i looked at formality lol

[cedar pebble]

same parity constraint applies to any odd cohomology group

there is the classic example of the Hopf surface which is a quotient of C^2-{0} by a free action of a discrete group

which is a compact complex surface but which cannot possibly be Kahler since its Hodge diamond is asymmetric (hence cannot possibly be algebraic either)

[unreal stratus]

Oh hot

One thing I mentioned in a talk was uh

[cedar pebble]

Hodge theory is great

[unreal stratus]

The Kodaira Thurston surface

(Surface?)

Like it is complex and symplectic but not Kähler

[cedar pebble]

yeah it's a surface and this is another good example

in this case its fundamental group has rank 3

[unreal stratus]

And you can check it isn't formal

Which was what I was doing as an application of stuff on formality

[cedar pebble]

nice

[unreal stratus]

Rather than going via hodge theory

But yeah

I think it's like lol a case where it is easy to compute everything explicitly

With de Rham

[umbral panther]

That “hence” sounds wrong. There are a lot of conditions like symmetry that apply to both algebraic varieties and Kahler varieties, even though algebraic does not imply Kahler

[cedar pebble]

I agree, the sentence order is a bit funny

but you know what I mean

[lusty trench]

Do we have anything similar to equivariant cohomology for spaces equipped with the action of a monoid?

(I don't mind if it ends up being something horribly n-categorical or something like that.)

[empty grove]

I imagine that Elmendorf and its stable analogue are no longer true in the monoid case (mostly because not all M-manifolds would be M-CW if we just naively define M-cells as we define G-cells) which probably kills the whole theory of coefficient systems and Mackey functors. Not sure what the correct analogue would be, but maybe a better place to ask would be the alg top server

[solemn oar]

https://mathoverflow.net/questions/131249/borel-constructions-equivariant-cohomology-and-homotopy-quotients-of-monoid-ac

I doubt the Bredon notion of equivariance has any chance for a general monoid action, so best to focus on the naive Borel notion.
You can of course calculate some quantities that look like equivariant cohomology, but I don't know of any generalization of the Borel model structure to a monoid.

You might try to build one based on https://ncatlab.org/nlab/show/Introduction+to+Homotopy+Theory#ProjectiveModelStructureOnTopologicalFunctors

[empty grove]

Yeah that certainly gives a model structure

Which generalizes the Borel one ofc because same construction

The projective model structure exists whenever the codomain cat is cofibrantly generated and the domain cat is small

[solemn oar]

So you certainly get something where you can control the weqs and the fibs

[empty grove]

Ye

[solemn oar]

But will the Borel construction still give you cofibrant stuff?

[empty grove]

What do you mean?

Like does it tell you what those are to be?

[solemn oar]

Yeah

[empty grove]

It tells you the cofibrant generators yes

You take those of Top and apply the free M-space functor to those

[solemn oar]

So for any M-space X you still have that EM x_M X is a cofibrant replacement

[empty grove]

Sounds legit

[solemn oar]

Okay then go ham I guess 😄

[unreal stratus]

Borel equivariance is fun

[limber umbra]

Maybe a nonimportant question about very basic model category: can the weak equivalence of the two bifibrant replacements (fib. repl. of cof. repl. vs. cof. rep. of fib. rep.) be made functorial?
The proof I know invokes the lifting properties, so I might guess no.

[paper wedge]

probably check out #category-theory or #foundations

[ebon galleon]

Assuming they are functorial fibrant and cofibrant replacements R and Q (e.g. a functor R: M --> M along with a natural weak equivalence id_M --> R), you will get a natural zigzag of weak equivalences connecting RQ and QR. For arbitrary (co)fibrant replacements, you won't be able to do it in one step, altho I guess in nice situations you can (e.g. if everything is fibrant, take Q = id so you have a natural map Q --> Q). It doesn't really matter that much that it's a zigzag as opposed to a single step, though.

[unreal stratus]

Model cats should be fine here

Not rly foundations lol

[ebon galleon]

b-b-but it has category in the name!

[limber umbra]

maybe it could be foundations if you do type theory cough cough

[empty grove]

You mean to tell me that model categories and model theory have nothing to do with each other?

[unreal stratus]

Oh maybe that wsa the confusion lol

I say "not really" because there is a set-theoretic element to the questions

but i guess it is just solved by being careful rather than being hard set theory

[empty grove]

Barely lol

[ebon galleon]

lol

[unreal stratus]

owo

how are you moldi

[empty grove]

Goofin

As I usually am

how r u

[unreal stratus]

Hopefully in a good way mlao

[empty grove]

Absolutely

[unreal stratus]

I am okay, finishing off a masters dissertation

Have uh run out of space (there's a word limit)

[ebon galleon]

wait you're doing a masters dissertation i thought you were UG

[unreal stratus]

Well like

It's an integrated masters

[ebon galleon]

i see

[empty grove]

I had a page limit and I did so many tricks to get it under

[unreal stratus]

