point-set-topology

[red yoke]

Are you looking at fundamental polygon

[grim knot]

in this case yes

[red yoke]

What does the whole thing become when you identify boundaries

[grim knot]

oh wait a circle as well, so I get S1vS1vS1

[red yoke]

Ye

[grim knot]

oh thats nice

[red yoke]

Then it's the same

[grim knot]

Can I proceed inductively and get that if I remove k points I have the wedge over k+1 S1?

[red yoke]

Yea

[grim knot]

okay wait but then finding the homology groups is really easy

why is it so complecated here?

[red yoke]

There are many methods

In general I think you could Meyer-Vietoris bash for puncturing objects

[grim knot]

https://math.stackexchange.com/questions/4852830/homology-of-the-two-dimensional-torus-with-n-points-removed
here for example

I don't know it seems much more complecated to do it like this(?)

[red yoke]

The first paragraph is more or less what you did i think?

[grim knot]

maybe he did it more rigorously, that would make more sense

[grim knot]

I'm really grateful for you always helping me though

[grim knot]

regarding what I sent above, why is H_0(T,U)=0? shouldn't it be Z or am I not seeing something?

I think he forgot the tilda for reduced homology there

[unreal stratus]

I was thinking about this recently and there's a fun way to do it iirc

let me try to work it out

[winged viper]

If I have two vector bundles E and F, is there a general way to go from “I have a canonical isomorphism E_b -> F_b for each b in B” to “I have a bundle map E -> F”, other than manually checking continuity on local trivializations? It seems that people will often go from one statement to the other with no further justification.

[unreal stratus]

I think usually the point is that the map you write down is nice enough that is easily promotes to one continuous locally

Sorry that isn't too helpful though

Just to check, is the answer just that you get the same as the torus but with an extra H_1 group for each point you remove?

I'll explain how I'd solve it - lemme draw a little diagram

[grim knot]

as an end result I get Z^k+1 for H1(T\A), 0 for Hi(T\A)=0 for i>=2 and I'm confuse now about H0(T\A), I assume it has to be Z by pathconnectedness, so wouldn0t give it too much attention

k is the number of removed points

[winged viper]

So it still is completely necessary to check things in trivializations right?

[warm quiver]

I think the rigorous way of saying this is that you have an automorphism of some grassmanian that twist the two calssifying maps for E and F in the sense that the automorphism takes the fiber E_b to F_b. This will induce an isomorphism between the pullbacks which are isomorphic to E and F, respectively. Automorphism of the entire Grassmanian seems too strong, but maybe there’s a way to work around this or weaken what I said above

[grim knot]

Moreover, is (T,A) a good pair? I'd say yes since we can always find neighborhoods that retract to each opint in A, but I'm not ocmpletely sure

[winged viper]

Hmm I see, I guess this would mean you need to know what the classifying maps of E and F are?

[umbral panther]

Except the first one

[unreal stratus]

Exactly

[unreal stratus]

Yeah sure since then you get no H_2

[warm quiver]

Yeah, which requires you to have an embedding of each vector bundle into some common trivial, which probably means you need to work with partitions of unity which feels like working with the local trivializations anyways so maybe it’s not that helpful. I would need to think more about vector bundle stuff I forgot about

[winged viper]

Hmm yea I see

[unreal stratus]

Anyway the nice way I had in mind which doesn't use any Mayer vietoris or excision or working out boundary maps or anything:

The torus minus n points deformation retracts onto the torus minus n disjoint open disks, call that X_n

Pick one of the boundary circles and call it A. Then X_(n+1)/A is homeomorphic to X_n

and (X_(n+1), A) is a good pair

Then just use the long exact sequence and win (by inducting on n, beginning with the easy case n=1)

@grim knot

as in, here is X_3, with A in red

and the quotient map just like seals the hole by bringing together the red stuff

Another way, which is more "space-level", should be that you can pick some point in the middle bit of the torus and join it to the corners using edges whilst avoiding the poitns you removed and splitting them up

like this

then that should deformation retract onto like

[empty grove]

That's literally me

[unreal stratus]

and now this is a graph

And now you can count loops

(I hope this works at least)

Is this correct lol

Should be

@ moldi

lol

[empty grove]

I just came here to shit post

[unreal stratus]

No it's fine actually

Cause the homotopy is constant on the boundary

so nothing to worry about regarding the quotient

[empty grove]

That is an interesting observation

[unreal stratus]

I just like

Wanted to prove directly that this is homotopy equivalent to a graph

And this i guess gives you one (almost) explicit graph it deformation retracts onto

[empty grove]

Ye that's how I'd do it

[unreal stratus]

Actually, for bonus points I should do it like this

[empty grove]

How about Mayer Vietoris, inducting on the no of holes

You want homology of the intersection

[unreal stratus]

sorry old photo

[empty grove]

Looked cute

[unreal stratus]

the advantage being that if i do it like, i still only have one vertex

The only thing is I'm not sure how to guarantee the lines don't touch lol

actually nah it's fine - you can arrange the circles so that they are all in a line and then it clearly works, so it must work more generally

[empty grove]

Ye

[grim knot]

oh okay that's actually kinda nice, but as you said you still have to work inductively to calculate it generally

[unreal stratus]

Yeah

[grim knot]

I intuitively wanted to solve it like this, since then you get something that deformation retract to the wedge of some circles

[unreal stratus]

ye this is exactly a wedge of circles

but tehn to be fair, to compute that you use induction

But I see no issue with induction lol

[grim knot]

hahah which method do you think is better?

[unreal stratus]

I think the quotient one is m favourite to compute as it is very explicit

like it is 100% rigorous

[unreal stratus]

But this one here is stronger in terms of what it proves

[empty grove]

Speak for yourself, I use the wedge axiom

[grim knot]

for this to be rigorous you should properly defined everything

still nice though that there are so many ways to prove it

[unreal stratus]

Yeah, explicitly defining paths would be a bit annoying

[grim knot]

I have to say algebraic topology really captured my heart (I'm not good at it, buuut anyway)

[unreal stratus]

I think if i were pressed, I would appeal to a theorem on homeomorphisms of manifolds which means you can just move around the circles

I think that would not be hard to prove from first principles here anyway

[unreal stratus]

It is fun hehe and keep going

[grim knot]

If I only where THIS advanced yet hahha

[unreal stratus]

Well okay maybe I am wrong tbf lol

The idea was just that like, you can reduce to just having to move one point within the closed disk (i think)

[grim knot]

one can really visualize stuff, it's nice, then I also discovered that the applied part is really really cool

[unreal stratus]

which applied stuff do you mean?

(but agree)

[grim knot]

like persistent homology and topological data analysis

I also read a bachelor thesis about how one could descrive the nervous system through algebraic topology

[unreal stratus]

o ye cool, i'd like to read more about that

[novel ember]

what do i do from here

(hausdorff paracompact ==> normal)

(i have that hausdorff paracompact ==> regular)

[hollow geyser]

I spent too long on this problem. I just want to verify that it is correct and looks fine.

(Apologies for all the definition and proposition insertions. I find them to be helpful when I review in the future)

[fading fern]

i would just skip the last sentence and then apply regularity

oh wait i misread proof

i think you made some typos

it should be $\bigcup_{y \in Y}V_y$ covers Y i think

[gentle osprey]

HSF

[novel ember]

nope

[fading fern]

oh you used regularity to construct those sets

i thought you just picked random neighbourhoods

[novel ember]

im reading the proof i wrote down for hausdorff => regularity

[red yoke]

T2 → T3?

[novel ember]

the space is paracompact

[fading fern]

i think you can just use the fact that ||union of sets S such that S disjoint to X is still disjoint to X||

i don't get how intersection stuff is important

[knotty vine]

Looks good, you got the conclusion in the final sentence the wrong way around though Now do it in LaTeX!

[novel ember]

we cannot do this

[fading fern]

and how you included natural numbers in the indexes when X may not even be in a R^n

wait why not

[novel ember]

wait a minute

[fading fern]

did you well-order it or something (the indexes)

[novel ember]

no, we cannot do this i think

[fading fern]

imma try rigourizcsing my proof

[novel ember]

you're not even using paracompactness

[hollow geyser]

Oh damn I did. Thank you. Also no. Equation editor is my life

[hollow geyser]

What a beautiful finite covering

[fading fern]

i think you misunderstood
i'm not sure if i even can say my proof

[novel ember]

if the space was compact, gg

but i have paracompactness

[fading fern]

ok fixed the proof

the union of neighbourhoods of every point in X is a good idea

now try using W_x then apply regularity

[novel ember]

OH

i figured it out

its just the same proof for paracompact hausdorff => paracompact regular

i also have the fact that if $T$ is locally finite

i also have the fact that if $\mathcal{T}$ is locally finite, then $\overline\bigcup_{T\in\mathcal{T}}T=\bigcup_{T\in\mathcal{T}}\overline T$

[gentle osprey]

ディクッシの王・ルイー・ジュウロク世

[novel ember]

gg i did it

[median sand]

What are some good non-Hatcher AT books? Hatcher's pretty cool, but many of his proofs are too informal and hand-wavy (attach this, shrink that, etc.), I'm not yet at a point where I'm comfortable with this. Is Rotman too basic? His presentation appeals to me, so I'd like to know how he's rated by the community.

[livid escarp]

I liked May a lot and Spanier for not skipping details on simplicial and semi-simplicial complexes (delta)

[shy phoenix]

Let $c:\partial D^2\to S^2$ be a constant map. I want to show that $S^2\sqcup_c D^2$ (i.e., $D^2$ attached to $S^2$ via the given constant map is $S^2\vee S^2$. I understand visually how this works, but how can I prove this rigorously?