Like this isn't a proper masters compared to continental Europe

[unreal stratus]

aha like wht

One i've found is that because equations don't count, you can just change your style slightly to use fewer words

[ebon galleon]

Gonna put entire paragraphs into equations

[unreal stratus]

But last year for my undergrad diss I just kept chipping away at the bit where I reviewed prerequisites and merely stated that you needed to know about xyz for some things

Which I think I will have to do this time too

[empty grove]

A 0.3inch reduction to margins, 1pt decrease in font size, slightly less math mode in some places, calling proof is obvious instead of including it, getting rid of entire sections by putting in ghost references, deleting the thesis etc

[unreal stratus]

Ghost references lmao

Oh page limit lol

Makes sense

[empty grove]

100 pages

[unreal stratus]

Bruh lol

[empty grove]

Decided to not acknowledge my parents, got a page off

[ebon galleon]

"deleting the thesis" 😭

[unreal stratus]

Lol

One thing that is sad is that things like examples are not as important to the skeleton of a thesis but probably more important in terms of grading etc lol

[ebon galleon]

Do an entire project without giving examples and then observe that there are no examples of what you had in mind

[empty grove]

I got lucky because the final result was basically a long mackey functor spectral sequence calculation so I ended up having to give lots of examples along the way because they showed up in the calculation

Didn't have to worry about which ones to include

[empty grove]

I am the pope

[plain raven]

#foundations includes foundations for homotopy theory

also Abelian categories, because those are a foundation for homological algebra

[empty grove]

Damn that's a big change from what it was

Its channel description seems a lot more specific than just "foundations for any field of math"

[unreal stratus]

Isnt dC joking

Lol

[pearl holly]

My man moldi

The one and only

[tidal lynx]

how to see that :
every point has a relatively compact neighborhood implies that every point has a local base of relatively compact neighborhoods

relatively compact = precompact = a set whose closure is compact

[whole matrix]

im confused on how i can use B to define the isomorphism K

my professor told me something like this but B takes the classes to R, so i probably misunderstood him

[tidal lynx]

Damn we really killed the channel @whole matrix

[burnt garden]

I had to ask this question but what is topologie? (weird shapes that changes isn't helpful) and why it is in Analyse[Calculas I think] and Algebra.

[prime elbow]

I need an example such that X is compact but it has a proper subset which is not compact.

If I take X = [0,1] and it's compact but (0,1) is not , is it correct? But I am not sure about how subspaces work.

[cedar pebble]

yes this works

[whole matrix]

looks good to me

[whole matrix]

to summarize, it's the study of properties that don't change with continuous deformations

[prime elbow]

And I want a hint that every proper set is compact but X is not compact

[prime elbow]

Okay, thank you

[foggy cairn]

Hi, I have three questions about the solution to this:
0. How should I think about this object geometrically? Is that even a good idea?

1. How exactly is van Kampen applied here? Shouldn't it be a free product?
2. I think it should be pretty straightforward to make a delta complex structure once there's a picture. But how does one come up with the picture on the left?
   Any help/insights appreciated 😢

Usually if it's something like X x I where X is a space I imagine it as some blob in R^2 lifted to a volume (like a cylinder) in R^3. So maybe (S1 v S1) x S1 is like a double loop -> double cylinder with ends identified -> double torus....??

The picture is also like a double torus I guess... but I'm not sure

[winged viper]

1. it’s two Tori stacked on top of each other (your logic is correct)
2. the first step in that equation doesn’t use van kampens, it just uses the fact the fundamental group commutes with products
3. S^1 v S^1 can be thought of as an interval with the two endpoints + the midpoint identified. S^1 is an interval with its two endpoints identified. So to get the product S^1 x (S^1 v S^1), take the product of two intervals to get a rectangle, and then add the appropriate identifications

[foggy cairn]

Ahhh thanks so much that's super clear🙏

[earnest ibex]

If I have two dunce caps and I glue them together along their boundaries, do I get something that is homotopy equivalent to S^2?
My idea is this: if I glue them along their boundary, I will get a S^2 whose equator is identified by three paths that rotate 2pi/3 degrees and whose endpoints are identified. I can deformation retract both hemispheres to the plane where the equator lies, and I get that this structure is homotopy equivalent to a dunce hat. Since a dunce hat is simply connected, it can't be homotopy equivalent to S^2

Also, the homology of this space is the same as the homology of S^2, so I guess this shows that same homology doesn't imply homotopy equivalence

[raw hornet]

i have question about toplogy actually about knot

im learning about knot

they are saying in the book that a knot is "a knot is an embedding of a circle into three-dimensional space

i do not even get what that suppose to mean

[cedar pebble]

It’s an embedding S^1->R^3

Sometimes R^3 is replaced with S^3

You should review what an embedding of topological spaces is

If you don’t know enough point set topology for this to make sense then I would learn that first

[raw hornet]

well we did not even get a toplogy course yet

knot theory is an extra subject that we can follow

so im doing that right now

[cedar pebble]

It’s hard to follow without some basic familiarity with topology

There are ways you can study knot theory in a slightly more elementary way but it’s a handicap to properly understanding the subject

[raw hornet]

i see thx alot

i find it really interesting as well and its extra

but now im reading in the book and there are so many terms regarding toplogy

[alpine nest]

Yeah, you won't get much out of a knot theory course if you don't know point set topology

[raw hornet]

i see

well i was reading something in the book about i an internsic toplogy and externsic toplogy

and idk what that even suppose to mean

i searched on google and it was like a really difficult explanation that i do not get

[alpine nest]

Again, the advice is "learn some topology before you attack knot theory"

[raw hornet]

well how did they even let us do this course

but sure

agree about that

[alpine nest]

I don't know, I'm not them

[raw hornet]

i see

[alpine nest]

You might want to talk to the person teaching the course

[raw hornet]

but now i take it and since i have many subject i do not even have time to learn toplogy

[alpine nest]

And tell them you don't really understand what an embedding or intrinsic topology is

And ask how you could work around it

[raw hornet]

sure

do not worry

well i mean im not really sure if its a difficult term

[whole matrix]

(repost) does anyone know a way to define the Isomorphism K

[swift latch]

Is it a perfect pairing?

[whole matrix]

im not sure im familiar with that term

[swift latch]

Oopsie got it wrong. That's what you want I think

Yeah seems to work

[whole matrix]

so like this?

[whole matrix]

so in my case with the (co)homology i could write K
K([\phi])([v]) = B([\phi],[v])?

thank you :)

[earnest epoch]

Hello, I'd like a clarification about the following statements that seem to be at odds with each other.

1. Every net has a universal subnet
2. A subnet of a sequence is the same thing as a subsequence
3. A sequence is universal if and only if it's eventually constant
4. The sequence a_n:=n does not contain a subsequence that is eventually constant.

[red yoke]

Not every subnet is a subsequence

I think I have a construction

Take any ultrafilter on P(N) and take the index set to be the collection of pointed large subsets

Take the partial order to be (S, n) ≤ (T, m) iff S ⊃ T and n ≤ m

Then (S, n) → n is a subnet of n → n

And is eventually in any large set

@earnest epoch

[earnest epoch]

yeah already got that

[earnest epoch]

very nice

so large in this context just means co-finite right

[red yoke]

It means inside the ultrafilter

Well nonprincipal ultrafilter

So in particular containing all cofinite sets yes

[shy laurel]

I came across this definition of "local finiteness" of a family of sets. "An indexed family of sets {Ai} is said to be locally finite if each point x in X has a neighbourhood intersecting Ai for only finitely many values of i." Can someone help me imagine this with an illustration?

[alpine nest]

For example the collection of intervals (n,n+1) in R is infinite, but locally finite

[shy laurel]

Nice example, I see. Consider the family (-inf, 0] U { [1/n, inf) | n >=1 }, isn't this family locally finite as well?

[alpine nest]

Indeed, good exercise! What do you think?

[round oyster]

can you guys charecterize set of integers congruent to 0 mod 2 in ur own way?

[iron kite]

What does it have to do with algebraic topology?

[raw hornet]

@alpine nest there is indeed lots of thing that I must know before getting into knot theory

I tried but I found a lot of terms that I never even heard of

I find it only rare that they provide knot without topology

[chilly lynx]

Guys, if i want to prove that compactness is a topological property and f: x ->y, is it necessary that Y be a Hausdorff space?

[gritty widget]

compactness doesn't require hausdorffness

[chilly lynx]

I say this because in Munkres it says that to be homeomorphic and it has to be a Hausdorff space

[gritty widget]

i'm not really sure what you mean. can you post a screenshot or a picture?

[gritty widget]

this is not saying "compactness is a topological property"

"compactness is a topological property" would be the statement "if X is homeomorphic to Y and X is compact, then Y is compact"

[chilly lynx]

oh

i see

So to prove that it is a topological property is this theorem enough?

[gritty widget]

yes

[chilly lynx]

ah thanks, I was confusing things

[tidal lynx]

This isn't exactly the statement that "compactness is a topological property", but I feel like there should be a similar description

What could that be?

[gentle girder]

what could what be

[tidal lynx]

the description

[gentle girder]

can you be specific.. I really am having trouble parsing your statement

[tidal lynx]

sure

[gentle girder]

you want words that say "[x] is a topological property" but rigorously?