[gentle osprey]

Beous

[shy phoenix]

like i want to show there is a homeomorphism $S^2\sqcup_c D^2\to S^2\vee S^2$, and I have a quotient map $S^2\sqcup D^2\to S^2\sqcup_c D^2$ and a quotient map $S^2\sqcup S^2\to S^2\vee S^2$, but I'm not sure how to proceed.

[gentle osprey]

Beous

[hidden crag]

Rotman is really good for a first pass

Though a lot of the exercises are too easy

[tribal palm]

wish i ever had that problem, ehehe

[hidden crag]

But I would recommend getting used to this kind of terminology, it’s common to say these things

Instead of writing down the maps every time

[floral bear]

is there a name for the quotient/sum of a knot and a braid like this?

[winter frost]

do any of you have any good books for beginning topology. I was thinking about doing my senior research in topology

[haughty cedar]

im not sure, how the final result follows.

Say that I have a topology T in C. I take a set B in T s.t. some point in L is in B. Then I come to some conclusion that all points in B are in T. Last sentence however claims that if B included in T, then T - open? How so?

[knotty vine]

I'm not too familiar with the subject, but I think this is composition of tangles

[knotty vine]

Munkres, Topology

[winter frost]

thanks ill look into it

[abstract saffron]

Is it well-defined tho, hmmm

I'm thinking the result depends on where you add the extra piece

in this case it doesn't matter per Reidemeister's moves, but Ig in general it does

[floral bear]

it does, i can show you the example i confirmed it with if you'd like

[abstract saffron]

oof, then it's gonna be a pain to define what the operation does

i can't see how it'd be interesting

[floral bear]

that's why i said "quotient" bc it can really be any topological quotient by a finite set of points

[abstract saffron]

so, if I understand correctly, you are constructing equivalence classes of knots induced by a particular braid?

oof, this's gonna be really ugly

afaik I haven't seen anything like that before, tho I'm def not knowledgeable about knot theory

it does seem interesting (and horrible!) to investigate tho

[floral bear]

not equivalence classes, honestly i'm not sure how i'd notate it but i was just thinking cutting the knot at n points and gluing the (2n) ends of an n-tangle to those cut points

[formal tide]

Interesting problem from my book

Let C be a countable subset of Rn, n >= 2. Show that Rn \ C is path connected.

very obvious intuitively but constructing such a path seems very hard

[shy phoenix]

how do you construct a cell structure on S^2 with 2 0-cells, 1 1-cell, and 1 2-cell?

[knotty vine]

Take the cell structure with one 0-cell p, no 1-cells, and one 2-cell. Then connect another line segment to p so it looks like a lollipop. Not really S^2 but it's homotopy equivalent

Otherwise, take two points with a line e between them, then glue a 2-cell along e . e^-1

[tidal cedar]

I don't know if this has an existing name but this is a variant of the connect sum

If you are interested in {oriented, framed} knots you need to keep track of more data than just a braid group element I'd believe

[knotty vine]

Let k be a cardinal number, let X be a space so that for any two points there exist (at least) k disjoint paths (i.e. the interiors of the paths are disjoint). Then, for any subset C of X of cardinality |C| < k, is X\C path connected?

[formal tide]

Cool! Love when more general problems seem easier than more specific ones

wonder if choice can be avoided somehow for the Rn case

[knotty vine]

idk if this is true

[knotty vine]

Is choice involved?

[unreal stratus]

No choice should be needed

Well

Like all you want is to show there exists a path between any two given points, and to do that doesn't need any choice

I suppose if you wanted to pick explicit paths between any two points and chuck them into a set you might need choice lol

But that isn't what you normally do

[formal tide]

It seems true to me, for your case. For any pair x y in X choose a path p not intersecting C and you're done. Such a path must exist: lemme think how to prove this nicely

[unreal stratus]

But you don't need to form a set of paths across all points

[formal tide]

oh right, depends on your definition

[unreal stratus]

You just need that for any given pair these exists a path

so it foin

[formal tide]

I agree under yours it is not needed

[unreal stratus]

I'm annoyed cause like

the proof I was gonna sketch is just exactly how Semer has set up his statement

But it's cool

[formal tide]

Continuing, given an element a c in C, there is at most one path p that intersects c (you can make a partial map if you like). Since |C| < k, then the partial map is a bijection onto its image, hence there is at least one path p that does not intersect any c

[hollow geyser]

What is the converse here (part B)?
If x is any point in a space X, and the intersection of all open subsets containing x is {x}, then X is Hausdorff?

Is part B asking me to find a space that is not Hausdorff, but still such that intersection of all open subsets containing x is still equal to {x}, for every x?

[formal tide]

yeah

[hollow geyser]

Oh fun. I have no idea then. Time to ruminate.

[formal tide]

https://tenor.com/view/cow-cows-chew-chewing-idgaf-gif-4536854

[hollow geyser]

It can't be a finite space, I know that.

[formal tide]

personally I find disproving conjectures in general topology an absolutely horrible experience

you usually have a couple of well known spaces that don't satisfy the conclusion

if one of them satisfies the premise, you're done, if all of them don't, you're going to have a hard time

for this conjecture, there is one somewhat well known non-Hausdorff example that comes to mind

luckily it satisfies the premise 🙂

[hollow geyser]

My first guess would be the co-finite topology

But let me see

[hollow geyser]

Yeah I hate it. Finding the hay in a needlestack

[unreal stratus]

Yeah I mean in the proof of A you won't use Hausdorffness, but something weaker

And cofinite topology in general is indeed weaker in exactly the way you want

:)

[ebon galleon]

b-but all spaces are Hausdorff!

Except the ones which are only weakly hausdorff

[gentle girder]

it's true

[marble steeple]

Hi can anyone help me figure out where my reasoning goes wrong? (This is for a first year algebraic topology class btw)
I know that $S^1 \vee S^1$ retracts to $S^1$ so the induced homomorphism $i_*: \pi_1(S^1 )\longrightarrow \pi_1(S^1 \vee S^1)$ is an injection. This makes sense, since $\mathbb{Z}$ is a subgroup of $\mathbb{Z} * \mathbb{Z}.$

But doesn’t $S^1 \vee S^1 \vee S^1$ retract onto $S^1 \vee S^1$? And that would imply that that $\mathbb{Z} * \mathbb{Z}$ is a subgroup of $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ - which is false.

[gentle osprey]

Maria

[ebon galleon]

But it is a subgroup....

If we think of Z*Z as {formal products of a, b} and Z*Z*Z as {formal products of a,b,c} then it's the obvious inclusion

And you should check that this is what's happening geometrically in terms of loops at whatever basepoint you choose

That is, if you picture S^1 v S^1 as a figure 8, oo, and S^1 v S^1 v S^1 as ooo, where you include oo into ooo in the first two loops, then your generators are:
a - loops on the first o
b - loops on the second o
c - loops on the third o
So the inclusion is taking loops in the first two o's to loops int he first two o's. You have many options for the geometric retraction. Probably the simplies is that you contract the last o to the single point it touches the middle o. What this would do it simply project/"forget" any c's in your product.

[marble steeple]

@ebon galleon I see ... but if Z * Z * Z is a subgroup of Z * Z (for ex: the subgroup generated by a^2, ab, ab^{-1} ) AND Z * Z is a subgroup of Z * Z *Z ... then wouldn't that imply they're isomorphic - which i know they're not?

[ebon galleon]

Nope, it tells you that their underlying sets are in bijection (since there are injections from one to the other in both directions), but the bijection you get from this won't necessarily be an isomorphism as groups. And indeed, it's not an iso here

Uhh not an example of groups, but in terms of spaces you have embeddings of [0,1] and (0,1) into each other, but they are not isomorphic as spaces

[marble steeple]

Thank you so much! @ebon galleon I was confusing the Cantor-Schröder-Bernstein property and using it for groups.

[steel glen]

if i have two topologies t and s on some space X, what are some ways to "combine" the topologies?
things that came to mind:

1. intersection of the topologies
2. the finest topology such that the identities from (X,t) and (X,s) to X are continuous
3. the coarsest topology such that the identities from X to (X,t) and (X,s) are continuous

it seems to me that 1 and 2 give the same topology, and 3 generates the smallest topology containing both t and s
how can i interpret the topologies from 2 and 3? are there any interesting ways to combine the topologies and keep some of their information

[kindred prawn]

Is it true that homeomorphism between subsets of $\mathbb{R}^d$ takes circles to circles ?

[gentle osprey]

spectrum

[steel glen]

what do you mean by circle

if you mean just a regular circle, then consider that there is a homeomorphism between the unit square and unit circle
if you are wondering if it keeps the homotopy type of a circle, then note that restrictions of homeomorphisms are also homeomorphisms

[ebon galleon]

Consider the collection of topologies on X, ordered by inclusion: τ ≤ σ if τ is contained in σ (τ coarser than σ). 2 states that this new topology on X is coarser than σ and τ, and is maximal in this regard. Dually, 3 states that this new topology on X is finer than σ and τ, and minimal in this.
So 2 gives meets in this poset, and 3 gives joins in this poset.
In fact, this generalized to arbitrary collections of topologies on X, so analogous topologies will give all joins and meets, forming a complete lattice.

[steel glen]

i know nothing about meets and joins, but i guess that's a good jumping off point

[ebon galleon]

Meet = infimum and join = supremum

[steel glen]

sorry for my lack of proper language, but given the complete poset of topologies, would the supremum be the "last common ancestor" of the too?

[ebon galleon]

Yeah, if you're drawing t above s whenever s <= t (or s is a "descendent" of t)

[steel glen]

ok makes sense, thank you

is there a nice interpretation of the resulting topologies?

or any canonical (illustrative) examples

hm maybe i should just look at some texts on lattices of topologies. this is a good start, thanks again!

[kindred prawn]

Let $P,Q$ be two points in the positive imaginary axis in the upper half plane . Why there is a unique Geodesic containing these two points ?