[tidal lynx]

no

[gentle girder]

oh okay I'll let you say

OH

you want a description of the thing

the picture

[tidal lynx]

yea

wait

no

[gentle girder]

oh

[tidal lynx]

kinda

[gentle girder]

okay, I will seriously let you say now

lol

[tidal lynx]

ok

[tidal lynx]

• We say that a property P is a "topological property" if it satisfies the following statement: "if X has property P and X is homeo to Y then Y has property P"
• Is there an adjective used to describe properties that satisfy the statement given in the image? I'm not exactly sure what that statement would look like for a general property P, but it would be something to do with the property playing nicely between subspaces and the larger space

[gentle girder]

oh okay great, uh i don't know any word for that. The property that you're looking for is probably just that open sets in X restrict to open sets in Y, and i have never heard someone use a specific word for it. Though maybe there should be a word for it.. maybe you can use some language surrounding pullback? idk

[tidal lynx]

hm

[gentle girder]

i am currently looking in the nlab for a name for this and shit is hitting the fan

i learned about another pair of adjoint functors though so that is nice

and also learned what sharp and flat modality are

[tidal lynx]

idt it’s a thing unfortunately

[chilly lynx]

Guys, maybe this is trivial, but I got stuck on this part.

I need to prove that arbitrary unions belong

in this topology

ignore the last one I was rambling

any suggestions?

[gentle girder]

what is $\mathscr{C}A$ here?

[gentle osprey]

smay

[gentle girder]

oh it's complement

happy midnight btw

[shy phoenix]

hey so im trying to compute the 1st homology group of the 3 sphere with 2 points identified using excision (call this space X). i already have that H_n(S^3,{p_0,p_1})=H_n(X) where p_0 and p_1 are the 2 points in S^3 which are identified to create X. So far this is what I have:

I let $A=S^3-{p_0,p_1}$ and $B=U_0\cup U_1$ where $U_0$ and $U_1$ are disjoint open nbhds of $p_0$ and $p_1$ respecitvely. Then $\mathring{A}\cup\mathring{B}=S^3$ and $A\cap B=(U_0-{p_0})\cup (U_1-{p_1})$.

[gentle osprey]

Beous

[shy phoenix]

Then by excision I get $H_1(S^3,B)=H_1(B,A\cap B)$

[gentle osprey]

Beous

[chilly lynx]

same

[chilly lynx]

yes

[shy phoenix]

So $H_1(B,A\cap B)=H_1(U_0\cup U_1,(U_0-{p_0})\cup (U_1-{p_1}))=\bigoplus_{i=0,1} H_1(U_i,U_i-{p_i})=\bigoplus_{i=1,2}H_1(D^2,\partial D^2)=\bigoplus_{i=1,2}H_1(S^2)=0$

[gentle osprey]

Beous

[shy phoenix]

this would imply the 1st homology of X is 0. i think this is wrong, although only cause i thought the 1st homology group would be Z and not 0

[shy phoenix]

ok i figured out an alternative and found H_1(S^3,{p_0,p_1})=Z

[warm hedge]

why the G_0(U,V) is nonempty?

[keen urchin]

Have I understood this right? If you take a 3-simplex [0123] and identify faces [012] = [123], and identify edges [01] = [12] = [23] then you necessarily also get the identification that all 4 vertices become the same point?

Have I understood this right? If
you take a 3-simplex [0123] and identify faces [012] = [123], and identify edges [01] = [12] = [23] then you necessarily also get the identification that all 4 vertices become the same point?
And you also get the identification [02] = [13]? And no other identifications?

[gritty widget]

True?

[iron kite]

Ext(A), A minus and A tilde are not standard. Maybe you meant “the interior of the complement of A” (or the complement of the closure of A) for Ext(A), and the closure of A for A minus (which should be denoted \overline{A})? But I cannot make any guess as to what tilde is.

If I guessed correctly, I believe the four first ones are correct.

[gritty widget]

thank you

[median sand]

Is free homotopy of closed paths the same as homotopy with fixed endpoints in general? I'm studying complex analysis atm and the homotopic version of Cauchy's theorem states that the integral of a holomorphic function over a loop that is freely contractible (not necessarily with fixed endpoints) is 0. Call a subspace of C freely simply-connected if every loop in it is freely contractible, then by the Riemann mapping theorem a freely simply-connected domain is homeomorphic to the unit disc and is in particular simply-connected in the usual sense. Does this generalise to other spaces or is this a peculiarity of C?

[fading vale]

Free homotopy of closed paths is the same as conjugation basically

So it's slightly weaker

E.g. in S^1 wedge S^1, a and bab^-1 are freely homotopic

But not homotopic

[coral pawn]

When reading homotopy theory papers, should categorical constructions always be interpreted in the infinity category setting? For example, when they say quotient or pushout in the paper, do they really mean those things in the infinity category context (i.e. homotopy quotients and homotopy pushouts)?

For example

In a paper I'm reading

The authors define a map out of

By defining it explicitly as subsets of the affine plane

But if it were the homotopy pushout, how would they know what it looks like?

[umbral panther]

In modern papers, that’s what they mean, but they should make it clear

In old papers, it is the point set operation and much of the paper is devoted to arranging so that the point set operation computes the derived operation

[coral pawn]

What would you consider modern?

[umbral panther]

My guess is that they mean the push out of sheaves, but it turns out to be the homotopy push out

[coral pawn]

This paper is from 2001 lol

Is modern after that?

[umbral panther]

Yes, that’s too early

If they never say anything about infinity categories, then they should not say things that are to be interpreted in an infinity way

[coral pawn]

Damn

it's kinda cool that the field went through a complete change in terminology after I was born

Thanks bw

[merry geode]

I cannot believe I forgor tautological bundle

[fading vale]

Homotopy theorists invent an entire new foundational calculus for their field on the same timescale as el nino and la nina occur

[merry geode]

How.. wow

[gritty widget]

Guys what is topology I have to take it next semester and I have no clue what it is

[quiet thorn]

sounds like you'll find out next semester :)

[gritty widget]

Describe it in 5 words

[umbral panther]

“you'll find out next semester”

[tidal lynx]

Are homology theories only used to help you determine if two spaces are homeomorphic, or are there other things that people do?

[solemn oar]

This is a good question, but an extremely broad one. One of the main uses is, as you say, to determine if two spaces are homeomorphic, or more generally, homotopy equivalent.
However, already at a basic level, they allow you to answer more fine-grained questions, like lifting and extension problems. For example, given spaces X, B, E, and cont. functions f: X -> B and g: E -> B, does there exist a cont. function h: X -> E such that the composition (g o h): X -> B equals f? This is an example of a lifting problem. Homology can give us obstructions to solving this equation. A famous example of this is the Brouwer fixed-point theorem.

So a general answer is that (co)homology gives us obstructions to solving certain problems.

To get an idea of how far this idea reaches, it turns out that the key obstruction to finding global solutions of differential equations and functional equations in smooth spaces is (co)homological. This is the field of De Rham cohomology.

The cohomology of groups has things to say about solving polynomial equations in fields. Etc, etc.

Here are some M.SE posts you can explore, and the links inside.
https://math.stackexchange.com/questions/4232919/why-is-cohomology-useful-and-in-which-way
https://math.stackexchange.com/questions/13627/applications-for-homology

[tidal lynx]

Thank you!

[wide lily]

Can someone help me visualize how gluing indicated sides of the pentagon on the left results in something homeomorphic to the handle on the right? This is from Schwarz's "Topology for Physicists" (p. 8 in my edition).

[tender halo]

imagine you collapsed AE to a point, then you will get a regular torus

then start growing AE, it will result in the torus growing a hole in the side and you'll get the thing on the picture

[grim knot]

can somebody give me some hints on how tosolve this exercise? I don't even get what I have to begin with

[prime elbow]

What is the immediate successor in order topology?

[alpine nest]

Immediate successor is not a topological notion, just an ordering notion: y is the immediate successor of x if y > x and there is no element u such that y > u > x (all inequalities strict)

[prime elbow]

Okay thank you

[keen urchin]

bump!

[rich cliff]

what's the immediate successor of 1/2 in Q? is it 3/4?

[tidal lynx]

what about 5/8

[rich cliff]

ok. then it is 9/16?

[tidal lynx]

17/32

[rich cliff]

Fair point

[alpine nest]

The existence of an immediate successor isn't always guaranteed.

As demonstrated above

[rich cliff]

But here it's probably just 33/64?

[tidal lynx]

65/128

[keen urchin]

whats the biggest number?

[alpine nest]

Depends on the set of numbers we're considering.

[keen urchin]

must be 10000 right

[tidal lynx]

10001

[rich cliff]

There's no biggest number

[frail coyote]

here is topology?

or not quite

its a really easy question but im quite dumb

[gritty widget]

this is indeed #point-set-topology

[frail coyote]

oket so
how do I prove that Set is all the inner points of the set and the boundary points

like in the end I need to prove that the closure is indeed the disjunct of both inner and boundary points but I do not know where to begin

I just did one part

rn im missing that the closure is also subset of the union like the U one of all inner points and all boundary points

[alpine nest]

Interior points of A and boundary points of A are all defined as elements of A, which immediately gives you an inclusion (the union of the set of interior points and the set of boundary points is all of A)

In the other direction you need to prove that every point of A is either an interior point or a boundary point.

Equivalently, you can prove that if a point of A is not an interior point, then it is a boundary point.

[frail coyote]

but how

im not that good with proofs actually

xd

[opaque scroll]

Usually the boundary is just defined as a point in the closure that is not in the interior. So then there isn't really anything to prove.

Are you working with a different definition?

[frail coyote]

mmm well its what you say but the definition ive been provided with is the following:

its a point for which for every r the r-enviroment the intersection between that enviroment and the set is non empty, and the intersection between that r-enviroment and the metric without your set is also non empty

wait im gonna try to write it properly

[opaque scroll]

I see, so for every neighborhood (r-environment?) of your point. Either they all intersect both the set and the complent, or there is some neighborhood that intersects just one or the other.

Then just look at those cases

[frail coyote]

yeah neighborhood

sorry im not studying this in english

[frail coyote]

you know what the main problem for me its I think that most of the time I can imagine why somethings are right or wrong but at the proving part i get lost, do you know how to get better at that?