[gentle osprey]

spectrum

[feral copper]

If they are vettically aligned, then there is a unique vertical line connecting them, and it's clearly the only geodesic. If not, then try to convince yourself that there is a semi-circle orthogonal to the real axis (you could even give an equation for it), and it's the unique geodesic too

(Assuming, of course, you mean the hyperbolic metric)

[gentle osprey]

Leno

[west spindle]

@bitter shell can you (re)post the definitions of all your notations here

[gentle osprey]

Leno

[west spindle]

so X is a set of ordinals?

wait so what is Omega then, is it not the first uncountable ordinal?

this is confusing.

did you get the problem from munkres?

[abstract saffron]

[west spindle]

can you screenshot the homework problem itself

your rewording of it feels confusing

if X is just a well ordered set with zero relation to ordinals, then how are we comparing anything in it to an ordinal alpha?

you didn't say what alpha was

what is alpha?

[knotty vine]

(A parameter)

[west spindle]

ok, so when you wrote "Given x in X" you meant "Given alpha in X".

what do you need to prove again?

[gentle osprey]

Leno

[west spindle]

hm ok

at least we are on the same page now

but i need to think about this

[tight ocean]

any idea why the underlying statement is true?

[hidden crag]

Which part

The first one is true because A is just a curved copy of D^2

And for the second part think about deforming the complement the same way you deformed A

[knotty vine]

Soft question: is there a sense in which defining a (semi)simplicial set as a contravariant functor is more "natural" than defining it as a sequence of sets together with face/degeneracy functions satisfying a bunch of identities? It is certainly more compact (if you already know cat.thy!) and more modular. More specifically, how could I convince someone who is not already very convinced of category theory?

[solemn oar]

The most general setting in which you can consider this kind of question is probably the theory of shapes. The nLab has some discussion here:
https://ncatlab.org/nlab/show/geometric+shape+for+higher+structures
https://ncatlab.org/nlab/show/test+category

Within topology it might be useful to consider in the context of nerve-realization adjunctions.
https://ncatlab.org/nlab/show/nerve+and+realization

Whether any of this will convince anyone who doesn't already appreciate category theory is another story.

[plain raven]

My answer echoes persistent Cloudberry; it is just my own personal choice of wording.
This is not necessarily a good answer to "how can I convince someone who doesn't know category theory" but there is some deep content to the reformulation, I think.

One of the fundamental contributions of category theory to mathematics is that it provides a clear, widely applicable theory of "gluing", which is a concept that arises everywhere in mathematics, through the notion of colimit.
The categorical theory of gluing does not just provide simple definitions, i.e., colimit, it also has certain deep theorems, such as the statement that the category of presheaves is the free cocompletion of a given category. Indeed, the whole concept of pointwise left Kan extensions belongs to this general theory of "gluing" in category theory -
it says, if I have a category C with a distinguished collection (diagram) of basic/fundamental/well-understood objects F : M -> C, and another category D with a similar collection G : M -> D, then to transport an object from C to D, we can first try to decompose it into the information of how it is constructed from basic objects (the nerve) and then use this information as a blueprint or schema to try and glue the same structure-shape together in M using the object in the diagram C.
An inspection of this argument reveals that the notion of weighted colimit is the more natural definition of colimit in this setting, and we understand that it is the notion which generalizes to enriched category theory, while the standard notion of colimit does not.
So in this categorical theory of gluing the central notion is that of weighted colimit, with weights in a presheaf, and this is the definition with which we should state and prove that the Yoneda embedding is free cocompletion, and so on.

Now, what does this have to do with topology? Because the concept of semisimplicial set comes to us directly from this general categorical gluing theory. It can be defined, understood, and motivated completely by appeal to "gluing theory."
What is a simplicial complex? Answer: A collection of simplices glued together along certain distinguished maps.
How should this be defined formally? Answer: One should give a category whose objects are the simplices and whose morphisms are the permissible gluing maps; then a presheaf on this category can immediately be understood as a gluing schema which explains how to glue together "simplices" in any arbitrary category with colimits which admits an interpretation of the simplices and face maps.

When we combine this with the obvious monoidal structure on the augmented (semi)simplex category, we get more interesting results:

• the augmented semisimplex category is the free monoidal category with distinguished pointed object
• the augmented simplex category is the free monoidal category with distinguished monoid
  And both of these can be combined to give monoidal categories on the presheaf structure: the category of augmented (semi)simplicial sets equipped with Day convolution (which is just the join of simplicial complexes, gemoetrically) is the free cocomplete-monoidal category with distinguished monoid (respectively, pointed object).

Again, this helps us to conceptually understand what simplices are: they are inductively generated by repeatedly taking the join of a point with itself. This concept makes sense in any monoidal category, i.e., one can talk about a "point in the n-simplex M^{\otimes {n+1}}" as any point in the iterated tensor product (say, generalized element if you want to be fancy, but more concrete examples work too)
Therefore the concepts of simplicial homotopy theory should be of broad consequence, i.e., they should have implications for all cocomplete monoidal categories where there is a distinguished monoid (pointed object). Understanding things at this level of abstraction gives us insights as to why homotopy theory (and homological algebra) seem to appear in so many areas of mathematics. Put simply, monoids arise everywhere - they are among the most fundamental structures of mathematics - and in any cocomplete monoidal category with a distinguished monoid, anywhere where we care about both tensoring and gluing, it is plausible that the canonical cocontinuous monoidal functor SSet -> (C, \otimes, M) may play a helpful role

[knotty vine]

This is a very nice perspective! I will try to see if I can convince my advisor (the person in question) along these lines

I feel like defining the simplex category as the finite ordinals with order preserving map is a disservice to simplicial sets, though it is the most economical way (in terms of ink used)

[kindred cairn]

what's an example of a space that has trivial homotopy groups for all n > 0 but that is not contractible

[unreal stratus]

Well, without path connectedness you can do something silly like a disjoint union of points. with path connected, there are examples like the long line

[kindred cairn]

the long line ?

[unreal stratus]

You can find stuff on it online, but basically it is like the real line except formed of uncountably many copies of [0,1) rather than only countably many

It's nice for counterexamples

So like, any map from S^n hits something which looks like R

and then you can just contract it to a point

But you can't just contract the whole of the long line at once to a point

Note of course that any example is likely to be a bit gross because it cannot be homotopy equivalent to a cw complex (in particular, cannot be a manifold)

[dusk slate]

I didn't get that....

[unreal stratus]

Maybe I am wrong lol tbf

[dusk slate]

Nice one though...

[kindred cairn]

yea, it's because my professor wrote that contractible <=> trivial homotopy groups holds if the space is a simplicial complex

[unreal stratus]

with the reasoning

[unreal stratus]

Yeah

[red yoke]

It looks good

Are there any ones that can fit in R^n

[hidden crag]

That is my argument

If you remove a point you can stretch that hole to a disk

If you remove a disk you can pull the the hole together to a point

[unreal stratus]

You can write more explicit stuff if you want - consider a homeo f: A -> D^2 and let g : D^2 \ 0 -> S^1 be the radial projection map. Then you get a nice map R^3 \ {x} -> R^3 \ A given by like uh

gf on A and identity otherwise

Which is a continuous retraction of the former onto the other

Since g is a deformation retraction, so is this map I just gave

[hidden crag]

Nerd super react this person

[unreal stratus]

Lol

[hidden crag]

Im not even sure if the equator is included here

[unreal stratus]

Me neither

Makes me unsure what to write lmao

I am assuming it not removed I think

But can be adapted ig

[hidden crag]

That’d make it easier to write down the way you suggested I think

Doesn’t really matter ig

[unreal stratus]

Wait nvm

Wait yeah

Lol

[hidden crag]

How many msgs did you delete in the last few minutes

[unreal stratus]

Eepy.

57

[hidden crag]

I also forget that it exists though

[stuck thicket]

Is a set such as a line in R^2 its own closure and its own boundary? Does it also not have an interior?
I just started learning about topology in my first year in uni so I don't really know if this is the right channel, I just know it is somewhat related to topology haha
On the other hand, is a set such as this one, composed of infinitely many segments that never touch its own closure and boundary?

[unreal stratus]

Yes those are all correct

For the first bit I mean

[slim grove]

is it smart to take intro to geometry and topology in second semester?

[steel glen]

composed of infinitely many segments that never touch its own closure and boundary
what do you mean

[stuck thicket]

I meant if this set was also its own closure and boundary

since it looks sort of like this if im not wrong

[rancid umbra]

does it include the line x = 0 for 0 <= y <= 1? if not, it’s not closed

if not, it’s not closed

[hollow geyser]

What's the weaker structure you were referring to here?

btw, in regards to the problem I asked that day...

Is this response okay?

[unreal stratus]

All you need is that given any distinct points p and x, there's an open neighbourhood of p not containing x, and also the other way round

I guess that is T1

[livid escarp]

Is this latex? It isnt, right? Seems familiar

[hollow geyser]

Microsoft Equation Editor

[unreal stratus]

This is good, though it could be simplified a little

[livid escarp]

Crazy. I know some prof who writes his papers in word too, in diff geometry

[hollow geyser]

Just the thinking-out-loud pragraphs? Or in the actual content as well?

[unreal stratus]

The actual paragraph

[hollow geyser]

equation editor is pretty sweet. It's just easier to use since I already have word

[unreal stratus]

You sort of overcomplicated it, like it suffices to find a single open set containing x which doesn't contain x'

and there is a candidate for that which is obviously cofinite

And if you do that, there's no need to introduce this U_x notation etc either, i guess

[hollow geyser]

Oh you mean like use $\bN$ explicitly?

[gentle osprey]

SWR

[unreal stratus]

And tbh in homework I would just omit the first two paragraphs but yeah

[hollow geyser]

omit? Yeah. But they're great notes for me.