[opaque scroll]

I don't know that I have very good advice, besides practice.

It might help to read proofs, and guess what should come next as you read.

But often times a proof is just about carefully writing out what the definition of something is and what you're trying to show. And perhaps split it into a few cases.

[frail coyote]

okey

yeah

thanks person

[wide lily]

That helps, but I guess it isn't intuitive to me that including edge AE results in a hole and not something else. Good starting point for me to think about it though, thank you!

[warm hedge]

i think i cant undesrant the " for comeager many " and " for nonmeager many "

[rugged canyon]

guys, why is topology in n=4 so weird?

(please explain to a person who didn't study it much, except for maybe some metric-spaces chunk of it)

like there are sooo many things that break in four dimensions (don't make me list the examples please )

[languid patrol]

It is big enough that things are complicated but not big enough that there are enough dimensions to do some dumb tricks

[potent elk]

How to characterize compact subset of sorgenfrey topology?

[umbral panther]

The specific dumb trick that you can’t do is the Whitney trick, where you embed a 2-disk and use it to move things around. The problem is that a generic map from a 2-manifold to a 4-manifold has self intersection. Whereas in any higher dimension it doesn’t

[prime elbow]

If T_1 is finer than T_2 then any set which is dense in T_1 is also dense in T_2, right?

[red yoke]

Yes

[prime elbow]

If I define basis B as [a,b), a and b are real numbers and b is a rational number and a<b.
Then let T_1 be the topology on R generated by basis B.

Then Q is dense in T_1, right ?

[lusty trench]

Your basis B isn't a single interval [a,b), but rather the collection of all such intervals.

And, yes, I think Q is dense in that topology. It suffices to check that, for any nonempty basic open U = [a,b), the intersection Q \cap U is nonempty.

[prime elbow]

Okay, thank you

[gritty widget]

what is dense set?

[naive trench]

Set which its clousure coincides with the whole space

[gritty widget]

alright what is in itself dense set?

[tender halo]

a set without isolated points

alternatively, a set that is contained in it's derived set

[gritty widget]

Mean A is not in itself dense set and not dense set

true?

[tender halo]

A is dense

but not dense in itself

A is contained in the closure of A (which is true for any set), but the derived set is not the closure

the derived of A is {2, 3, 5, 6}

[gritty widget]

hm yes but not everywhere dense

[tender halo]

thats the same thing

dense means every open set has a point of A, alternatively - closure of A is X

[gritty widget]

hmm okay ı understand thanks my brother

[gritty widget]

A not dense but A in itself dense @tender halo

I understand correctly now

[sterile fossil]

8 is not isolated point of A

Is $A^\sim$ the derived set of A?

[gentle osprey]

Leu

[gritty widget]

8 isolated poin of A because 8 ∈ A and 8 ∉ accumulation point set

it is rule written in my book

[sterile fossil]

8 is an accumulation point of A

X is the unique neighborhood of 8

and 9 is element of X

[gritty widget]

oh ım not see

ı false writer

ı sorry

[sterile fossil]

no problem

[fallow marsh]

I have a project where my task is

Produce complete list of K_6 free graphs of order less than or equal to 13.
Does anyone know what they might mean by k6 free graphs? Do they mean all graphs of order 13 that do not contain a k6 subgraph?

[languid patrol]

Yep

[ebon galleon]

thats a lot of graphs

[fallow marsh]

I am just now realizing that most of the darn things are gonna be isomorphic

[fallow marsh]

Yeah, this code is gonna be ran on a super computer

im currently looking for a way to generate all non-isomorphic graphs of a particular order

or else this search space is gonna be BIG

[lusty trench]

Let $X$ be a nice topological group, such as a Lie group. Is there some way to construct a model of $X$'s Postnikov tower $(X_n){n \in \mathbb N}$ such that each space $X_n$ is a topological group and each map $X_n \to X{n-1}$ is a continuous group homomorphism?

[gentle osprey]

Eduardo León

[lusty trench]

Even special cases (beyond the trivial case X = R^m x (S^1)^n, of course) would be much appreciated. Thanks.

[umbral panther]

Take any postnikov tower of BG and apply a loops functor that lands in topological groups

[lusty trench]

I see, thanks!

[umbral panther]

This might fail to get you a homomorphism from X to X_n, which you didn’t ask for but obviously want. But I think it is pointing the right direction which is having a loops functor that lands in groups

[grave solstice]

If X is noetherian then are there only finitely many connected components?

[gritty widget]

yes

[grave solstice]

it follows immediately from the finiteness of irreducible components right

like irreducible components are connected, and each point belongs to some irreducible component

[granite slate]

Does this generalize into some nice space(s)? for n=2 it should be RP^2 right?

[unreal stratus]

I'm not sure what you mean by generalising. You can form the mapping cone of any map.

[unreal stratus]

Yes for the RP^2

[grave solstice]

Let X be connected and noetherian. If the irreducible components of X do not intersect each other, is X necessarily irreducible?

[languid patrol]

yes

[languid patrol]

yes

[unreal stratus]

Noetherian is the key thing; it means ||you have finitely many irreducible components, which means they're all clopen||

[granite slate]

You can different spaces for different n, so I was wondering if you get some nice pattern like idk rp^n for even n and S^n for odd n or something like that

[unreal stratus]

Ah

Well it's always going to be a complex with 2 nontrivial cells so it wont be something higher dimensional like RP^n for n > 1

[umbral panther]

It is 2d for all n

[quartz edge]

define d

[umbral panther]

Dimensional

[wooden glen]

hey I'm being stupid but how do I start proving the reverse inclusion?

(excercise at the end)

[lusty trench]

Recall that products of intervals are the open balls in R^n with the infinity-norm. Do you remember how to prove that the 2-norm (i.e. Euclidean norm) and the infinity-norm generate the same topology on R^n? The proof here is essentially the same.

[rancid umbra]

for each i, you get a radius r_i such that B_{r_i}(a_i) is contained in U_i

[wooden glen]

Yeah I got it

tysm

[gritty widget]

where can I find a proof of the fact that for every subgroup H of the fundamental group of a graph there is a covering space of the graph whose fundamental group is isomorphic to H

[lusty trench]

When in doubt, default to dumb. Construct the universal cover and then quotient it by the action of the subgroup. Works for any space satisfying the usual hypotheses of covering space theory (not just graphs): Hausdorff, locally path-connected, semilocally simply connected, blah blah blah. (I might be misremembering the actual conditions.)

[gritty widget]

what is set of the first category?

[tidal lynx]

Why do we require homomorphisms between topological vector spaces be open mappings onto their image? I feel like continuous linear map is enough

[tender halo]

countable union of nowhere dense sets

baire category theorem says that sets of the first category are co-dense

[gritty widget]

alright what is set of the second category?

[tender halo]

all the other sets

[gritty widget]

thank you

[gritty widget]

mean intA¯=empty set, A set of first category

[obtuse meteor]

In analogy with the open mapping theorem@for banach spaces

[tender halo]

yes

[gritty widget]

thanks again

[tidal lynx]

Ohh like if you remove the condition it’s still equivalent

[plush folio]

Is Introduction to Topological Manifolds by Lee good for learning general topology? I'm looking for something that covers homotopy

(I already have Munkres, but there's no homotopy AFAICT)

[obtuse meteor]

For Banach spaces yes but not in general

I assume there are many distinct ways to define a category TVect

If you’re familiar with Banach Spaces and trying to make analogies with the category Ban then you would include the open mapping as an assumption

[gritty widget]

munkres definitely talks about homotopy

most of the latter end of the book is dedicated to algebraic topology

half of lee's book is dedicated to algebraic topology, too

i think both books are good, but i like lee more. i think his writing style is less dry

[plush folio]

Oh right, just found it 👍 hate that munkres doesn't have subchapters in the ToC

Thanks for the recommendation, I might get Lee as a supplement

[gritty widget]

download it and check it out first!

[plush folio]

Yep, I will! 👍

[tidal lynx]

Right sorry

[heady girder]

Hello! Does anybody know how to describe a covering map given a subgroup of the fundamental group? I mean: Let $X$ be a topological space (with good properties). Let $H$ be a subgroup of $\pi_1(X,x_0)$. Then there exists a covering map $\hat{p}:(\hat{X},\hat{x}0)\to(X,x_0)$ such that $p\ast(\pi_1(\hat{X},\hat{x}_0))=H$. So, if I have $H$, how can I describe $\hat{p}$?

[gentle osprey]

Gkn

[fading vale]

intuitively you build X hat by unwrapping the loops in X contained in H

explicitly H acts on the universal cover of X. X hat is the quotient by this action

[spring trellis]

hello, I'm learning about coverings, and I want to know why when we have a normal defining subgroup (i.e the covering is regular) we can say loops lift in loops, Thanks 🙂

[lusty trench]

The precise statement is that, given a covering map p : (E,e) -> (B,b), the following statements are equivalent:

a) The image of p_* : pi_1(E,e) -> pi_1(B,b) is a normal subgroup of pi_1(B,b)
b) Given a loop f : (S^1, 1) -> (B, b), either every lift of f to E is a loop or every lift of f to E is an open path.

And, when these statements hold, we say that p : (E,e) -> (B,b) is a normal cover.

[lusty trench]

Oh, nice. This was the proof I gave in my undergraduate algebraic topology midterm:

Let a : x ~> x be a loop in B that lifts to a0 : x0 ~> x0 and a1 : x1 ~> x2 in E. Let b : x ~> x be a loop in B that lifts to b0 : x0 ~> x1 and b1 : x3 ~> x2 in E. Notice that b * a * b^{-1} lifts to b0 * a1 * b1^{-1} in E. By the uniqueness of lifts, b0 * a1 * b1^{-1} is a loop iff x0 = x3 iff b0 = b1 iff x1 = x2 iff a1 is a loop. In other words, every lift of a is a loop iff p_*'s image contains [a]'s whole conjugacy class. Therefore, p is a normal cover iff p_*'s image is a normal subgroup of pi_1(B).