[unreal stratus]

Not really, saying arbitrary X is fine

I meant an explicit open set containing x but not x'

[hollow geyser]

I come back to old hw all the time and am like "how the hell did I ever think of this?!"

[unreal stratus]

Ah then yes good reasoning

Well a good exercise would be to show that A is equivalent to being T1 i guess

That is, singletons are closed sets

Then, for example, the space cannot be finite, as every subset would be a finite union of closed sets (and hence closed)

But yeah I should reassure you that your argument is completely correct and that your thinking out loud is all good thinking aha

[hollow geyser]

I guess I'm a little confused by this. I thought I did this with U_x/{x'}.

[unreal stratus]

Yeah but you just said "pick any U_x" and I think it's nicer just to write one down

More conceptual

Well okay what I mean is just X \ {x'}

Maybe this is more aesthetics lol

But like, this is obviously cofinite and contains x but not x'

[hollow geyser]

ah

I think I see

yea

that is easier

[hollow geyser]

What are A and T1?

[unreal stratus]

Oh sorry, saying A was dumb lol

Show that the converse of the statement in part A is true if and only if X is T1

[hollow geyser]

ah ok

[unreal stratus]

"T1" is an adjective meaning like, points are closed

which is weaker than T2 (Hausdorff)

[hollow geyser]

Are T1 and T2 common terms?

[unreal stratus]

Yeah they're completely standard

[hollow geyser]

okay cool. My book has betrayed me then

[unreal stratus]

There's a long list of "T_n", which are known as the separation axioms

[hollow geyser]

But the book is basically topology for infants, so I'm not surprised

[unreal stratus]

They sort of describe how well you can separate stuff

So like

T0 is pretty crap and just means: "for every distinct x and y, there is an open set containing only one of them"

T1: "for every distinct x,y, there's an open set containing only x (and one containing only y)"; equivalent to "points are closed"

A good exercise is to find a space which is T0 but not T1 :)

And then a space which is not even T0

[hollow geyser]

indiscrete is obviously not T0

[unreal stratus]

Yup yeah

[hollow geyser]

(if space has at least two elements)

[unreal stratus]

Provided the set has more than 1 element aha

Yeah okayy nice you sniped the pedantic comment :)

[hollow geyser]

I'm somewhat of a pedantic commenter myself

[unreal stratus]

I guess for pointset topology it can pay off

T0 but not T1 is kinda more interesting

But actually important for some areas of maths

[hollow geyser]

So T0 means we can have, say, an open subset U such that x in U, y not in U, but we don't require a subset V such that x not in V but y in V?

[unreal stratus]

Yeah exactly

There isn't the same level of symmetry

[hollow geyser]

okay like the Sierpinski space?

I think?

[unreal stratus]

Is that like

{{},{0},{0,1}}

on {0,1}

[hollow geyser]

yep

[unreal stratus]

then yup exactly, very nicely done

in fact my hint was going to be to consider finite spaces lol

A nice way to do it is just like, force every open set to contain some fixed point P

As you did there

Such a point P is called a "generic point"

[hollow geyser]

It's becoming clear I'll need to buy a second topo book in the future.

But I just wanted to get my feet wet anyway

Okay and T1: $\forall(x, y\in X)(x\ne y\to\exists (U\in T_X)(x\in U\wedge y\notin U)\wedge \exists(V\in T_y)(x\notin V\wedge y\in V))$?

[gentle osprey]

SWR

[heady skiff]

does continuous in each variable separately imply continuous

[hollow geyser]

e.g. $(f\times g)(t)=(f(t), g(t))$ is continuous if $f$ and $g$ are continuous?

[gentle osprey]

SWR

[unreal stratus]

Nope

Wait

It depends what you mean lol

[hollow geyser]

In the meantime, did I write T1 correctly here?

[unreal stratus]

Sure, though you dont need the bit with V by symmetry

[hollow geyser]

No, I did not, but it leads to my next question

Is T0 not T1: $\forall(x, y\in X)(x\ne y\to(\exists (U\in T_X)(x\in U\wedge y\notin U)\vee \exists(V\in T_y)(x\notin V\wedge y\in V)))$?

[gentle osprey]

SWR

[hollow geyser]

(err.. I could clean up the exist quantifiers)

[unreal stratus]

Tbh it is hard to read when it is in logical notation

[hollow geyser]

My root question: How can a space be T1 but not T0

[hollow geyser]

@heady skiff could you elaborate on your question?

[heady skiff]

like does the converse of this hold

[unreal stratus]

Basically like this comes down to facts about relations

Write a ~ b to mean "there's an open set about a not containing b"

T0 says like, for any pair of distinct points a,b either a ~ b or b ~ a

T1 says for any pair of distinct points a,b, both a ~ b and b ~ a

Now of course in the latter you can just swap b and a, but in the former it can just be that a ~ b holds and b ~ a doesn't

so that when you swap a and b, you cannot conclude the other

I can give an analogous equivalence relation for example

Let ~ be <=, the order on R; it satisfies the analogue of T0 but not T1

[hollow geyser]

Yes that I get. But you said a space could be T1 but not T0

[unreal stratus]

Then that was a typo, sorry

[hollow geyser]

ohh okay

I feel less crazy now

[unreal stratus]

Did I sa that anywhere though?

[hollow geyser]

oh derp

[unreal stratus]

.

[hollow geyser]

you never said that

[unreal stratus]

Dw lol

[hollow geyser]

I lied to myself

okay then sierpinski space was the answer all along

[unreal stratus]

Indeed

Or uhhh like

[0,oo) with open sets those of the form [0,a)

[hollow geyser]

I'm guessing this is for a topology class, and you understand the definition of continuity from open subsets?

[hollow geyser]

oh, another clever one

[unreal stratus]

Then no the converse doesn't hold

Though I'm not sure how easy it is to concoct a kinda natural example

[heady skiff]

yea i understand taht definition

this isn't for a class i'm just self-studying from munkres lol

[hollow geyser]

(oh, ignore me. I was thinking of continuity from $Z\to X\times Y$)

[gentle osprey]

SWR

[unreal stratus]

Ah yeah, one way to do it is like

You can have functions R^2 -> R which are only continuous in certain directions

e.g. ||(x,y) |-> xy/(x^2 + y^2), with 0 at origin; there is no limit as (x,y) |-> 0 ||

Caveat: I did haveoto Google that lol

But that function is useful

There are even funnier examples

Like functions continous away from origin which are continously along every line towards origin

[hollow geyser]

I DM'd it to you too.

[fair idol]

Hey I'm confused about what the open sets in the disjoint union topology S disjoint union T.

I know it's the topology that makes the canonical injections continuous. Is this exactly the same as saying the open sets are sets in the disjoint union which are unions of open sets from both?

Or is that a bit different from saying the preimage of each injection is open

[lime sable]

yes, each open set can be written as the union of an open set from S and an open set from T

if U is open in S + T, then U = (U ∩ S) ∪ (U ∩ T)

conversely, if you have an open set from S and an open set from T, you get an open set of S + T

this is a bijection

[heady skiff]

oh ok

wait so i'm blanking

what's the easiest way to see functions like multiplication from R^2 to R are continuous

[hexed steppe]

verify the definition

[ebon galleon]

epsilon delta

[gritty widget]

Someone pls explain what is Monge cone?

[honest shore]

Guys I just started math in English with algebraic topology I don't get what's a cell

Like they basically explain how a cw complex is created by inductance while u "glue" k cells to X^k-1 so you get X^k and then u consider their union

[steel glen]

the n-cells are the n-disks
so points, lines, disks, balls, etc.

[honest shore]

Oh ok

In wikipedia it's said that they are homeomorphic with the n balls so it's should be a different notion, am I wrong ?

[steel glen]

since they are homeomorphic they are interchangeable

the theory is the same either way

[honest shore]

Yeah why wouldnt they just say n balls, I just don't get this notion

[steel glen]

how does hatcher define the cells

he uses D^k right?

[honest shore]

He doesn't that's my issue 😭😭

[steel glen]

oh

[honest shore]

Maybe I missed it

[steel glen]

this is from chapter 0 where he is pretty loose (moreso than usual)

also i just noticed he uses open disks, but you can just close them by attaching (k-1)-cells inductively

just be aware that he uses the open disks

[honest shore]

Ok, would you suggest me another book or is it a good one

[steel glen]

i think this is a good first book, especially if you prefer geometric thinking

but im not the most knowledgeable so i cant really say!

[honest shore]

Ok ty !

[prime elbow]

Is it a coincidence that in topology we say the space X is Discrete topological space if all subsets of X is open set and in metric space we know that if we (X, d_disc) then all subsets of X is open set

[red yoke]

Metric spaces are topological spaces

The discrete metric yields the discrete topology

(Though there are some other metrics that also induce the discrete topology)

[prime elbow]

Example

[red yoke]

Half the discrete metric

[prime elbow]

Yes

But the idea is the same is there other metric which does not have structurally discrete metric and form discrete topology

[red yoke]

Exercise: find all the metrics that induce the discrete topology

[prime elbow]

Okay

In this question what is meant by complement of all finite subsets? Complement of each finite set?

[gritty widget]

First time seeing discrete metric

[pallid delta]

Yes
{R \ A | A is finite}

[gritty widget]

Is it like, any countable set we can have discrete metrics

[unreal stratus]

Any set

[prime elbow]

It is a topology, right?

[pallid delta]

Called the cofinite topology

[prime elbow]

Okay thank you

[pallid delta]

You also have the co-countable topology

[pallid delta]

Generally, co means we define the topology by the closed sets, like we did here

[unreal stratus]

Well also empty set

[pallid delta]

Ofc

But there was no confusion about that

[unreal stratus]

Feels pedantic but the way you write it makes it seem like that's everything

[prime elbow]

In co-countable contain uncountable and X sets ?

[prime elbow]

From we conclude that R is also in that topology

[pallid delta]

Replace finite by countable

[unreal stratus]

No I mean the empty set itself

Not its complement

[prime elbow]

Its is given that empty set belonging to topology

[unreal stratus]

I know

I was just pointing out that what was written wasn't a topology

[prime elbow]

In that I use intersection of countable set is countable and arbitrary union of countable set is countable right

[prime elbow]

Okay

Why we take a finite intersection why not infinite ?

[pallid delta]

For open sets ?

[prime elbow]

Yes

[pallid delta]

We want it to match what holds for metric spaces

[prime elbow]

Wait this is wrong

[pallid delta]

Hence my reaction

[unreal stratus]

A countable union of countable sets is countable ||assuming axiom of choice||

[prime elbow]

Arbitrary intersection of countable set be countable and finite union of countable set is countable

[pallid delta]

Ok, I shall never argue against AoC again

[prime elbow]

I don't know about it

[unreal stratus]

Note it would fail for R for example

[pallid delta]

You've never talked about distances and metric spaces before doing general, point-set topology?

[unreal stratus]

since then ${0} = \bigcap_{n=1}^{\infty} (-n,n)$ would be open

[gentle osprey]

Süßkartoffel

[unreal stratus]

and then every subset would be open

[gritty widget]

I see, okay

[prime elbow]

I am doing it together

Yeah but where is problem in construction of that proof because counter example comes later

[grim knot]

Guys am I flippi, ora H_2(X) here should be Z_2 + Z?

[unreal stratus]

No, they are correct

More generally this follows from the Künneth theorem

[grim knot]

we didn't do that

How does the quotient simplify?

[sick wharf]

I have a question about the contractibility (homotopy) of the unit sphere in R^omega.
The homotopy linearly interpolates between identity, and right shift : (and then maps the codimension -1 sphere to the unit vector (1,0,0,...)

https://math.stackexchange.com/questions/282268/unit-sphere-in-mathbbr-infty-is-contractible

I wonder if we can add constraints, so that things like the right shift operator are not allowed. The right shift operator sounds like a dirty trick to me.
Something like requiring the tangent planes of the continuous function to have nice properties.

This is so frustrating. I cant put my finger on the problem.

[formal tide]

Munkres' definition of locally compact (each point has a compact neighborhood) seems to be the most common, yet I've never seen it mentioned in any theorem without the addition of T2

is there any interesting theorems that only require that definition, and not the stronger "each point has a local base of compact neighborhoods"?

[gleaming warren]

I guess you can build the one-point compactification with only knowing about one compact neighborhood

[unreal stratus]

Idk, I mean using the right shift is a sort common trick with infinite sets and stuff

i'm not sure what you mean by "adding constraints"

[grim knot]

I'm doing c of this exercise and I don't find solutions of it. Could maybe somebody tell me what the homology groups are, just for me to compare the result. Also if somebody knows where I coul find a solution, it's highly appreciated

[sick wharf]

I mean usually when we generalise something to infinite dimensions we add additional constraints E.g. boundedness, element-wise square summable (for trace class operators), …

I think if we define a equivalence relation like homotopy with additional constraints on the continuous function, then the infinite dimensional sphere wouldnt be equivalent to a point.

The way It looks to me, is that homotopy „doesn‘t work“ on holes holes in infinite dimensional space. (Except the pi_n homotopy groups.)

[umbral panther]

Sure, you can work with proper homotopy equivalence

Or the extra data used in semi-infinite cohomology

[sick wharf]

But is the infinite sphere contractible using proper homotopy equivalence? If so, we have still have the same problem.

[umbral panther]

No

[sick wharf]

Then nvm. I'm gonna look into it.

[empty grove]

S^infty is strongly contractible

[unreal stratus]

Hm why should that mean it "doesn't work"

Like do you think the sphere shouldn't be contractible?

[umbral panther]

It’s not contractile by a proper homotopy. Nor a bounded homotopy, which is a better choice

[sick wharf]

yeah, because if it is contractible, then we cannot tell if a set has a hole or not, based on its homotopy groups.

(but ofc using homotopy it is contractible, I'm not doubting that statement. Just that this renders homotopy useless in this setting.)

[umbral panther]

But it doesn’t have a hole in it. It’s homeomorphic to an infinite dimensional vector space

[sick wharf]

I see.

[sick wharf]

But if we have e.g. R^omega / {(0,0,...)}. Is there a proper homotopy between the unit sphere and a constant?

(I'm confused rn.)

I mean we still have to capture the notion of a "missing point" "hole" somehow.

[umbral panther]

I’m not sure, but I’d guess that there is a bounded homeomorphism between the unit disk and the unit disk minus a point (and the unit sphere)

(Bounded = Lipchitz)

[sick wharf]

I'm just gonna speculate, that there is (or we can define) some "homotopy relation" for which we can define a chain of homotopy groups pi_0,pi_1,pi_2,... but also extending to homotopy groups for spheres of a finite codimension. (Which then correspond to the "holes" that are points, lines, 2d-planes, ...).

Gonna have to do more reading though.

[plain raven]

yeah but who ever took a union really
You do category theory you only consider disjoint unions

[unreal stratus]

Based?

[plain raven]

Ok i guess wedge sums of basepointed spaces are not technically disjoint unions

but still, the wedge sum of countably many countable pointed sets is still countable

[ebon galleon]

lol i was like why would that require (countable) choice and then it hit me.

[grim knot]

I know that I'm spamming you guys, but do you know how I could compute this? I thought about doing a CW complex structure on X which would consist of one i-cell for each i in {1, ..., n-1} and two n-cells and then computing it through cellular homology, but I think something's fishy

[unreal stratus]

What hit you?

Jk

[fading vale]

Your CW structure does not have 2 n cells

Or at least the natural one doesn't

Oh wait I misread

[unreal stratus]

I misread it too lol

[fading vale]

Yeah this is fine

Your n-1 skeleton is a pretty recognizable space if you stop to think about it so most of the work may be already done for you

[grim knot]

it's RP^(n-1) right?

my problem is that me and a friend of mine have two different solutions

[fading vale]

Yes

Well

One of you is definitely wrong

[grim knot]

yeah or both

[unreal stratus]

Assuming consistency of ZFC

[fading vale]

Ew

[grim knot]

I got homology group Z/2Z for k = n-1 and and Z for k=0,...,n-2 and n and 0 otherwise

and it made sense to me cause the boundary homomorphisms I got are all 0 (?), besides from d_n

[unreal stratus]

I never know how you are meant to do geometric things like this without being able to visualise lol

Other than ig breaking stuff up lol

[grim knot]

I don't know, we don't even have solutions and I don't find anything on the internet

[fading vale]

Do you know what the homology of RPn is

[grim knot]

yes it depends if n is even or odd

gimme a sec

[fading vale]

Yes and it also depends on if i is even or odd for H_i

[grim knot]

this thing

[unreal stratus]

I don't see how this is RP^{n-1}, isn't it that with two more n cells?

Oh lmao

[grim knot]

guys I am so confused

[unreal stratus]

Okay I misread this

Sorry ignore me

Yes, the n-1 skeleton is RP^{n-1}, we are all agreed on that

[fading vale]

The boundary maps for i < n are not changing so your homology should agree with RP^n-1 for i<n-1

[unreal stratus]

So yeah the way I would think about it is that we understand the n-1 skeleton Y and the quotient X/Y very well

[grim knot]

I thought that the boundary maps are all 0

[fading vale]

No

Reread a description of the computation of the cellular homology of RP^n

In Hatcher or something is fine

[unreal stratus]

The normal CW decomposition of RP^n has a single cell in each dimension (up to n). You may be thinking of CP^n

[grim knot]

wait, do we go around it twice here as well?

[unreal stratus]

Where it's one cell in each even dimension, and so the boundary maps are obviously 0

[unreal stratus]

What do you mean

[grim knot]

when we look at the generator for even degrees, than the boundary map is multiplication by two?

[fading vale]

Yes

[unreal stratus]

Yeah it alternates between 2 and 0

[grim knot]

but why though? If I only have one cells representig the skeletons

[unreal stratus]

I don't see why that helps

You can still get arbitrary boundary maps

Like if you just take a wedge S^1 v S^2 v S^3 v ... then all the boundary maps would be zero

[grim knot]

I see this, but I don't understan why it has to alternate

[unreal stratus]

I mean,

that is a computation you have to do

It isn't entirely obvious

[fading vale]

Again this is laid out in e.g. Hatcher

[unreal stratus]

But basically it comes from the fact that you can get a cw structure on S^n with one cell in each dimension, and then quotient that out b antipodal thing

[fading vale]

You should just look at it there

[unreal stratus]

and the antipodal map on S^n has degree (-1)^(n+1)

That's at least my intuition lol if you go through the cellular stuff

[grim knot]

I will come back if I don't get it

[unreal stratus]

Well

[grim knot]

that makes sense cause it changes the whole time

[unreal stratus]

the computation of RP^n is like

kind of a prerequisite for this question i guess

[grim knot]

It makes sense I guess, I just didn't see how my construction leaded directly to RP

[unreal stratus]

Well tbf you could blackbox it but it is good to be familiar w it i suppose

[unreal stratus]

Well the equator is quotiented out bythe anitpodal map, which is exactly how you form RP

[umbral panther]

Have you tried n=1?

[grim knot]

yeah and two as well

[unreal stratus]

Well if you've proven something for 1 and 2 then by induction it holds for all n

[umbral panther]

Does it match this?

[grim knot]

unfortunately yes, this is why I got confused

[hidden crag]

Taking this out of context

[grim knot]

I see it better now, thank you patata dolce

[umbral panther]

Unfortunately…no?

[hidden crag]

Oh wait that was a joke

[unreal stratus]

It was lol

[hidden crag]

I thought it somehow fit the context because I didn’t read the rest

[unreal stratus]

Probably a good heuristic though

Lol

[ebon galleon]

So true!