[scenic trellis]

i have a question on simplicial homology: when computing the boundary map from 2-chain to 1-chain for e.g. torus, RP2, how exactly do we pick an orientation for each 2-simplex?

[scenic trellis]

oh ok I see thanks!

[lime sable]

actually the standard way to pick an orientation is to order all the vertices, which gives an orientation on every subset and thus every simplex

[lime sable]

my bad, you do need the orientation to be consistent otherwise weird things happen, like boundary^2 not being 0

i misread, the global orientation doesn't matter

a careful proof can be found here https://math.stackexchange.com/questions/1565192/simplicial-homology-does-not-depend-on-the-orientation

[scenic trellis]

i see, so for a 2-torus, where there is only 1 vertex, how do i get the induced orientation for 2-simplices?

[lime sable]

are these the delta complexes from hatcher

[scenic trellis]

yes

[lime sable]

basically you draw a picture

the drawing is like a blueprint showing the domains of those σ_α in the definition and how they restrict to each other

the picture should work if you think of it as a oriented simplicial complex bunch of disjoint oriented simplices and you identify simplices (by writing the same letter next to them in the drawing) while preserving their orientation

More generally, ∆ complexes can be built from collections of disjoint simplices by
identifying various subsimplices spanned by subsets of the vertices, where the iden-
tifications are performed using the canonical linear homeomorphisms that preserve
the orderings of the vertices.

[lime sable]

in practice, you take these pictures and draw arrows to show the orientation/order, and then just do the calculations

[scenic trellis]

thank you! i think i understand on edges and vertices now (particularly why the cyclic one doesn't work), i am still not sure on 2-simplices, for example on Hatcher's RP2, how can I tell if i should count clockwise or counterclockwise on U/L?

[lime sable]

you can think of a picture of a simplex with direction labels on its edges as showing you an order on the vertices: the vertex with all adjacent edges coming out is the first. removing that and finding the next vertex with all edges leaving is the second vertex

another way to think about it is that, when the edges are labeled by directions, looking at the direction of the edge between any two vertices shows you which one comes earlier in the order

[scenic trellis]

aaaa i see, so that's how we order the edges

[scenic trellis]

yayy thanks! i got the homology for RP2 correctly

[heady girder]

How the action look like?

[proper fiber]

Is any infinite product of complete metric spaces complete

Like i think there will be metrizability issues

But like what can we say in general about this

[proper fiber]

I looked it up online I am getting results on polish spaces and finite or countable product of complete metric spaces

[tender halo]

countable yes

non countable is not even metrizable

cuz there are no points with a countable base at that point, much less first-countability

exercise: prove that you can embed a cantor cube of weight m (a product of discrete topology on 2 points with itself m times) into a product of m metric spaces with more than 1 point, deduce that the product is not metrizable

bonus: deduce that the weight of Tychonoff cube of weight m (a product of I with itself m times) is m

[fading vale]

If [gamma] is a loop it sends the class of paths [p] to [p * gamma]

[heady girder]

Hahaha ok ok thx!

[gritty widget]

how do i find

[unreal stratus]

Consider their cohomology rings

presumably that is the topic of the section

[gritty widget]

okay thank you

[unreal stratus]

But yeah this is like the first thing you should try

(deleted lol yes the rings are indeed what you're meant to do)

[empty grove]

Stop deleting messages

[unreal stratus]

Lol

[unreal stratus]

What I confused it with was uh

||A common example is showing CP^2 is not equivalent to S^2 wedge S^4. So I got confused by keeping that in mind||

[gritty widget]

Hello I want find a Hausdorff space. M= {1,2,3} and T= {M, {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}}.And V,U∈ T, p∈V q∈U p≠q and U ∩ V = empty set. What is U and V, q and p?

[limpid fern]

wdym you want to find a hausdorff space

the question is confusing

[tawny juniper]

What would be the right concept/definition to make rigorous the idea of a family X_t of topological spaces continuously parameterized by the real variable t?

Would that be a fiber bundle?

Or could I just get away with it using a product space?

[umbral panther]

A space with a continuous map to R, denoted t

[tawny juniper]

And what would be the topology on {(t,X_t): t is in R}?

[balmy field]

You should look into Stochastic processes as this seems like a really similar construction

[tawny juniper]

Thanks.

[prime elbow]

I want to prove that (-∞,a] is homeomorphic to (-∞, 0] (a is a real number).

If I define f(x) = x-a then does it work?

[tawny juniper]

Yep.

You just need to show that f is bijective and continuous.

[alpine nest]

And that its inverse is continuous

[tawny juniper]

Yes.

Though the inverse is x + a, which is covered in the case a —> -a . :3

[alpine nest]

Yeah, I didn't say it's particularly harrowing work