[median sand]

1. Are homo-topy/logy groups of compact spaces finitely generated? Feels like "morally" it ought to be so, in Euclidean space at least, although knowing topology there's some fucked-up counterexample.
2. How does the classification of fingen abelian groups enter into AT? If H_n(X) is fingen, can X be decomposed somehow in terms of the free/torsion parts? What do the rank and the invariant factors represent? Do they determine a space uniquely (I expect not)?

[umbral panther]

Compact is not a good hypothesis. The Hawaiian earrings is compact. So is the infinite product of circles

I’d guess that the Czech cohomology of a compact space is compact

The homology of a finite complex is finitely generated. Its homotopy groups are modules over the group ring. The wedge of a circle and S^n has pi_n equal to the free module over the group ring Z[Z] = Z[t,1/t]. That’s not finitely generated as an abelian group. Since the group ring is rarely noetherian (eg, a free group), finite generation as a module over the group ring is brittle and there are finite complexes with homotopy groups that are not finitely generated as such modules

If a simply connected finite complex has a homotopy group that is nontrivial, then it has infinitely many homotopy groups that are nontrivial. But they are individually finitely generated. I think this was proved by Serre in his 1954 thesis. He built several machines to prove this. Someone described this as “He took the bull by the horns.” The argument requires as input that the homology of K(pi, n) is finitely generated in each degree. I’m not sure how hard that is to prove, but the complete answer was computed before Serre

(Open problem: if a group has a noetherian group ring then the group is virtually nilpotent)

[median sand]

Thanks @umbral panther, very informative (if not entirely clear, for lack of requisite knowledge on my part) as always.

[plain raven]

I looked into this and in "Two-dimensional monad theory" it is proven that, if K is a 2-category and T is a 2-monad, then the category of T-algebras and strict T-morphisms is a reflective subcategory of the category of T-algebras and lax T-morphisms, under the assumption that T has a "rank", i.e., that it preserves filtered colimits up to some regular cardinal alpha.

This seems to generalize the construction of the simplex category from the one object category. If I am not mistaken one should be able to apply this to, for example, 2-categories and bicategories, and get that lax (resp. pseudo) functors between bicategories C -> D correspond to strict functors between bicategories C' -> D for some C'. I have not checked that the monad in this setting preserves filtered colimits, mind you, so the theorem may not apply.

[unreal stratus]

How do you show the first bit / do you have a reference?

Or does this basically follow from a computation for S^n

Oh okay, Serre still lol

[prime elbow]

I think it is topology, right?

[radiant walrus]

@paper wedge I managed to prove that theorem I told you about a few days ago!

It's a lot longer than I expected

[paper wedge]

That's great

I don't think it's easy to solve on your own ,imo

you should be proud

thats cool

[paper wedge]

check closure under unions

and intersections

[radiant walrus]

It took me the whole night

I wrote it on notepad but

[paper wedge]

did you figure out the diagonal argument

[paper wedge]

to prove completness?

[radiant walrus]

Completeness is proved in (8)

I proved that W is dense first though, which is (7)

And used that fact to prove completeness

[paper wedge]

im too busy rn to check it out but overall its great that you managed to do this much progress

good job

imo

[radiant walrus]

Ahh thank youu ❤️

It was really difficult

[paper wedge]

yeah it isnt easy imo

[radiant walrus]

I was struggling to prove that it was complete

But then I thought

"If I prove that W is dense, can't I just reroute the Cauchy sequence to a Cauchy sequence in W so that I have more information?"

And that's what I did

And it worked

[paper wedge]

yea ig this should work too i think i know what u mean

[radiant walrus]

Yeah like

I can always pick points in W arbitrarily close to any point in X^

[paper wedge]

if you have a metric space, call it X

[radiant walrus]

So I just made a sequence in W such that each xn is arbitrarily close to each one of the X^

[paper wedge]

and a dense subset such that this dense subset is complete then X is complete

[radiant walrus]

W is not necessarily complete though

In fact, I didn't even assert that or used that

[paper wedge]

hmm then ig hopefuly someone can check (8)

in ur proof

cuz idk how u would do it without either diagonal argument or proving that T(X) is a dense complete subset

[radiant walrus]

Well, I can write my (8) here

(8) Now we will prove that X^ is complete. Let (x'n) be a Cauchy sequence in X^. Then, given e > 0, there is an N such that n,m > N => d^(x'n, x'm) < e/3. Since W is dense in X^, let (x"n) E W be such that d^(x'n, x"n) < e/3 for each n. Then, n,m > N => d^(x"n, x"m) <= d^(x"n, x'n) + d^(x'n, x'm) + d^(x'm, x"m) < e/3 + e/3 + e/3 = e, so (x"n) is also a Cauchy sequence. Let (xn) E X such that, for each n, Txn = x"n. Using the same reasoning as in (7a), (xn) is also a Cauchy sequence. Thus, there is an x' E X^ such that x' = [(x1, x2, ...)]. Since (xn) is a Cauchy sequence, then, by increasing N if necessary, n,m > N => d(xn,xm) < 2e/3. By taking n > N, we see that d^(x"n, x') = d^(Txn, x') = lim m->infinity d(Txn_m, x'_m) = lim m->infinity d(xn,xm) < 2e/3. Thus, we have that, by taking n > N, d^(x'n, x') <= d^(x'n, x"n) + d^(x"n, x') < e/3 + 2e/3 = e, thus (x'n) -> x', making X^ a complete metric space.

For context
(1) Let Y be the set of Cauchy sequences of X. We'll create a relation ~ on Y, in such a way that x ~ y <=> lim n->infinity d(xn,yn) = 0.
(3) Let X^ be the set of all equivalence classes of Y with ~. Let d^: X^ x X^ -> R be defined by d^(x,y) = lim d(xn,yn), where (xn) E x and (yn) E y.
(6) Let T: X -> X^ be defined by Tx = [(x, x, ...)]

It cites a reasoning in (7a) in the middle. This is the reasoning
(7a) We'll now show that W is dense in X^. Let x' E W-, then, there is a sequence (x'n) E W such that lim x'n = x'. Since (x'n) E W for each n, there is an xn E X such that Txn = x'n. We'll show that (xn) E X is a Cauchy sequence. Since (x'n) is a convergent sequence, then it is a Cauchy sequence, as shown by Theorem 1.4-5. Thus, given e > 0, there is an N such that n,m > N => d^(x'n, x'm) < e => d^(Txn, Txm) < e => d(xn, xm) < e, which means (xn) is a Cauchy sequence

[prime elbow]

Yes it will be closure under arbitrary union and finite intersection

[radiant walrus]

Idk I don't see any flaws with my logic 😭

Like

(x"n) wants to converge to x', but you can't assume x' is in W

Otherwise you would have W = X^

[radiant walrus]

Like, being complete means that it's also closed, so W = W-
But also, being dense means that W- = X^
So you're basically saying that W = X^

[slate bane]

Hey, where could i learn very well about p-adics numbers, p-adic norm/valuation, p-adic topology.. in the context of a point-set topology course?

[merry geode]

That seems lengthy, are you constructing completion? what is W here?

[merry geode]

Hm, idk if point-set topology courses introduce these. Do you want some introduction on these?

[slate bane]

Yes, they're included in the program of our first topology course, but since I have almost no intuition or clue about them I'd like to understand their picture in the literature: where and in what context are they canonically introduced? What books treat them, for me to possibly rely on?

[merry geode]

Wait, is it included in the course?

Sorry, I have no idea what kind of material would it be.

But p-adic integers do have a metric, which gives the topology

So I think most reasonings about metric topology will work.

Ofc, there is one more thing - the p-adic metric is ultrametric, that is,
d(x, y) <= max (d(x, z), d(z, y))

[slate bane]

Yes, if I understand it correctly there will be a section about field norms and this is a prominent argument in that section

We'll work with the p-adic metric I suppose, yes. So is it just an example of a metric like any other, not of particular interest in the most common books?

[merry geode]

Yeah, at least I know munkres (and other similar books) does not deal with them much

[slate bane]

Thank you. And what about category theory? The corse also offers an introduction to it and shows its use in a topological setting (relations with fund. groups, for instance, I guess, but this is the only example I know of in advance, any others? [Question 1]).
Some resources that could be helpful in this regard? [Question 2] (That is, I start from the basis, and the category theory we'll deal with is topologically oriented)

[merry geode]

Well, in my algebra course prof talked about how category theory is usually taught in an ad-hoc manner

You can perhaps look into Weil's homological algebra, if you are interested in algebraic topology. But there is also the famous "Category theory for working mathematicians".

[slate bane]

Mm, that may be too advanced, the second I saw quoted on wikipedia, so I guess I'll give it a look, thanks!

Anyway, I also saw "Topology: a categorical approach" in the #books channel. Do you think it could be helpful as well or just too focused and overkill for my purpose?

[grim knot]

Guys question: what is the homology group of two disjoint points? like $H_p({N} \cup {S})$ for every p? I thought that it has to be $\mathbb{Z} \oplus \mathbb{Z}$ for p=0 and 0 otherwise. Now my question is, what is it's reduced homology?

[gentle osprey]

damn_guuurl

[red yoke]

reduced H0 = Z

[merry geode]

Good luck!
I cannot access the channel, so I cannot help with that

[red yoke]

A 0-hole is a gap between components

[grim knot]

why is it Z and not 0? I have a disjoint union right? so theoretically I could split the union(?)

I'm doing something wrong in my reasoning

[red yoke]

2 points, 1 gap

[red yoke]

H(topologically disjoint X ∪ Y) = H(X)⊕H(Y) holds for nonreduced homology

[grim knot]

I see what you mean, but I am doing something wrong in my calculation. I wrote:
$\tilde{H_0} ({N} \cup {S}) \cong \tilde{H_0} ({N}) \oplus \tilde{H_0} ({S})$

[gentle osprey]

damn_guuurl

[grim knot]

ohh that makes sense then

[red yoke]

The reduced version is wedge sum

At some sufficiently nice points on X and Y

[grim knot]

that makes obviously much more sense, thank you very much

so I can kind of erase one Z right?

[red yoke]

Ye

[grim knot]

thank you very much!

I have a related question to the above, If I have D a disk with k points removed, then why is $\tilde{H_0}(D)=0$? Is it because D is pathconnected?

[gentle osprey]

damn_guuurl

[hidden crag]

Yes

[grim knot]

thank you

[radiant walrus]

W = T(X)

Sorry for forgetting to say that

[merry geode]

Ah, so T is the natural embedding?

[radiant walrus]

I don't know what that means exactly

[radiant walrus]

I explain here what T does

[merry geode]

Sorry for jargon, I meant T is an injective continuous map

[radiant walrus]

Not necessarily continuous

[merry geode]

Which lets you identify X as subspace of Y

Ah perhaps continuity should be shown separately, yeah

[radiant walrus]

T: X -> X^ is an injective isometry

So X and T(X) are isometric

And I call W to be T(X)

[merry geode]

IIRC an isometry is continuous, but yeah that is not the point

[radiant walrus]

Didn't need to use that fact here, but good to know

Yeah I just saw this in the exercises

[merry geode]

How did you deduce that d^(x'n, x"n) < e/3 for each n ?

In (8)

[radiant walrus]

W is dense in X^, so, for any point in X^, in this case x'n, I can set up a sequence x"n in W such that the distance from x'n to x"n is less than e/3. It doesn't matter what the actual x"n is. What matters is that it exists, so you can pick any that satisfies it

[merry geode]

So, (x"n) depends on e, right? Because the choice is based on distance being < e/3

[radiant walrus]

Yeah?

[merry geode]

This poses issues in showing that the sequence is Cauchy

[radiant walrus]

Why?

[merry geode]

Because you need to be able to give arbitrary epsilon there

[radiant walrus]

I don't see the problem?

I got d^(x"n, x"m) < e

[merry geode]

Let's consider the real numbers, for instance.
In current state, you can make it so that
x"n = x'n + (-1)^n e/6 for all n.
Say, x'n = 0 for simplicity.
So (x'n) is trivial Cauchy sequence, but (x"n = (-1)^n e/6) is not Cauchy.

The issue here is that, to show that a sequence is Cauchy using epsilon-delta, the sequence should not depend on the epsilon.

[radiant walrus]

What if instead I do this?
Let (x"n) E W be such that d^(x'n, x"n) < 1/n for each n. Then, by increasing N to N > 3/e if necessary, you have n,m > N => d^(x"n, x"m) <= d^(x"n, x'n) + d^(x'n, x'm) + d(x'm, x"m) < 1/n + e/3 + 1/m < e/3 + e/3 + e/3 = e

It doesn't depend on epsilon now

And the distance keeps decreasing as n increases

I should've done that instead tbh. By not making the distance approach 0, you could have the sequence hover around in a circle

That part is fixed now though, yeah?

[merry geode]

Yeah, this works!

Otherwise, it looks good to me!

[radiant walrus]

Is the rest okay?

Ngl, on both (7a) and (7b), I said that a sequence converges to a point when that point might not be in the set. I just couldn't be bothered to differentiate from converging to "wanting to converge to"

Like, the sequence is clearly converging to a specific point, but that point might or might not be in the set, which is usually what can happen in non-closed sets

Thankfully the result is the same. I just wanted to spare myself some lines of writing, since this already took me 3 pages to write out

[merry geode]

You did prove that it converges in some element in the bigger set, right

[radiant walrus]

Yeah

It's just that this element in the bigger set might not be in the smaller set

So like

It "wants to converge there"

[merry geode]

Yeah, you can simply say it converges

[radiant walrus]

That's good

Now I gotta edit this thing you pointed out, and hope that I can make the handwriting small enough so that all of it fits

[merry geode]

Good luck!

[radiant walrus]

Thank you ❤️

It worked

But damn

Getting the logic of all of this proof was tiring

There were so many steps..

So many things I needed to prove

Hopefully the exercises should be easier

[merry geode]

Maybe it helps to shorten & summarize your approaches, and perhaps looking into other people's proofs.

[knotty vine]

I first learned of it here https://golem.ph.utexas.edu/category/2018/05/laxification.html

[foggy cairn]

I'm trying to prove that R^3\S where S is finite is simply connected. I can show path connectedness, but struggling with trivial fundamental group. I thought of two approaches and my friend suggested a third:

1. (probably bad) Make epsilon balls around each point in S, and argue the straight line homotopy can be modified to continuously pass over these balls.
2. Use induction and van kampen: wlog set basepoint as 0. Then assume R^3 with k points removed is simply connected. Now choose A1 to be R^3 with an additional point removed, and A2 to be a tube connecting 0 and this point. But problem is I'm not sure how to argue R^3 with an additional point removed still has trivial fundamental group.
3. Make balls around each point, join these balls with tubes and map R^3 onto the surface. Now this is equivalent to several S^2 balls joined together and since each S^2 has trivial fundamental group the result follows. The problem I have with this is I don't see why such a map onto this surface should necessarily exist. If I have just one ball I can map using e.g.x/|x| but I can't see any clear way that this can be defined for multiple connected balls in this case

Any help is appreciated!

[unreal stratus]

I think the second one is perhaps cleanest/quickest

So we can basically (by moving stuff around) pick some plane in R^3 with one point removed "above" the plane and all the other removed points (possibly none) below that plane, then fatten that plane out so we get van Kampen scenario

So all you have to do is to show that the "top bit" is simply connected, and this is just R^3 minus a point

[umbral panther]

Can you prove that the choice of S doesn’t matter? That R\S is homeomorphic to R\T if S and T are finite sets of the same size? And is a particular choice of S easier?

[high hill]

Interesting... is there a (interesting) larger class of S this holds for? I feel like compact works??

[red yoke]

S¹ fails

[high hill]

huh

[high hill]

i misread as R^2. but anyways, what loop fails to contract?

[red yoke]

A loop through the S¹

[high hill]

wait thats a set minus?

yeah so I originally read the Q as R^2/S

amazing

[foggy cairn]

Thanks! If I can prove this then the points can just be e.g. integers, and I can think of a map that takes R^3 to balls around them (and then use method 3). But is this statement itself actually easy to prove...? I might be missing something

[umbral panther]

I don’t understand your method 3

[umbral panther]

I like this

Rotate R^3 until there is a unique point with largest z coordinate

[foggy cairn]

This is like projecting R^3\{0} onto S^2, but with multiple S^2 adjacent to each other

[foggy cairn]

Right.. do you know what they meant by 'flatten the plane out'?

[umbral panther]

The van Kampen theorem requires that the overlap be open. A plane is not open. But an epsilon neighborhood of a plane is open

[foggy cairn]

Ah

Ok I see. Yeah that's really helpful. Thanks!

(don't know if I should ping subkartoffel since they're a mod : P)

[unreal stratus]

Yeah I said fatten, like

Often you have a decomposition of a space into a couple of parts and then you can just force a little bit of overlap in order to be able to apply van Kampen :) So for example if you're computing the fundamental group of S^2, you can't just take the north/south hemisphere (in the normal statement of van kampen anyway) but you can just take a bit more of each

[unreal stratus]

Feel free, dw

[grim knot]

If my exam goes well tomorrow, it is because of you guys. Thanks for helping me through all my problems!

[hidden crag]

Viel Erfolg

[grim knot]

Danke dir✨

[novel ember]

i wish i can say that there is some fixed epsilon s.t every point in the line has an open ball of radius greater than eps

[knotty vine]

Use compactness?

The image of the path is compact

[novel ember]

wait how is it compact

[knotty vine]

Cause it's the image of the compact space [0,1]

[novel ember]

oh my god

why the fuck was i thinking homotopy shit

i was somehow thinking of [0,1]xR^n

instead of [0,1]

[quartz horizon]

Actually could you use this result plus induction to show that R^3 \ S for S finite is simply connected

You reduce it to showing closed polygons are homotopic to a point or smth

Hmm might need a bit more thought than that

[novel ember]

i remember doing like

a handwavey version of this

choose 2 paths, create epsilon balls around the finite points, create a surface with the 2 paths and then if at any point we have an intersection we just mold the surface around the ball

we don't need to worry about any of the balls intersecting because R^3 is hausdorff

[quartz horizon]

Like i think this works

Maybe

[radiant walrus]

Why the need to specify that it is a subspace? Why not just say that it is a metric space?

I feel like this is unecessary information

[gentle girder]

it is, any finite metric space is complete, since for any sequence x you must have some N such that for n, m > N d(xn,xm) < min(d(a,b)) (a neq b in Y)

[radiant walrus]

Yeah it's not hard to prove that they're complete

[gentle girder]

okay so yeah it’s unnecessary info

if you prove it for finite metric spaces, then you can just apply it to Y knowing that subspaces of metric spaces are again metric spaces

[radiant walrus]

If a metric space is such that there is a positive real number c such that d(x,y) >= c for all x != y, then that metric space is complete

Because then you can pick a small enough epsilon so that the Cauchy sequence eventually becomes constant and thus convergent

So basically, this works for vector spaces with finitely many points, with the discrete metric space, with the integers with its regular metric

[unreal stratus]

I agree lol it is needless

Being complete is a notion internal to a metric space, rather than being defined for subsets etc

[radiant walrus]

Mhmm

[radiant walrus]

Isn't it itself since it's already complete?

[ebon galleon]

Correct

[radiant walrus]

Alright then

[prime elbow]

I have a doubt, in standard topology on R( Real line ) we take the basis as a collection of all open intervals , my doubt is if I take two basis elements (a,b) and (c,d) and they are disjoint then how can I say that their union is open set because both are open set, right?

[gentle girder]

their union is open because unions of open sets are also open, remember the definition of a topology

[prime elbow]

I think I am getting it now

[prime elbow]

Actually I was confused with open intervals I thought how a union of disjoint open intervals are open intervals

[gentle girder]

it's not an open interval. The topology isn't just "elements of the basis" and that's it. You get new open sets by unioning your open sets in the basis together.

[prime elbow]

Yes thank you

[prime elbow]

If I am correct then the intersection of the finite open intervals is open interval?

[gentle girder]

yes, the intersection of two (then by induction finitely many) open intervals is an open interval, but you shouldn't let that trick you into thinking that finite intersections of basis elements are again in the basis in general (for example in R^2, take two open balls and take their intersection, you'll get an open set, but not something in the basis itself)

[prime elbow]

Yes, thank you and the arbitrary intersection of the interval is the interval itself?In R^2 what will be my basis ?

[red yoke]

The intersection of rectangles be rectangles though

[prime elbow]

Intersection of circles is not a circle, right?

[alpine nest]

Not in general

[radiant walrus]

It's hard to think of an homeomorphism between an incomplete metric space and a complete one. I know that it can't be an isometry, since it would imply that both are complete. Hmmm...

[prime elbow]

Can you please explain it

[knotty vine]

The intersection of two open disks is a disk if and only if one is included in the other. (Or disjoint if we allow the empty set to be an open disk of radius 0)

[woven sinew]

A good example is to think of the open interval (0,1) and the real line (with their usual metrics).

[prime elbow]

Yeah I get it, thank you

[radiant walrus]

Why didn't I think of this

I kept thinking about stuff like "R and Q" or "R and R\Q"

[alpine nest]

I meant what Semer explained above, it sometimes is but only under specific conditions

[prime elbow]

Got it, thank you

[merry geode]

Homeomorphism between R and (0, 1) is one of the important recurring examples, good to memorize that.

[bright acorn]

Do pontryiagin classes satisfy the similar property of chern classes that any characteristic class of a real vector bundle can be expressed as a polynomial on pontryiagin classes?

[woven sinew]

My brain also goes CaNtOrSeT, whenever it needs a counterexample for anything.

[red yoke]

Is the Euler class a polynomial of pontryagin classes

Should be no by looking at degrees

[prime elbow]

If X has more than one element then there always exists a topology where I can find a set which is neither open nor closed , that topology may be indiscrete, is it correct?

And in co-finite space , the closed set are which are finite set and also the set X

[ebon galleon]

Yes you can take indiscrete

[prime elbow]

Thank you

In this question topology is not specified, I think it is Standard topology

[ebon galleon]

Unless otherwise specified, you can always assume R has the standard topology

[prime elbow]

Thank you

[prime elbow]

Closed sets concepts are different in Topology and metric space, right?

In metric space X, any singleton set is closed but in topology it is not true in general

[plush folio]

A metric is a topological space, so the concepts are the same, but more restricted in a metric space. But what you're saying is correct

Are there any beginner books on topology that focus on universal properties? Or do I need to learn point-set topology first, then universal properties crop up later in algebraic topology?

[prime elbow]

Thank you

[radiant walrus]

Honestly I didn't think of one about that. I just thought of one about R and (-pi/2, pi/2) by doing arctan x. But I guess you can shift it to (0,1) by doing arctan(pi/2 x)/2 + 1/2

[merry geode]

Yeah, that is also a good one

Shows that completeness is not an intrinsic property to a topology.

[thorny agate]

last part of this problem, showing closure of irred is irred

is X assumed to be irreducible?

or is X just some arbitrary topological space necessary for closure to make sense?

[ebon galleon]

X is any space. Y is irreducible in the subspace topology. Show that Ybar is irreducible in the subspace topology.

[thorny agate]

Aight

[grave solstice]

How big is knot theory compared to other subfields of topology?

[fading vale]

I don't really know how to quantify that but it is a significant subfield

Also it has a lot of overlap with other parts of topology (especially 3 manifolds and contact topology)

[merry geode]

Knot theory seems simple and easy on the surface.

That said, likely this first impression is deceptive.. (like how I struggled with algtop despite even easier first impression)

[grave solstice]

why? It doesn't look like that to me (not that it looks particularly complicated)

[merry geode]

Because it deals with knots, which are visible. You can (crudely) make a knot in reality, and check what it's like.

[grave solstice]

mmh

[merry geode]

e.g. Isn't this soo simple? https://en.wikipedia.org/wiki/Bracket_polynomial

[coral pawn]

What do they mean by the homology of E_r with respect to d_r? Is it just Ker d_r / Im d_r?

Also, this definition, when applied to the category of bigraded modules, doesn't agree with the usual definition of spectral sequences, does it?

[umbral panther]

Yes, exactly, a special sequence is a sequence of chain complexes, each of which is the homology of the next

What do you think is the difference between the two definitions?

[coral pawn]

Well, the d_r shift degree by (+r,-r+1) in the usual definition, but in the above definition, if d_r are required to be maps of bigraded modules E_r --> E_r, we would require the grading to stay the same

[umbral panther]

A spectral sequence doesn’t need any grading at all. If there is a grading, the differential changes degree by 1. Cohomological means that it increases degree by 1 (just like in a chain complex). In the bigraded setting there is an additional preserved degree

[coral pawn]

Wait so are you saying is that if our objects in the category have a grading, then we additionally require the differentials to jump grading appropriately?

[coral pawn]

But requiring a jump in grading would mean that the d_r are no longer morphisms of graded objects, which we require here

[umbral panther]

You can renumber to make it look like it respects the grading

[coral pawn]

Sure so doing something like $d_r: E^{x,y}_r ---> E^{x+r,y-r+1}_r$

[umbral panther]

Algebraic topologists often shear the grading so that one term is preserved

[gentle osprey]

Finitely Many Bananas

[coral pawn]

But then the differential is no longer a map of graded modules from E_r to itself

It's a graded nap from E_r to (E_r shifted by (r, -r+1))

[umbral panther]

It is what it is. You have some examples and they preserve what they preserve. The grading is a cosmetic detail compared to the spectral sequence which doesn’t need to name the grading at all

(Unless you have some multiplicative structure and you want signs in a Leibniz rule or something)

[coral pawn]

Got it. Thanks

Also

I've heard people say that derived stuff has replaced spectral sequences in modern approaches. Is this true?

[umbral panther]

If you use a spectral sequence to compute, you can’t replace it by a magic box where everything works, because eventually you have to actually compute

But if you have something like a Grothendieck spectral sequence that says that RF o RG => R(FoG), then maybe you just want to say that derived functors compose and you can get away without computing anything

[coral pawn]

Is it like model categories vs infinity categories where the latter is fine for developing theory, but is useless for computations?

[umbral panther]

You can say that right derived functor compose in model categories. Or triangulated categories. Maybe it’s not a good example. But then I don’t know what the claim is

[coral pawn]

Fair enough. It probably wasn't a well formed question. Thanks for the help!

[coral galleon]

How to define "finite spectrum" as excisive functors?

(which definition of spectrum is more oftenly used: the excisive functor one, or the sequencial one?)

[stuck geyser]

If you were to try to describe a 2d manifold to someone, would you say it's like a quilt of 2D spaces, i.e overlapping "patches" of 2D space in the homeomorphic sense

Classic example being the 2-sphere with the top and bottom hemispheres, or just, use the Reimann Sphere example

[solemn oar]

This sounds like Goodwillie calculus.
See also here and the references given: https://ncatlab.org/nlab/show/excisive+(∞%2C1)-functor#SpectrumObjects

Lurie uses the excisive functor definition in Higher Algebra.
Which description of spectrum objects is more convenient depends on what you are planning to do with them.

[stuck geyser]

Is the horn torus a 2D manifold even with that cusp at the center

[umbral panther]

No

[stuck geyser]

In that case, FUCK THE HORN TORUS

IT SUCKS

[cedar pebble]

singular spaces are bad

[lime sable]

hey come on, some of my friends are singular

[safe torrent]

singular cohomology

[west brook]

The spindle torus isn’t either

[prime elbow]

In standard topology on R the singleton set is closed? I think yes it is closed

[serene ferry]

Yes, they are closed.

[gentle osprey]

Volkenborn

[serene ferry]

Alternatively recall that singleton sets are compact (use standard open cover argument), and Heine Borel requires that every compact subset of R^n be closed and bounded so that also implies what you want

[prime elbow]

Thank you

Does indiscrete topology have only one basis or another?

[viral atlas]

Hint: every set in your basis is an open set in the corresponding topology. How many open sets can the indiscrete topology have?

[prime elbow]

2

[coral galleon]

thank you for your help!
I can't figure out how an excisive functor is related to a sequential spectrum, and I find it in Higher Algebra.
I was trying to show that any connective spectrum is a sequential colimit of finite spectra, and I found that the sequential definition is more convenient here.

[merry geode]

I wonder, can knot theory help untying this knot?

I guess this is a tangle rather than a knot, but still. Anyway

[paper wedge]

first compute the floer homology

[knotty vine]

Topologists literally begging to make themselves useful
"Can I pleeeaase untie your christmas lights?" "Pleaseee let met hang this picture for you!" "What's that, you want a pair of pants with precisely three holes in them? Coming right up!"

[woven sinew]

Knot theorist be like: "This invariant proves that your knot cannot be untangled. You're welcome."

[unreal stratus]

Stable homotopy theorist: have you considered just going to higher dimensions?

[viral atlas]

(it can be one, if you allow your underlying set to be empty)

In any case, the basis for the indiscrete topology on a set would be quite trivial

You want the entire space to be in it: no more, and no less

Although you could throw in a redundant empty set