point-set-topology

[gentle osprey]

qrno

[red yoke]

Yes

G/N → (G/N)/(kerψ/N) = G/kerψ → D

[worthy nebula]

Oh, this is the third isomorphism theorem, right?

[red yoke]

Yea some of the iso theorems

[worthy nebula]

Sure, thank you!

[shadow charm]

Again you’re taking the wrong implication: they’re never saying that U in T’ implies U in T. They show that for U in T’, one has x_n in U for n large enough. In particular, if U is an open set in T (and thus in T’ since T c T’), then it is also true that for n large enough x_n is in U

[real granite]

Oh I thought when they went from talking about U in T' to U in T contained in T', they were talking about the same U

I think it would have been clearer if they had just started with in U in T contained in T'

[plush folio]

Can you explain why you assume f in unbounded? I'm not super familiar with Bolzano-Weierstrass, but I thought that if you do a proof by contradiction then you have to assume that f is continuous and therefore bounded?

[gaunt linden]

My goal here was to prove that if a function is continuous, then it is bounded. I did that in contraposed form: assume it is not bounded, and prove that then it cannot be continuous.

[plush folio]

Aha, so it wasn't a proof of the original question, but just a lemma?

[gaunt linden]

Yes.

In particular the lemma Potato introduced here.

[unreal stratus]

Btw, it is actually very quick to prove the IVT from this fact

which I find amusing

I think it's like

eh okay i can't remember rip

[plush folio]

Can you prove it (the original question) using the extreme value theorem like this? Assume f is continuous and f(a) = f(b) = 0 and a < b. As SWR showed, f(c) != 0 for a < c < b, so assume WLOG that f(c) > 0. Then f attains a maximum M = f(m1) > 0 on [a, b]. By assumption there must be two values m1 != m2 such that M = f(m1) = f(m2). But then there exists an ε such that all of f((m1-ε, m1+ε)) is <= M and all of f((m2-ε, m2+ε)) is <= M, so by the intermediate value theorem there are four points around m1 and m2 mapping to the same value, which contradicts the hypothesis. Missing some details, but I think it works

[unreal stratus]

ah yes extreme value was what i wanted lol

[sullen nimbus]

is this true

i have been assuming its true and trying to prove it

and havent succeded

[dim perch]

I think this is probably not true, but I'm not a point-set person

e.g. you can prove that if such a topology exists, then for all x, at least one of {(-∞,x),(x,∞)} needs to fail to be open, since complements of 1 point in S^1 are connected

(I'd guess that there's a much simpler obstruction, but idk)

[unreal stratus]

Hm

this is tantamount to asking if there is a continuous bijection R-> S^1

[tender halo]

yeah

[unreal stratus]

and that is an easier thing to think about

[tender halo]

there obviously isnt one

[unreal stratus]

indeed

e.g. would be "increasing" in charts

[tender halo]

yeah

[tender halo]

and that would imply that its open

and therefore a homeomorphism

[unreal stratus]

sure, i had in mind more that like uh

you'd get a continuous bijection R \ {c} -> R and that is also impossible like

[tender halo]

also true yeah

i like yours better

[unreal stratus]

thanks though urs is a nice strong thing

[dim perch]

Is it obvious that this can't happen for some nonstandard coarse topology on R?

[unreal stratus]

this is the standard top on R

[dim perch]

The question asked for some tau' sub tau

[sullen nimbus]

continous bijection or an embedding of S_1 in R?

[unreal stratus]

like here i mean usual R

well like

say you find τ' with (R,τ') homeo to S^1

[dim perch]

ok yeah fine

I see your point

[unreal stratus]

the "identity" (R,τ)->(R,τ') is a continuous bijection and, par transport de structure, ...

[sullen nimbus]

oh okay

okay thanks @dim perch and @unreal stratus

[bold knoll]

How do I pass a topology exam where professor pulls problems out of his ass

[tender halo]

pull answers out of yours

[bold knoll]

You mean out of his

[tender halo]

that would be rude now wouldnt it

[bold knoll]

Let X be the real line R with the topology whose open sets are all subsets
of R that do not contain 0. Is X a T1 space?
Is this guy even real?

How is this even a topology my friend

[tender halo]

i mean its a pretty easy question

[unreal stratus]

Well if you have a collection of sets and none of them contain 0 the same is true of their union and intersection

Ofc you need to chuck in R as well though

[tender halo]

(the answer is no btw, any closed set contains 0 so one point sets are not closed besides 0 itself)

[unreal stratus]

Yeah

I guess it is T0 at least

Lol

[tender halo]

it is T0 yeah

its some kind of order topology uhh

its left order topology on (discrete order on C) + 1

left and right order topologies are all T0 but generally not T1

(not to be confused with linear order topologies)

[bold knoll]

but how it is a topology on R if it doesn't contain R

or we just pretend that R is there even if It wasn't explicitly mentioned?

[tender halo]

yeah

usually we dont mention the whole set and empty set when describing a topology

[limpid fern]

the question would be rather boring if the "topology" is not a topology in the first place no?

[tender halo]

its an understandable question for a newbie

[limpid fern]

perhaps the professor should've been more explicit in including the other sets i suppose

[bold knoll]

this guy still goes into pointing out the question is easy without even reading what I wrote

the same way the prof gives us an equivalence relation on s^2 by identifying "opposite" points - like opposite to what man

[gritty widget]

antipodal?

[opaque scroll]

Like on the opposite side of sphere

[plush folio]

Not all professors are good lecturers. I recommend just reading a book or finding lectures on youtube if you don't learn anything from your prof

[final grotto]

extremally disconnected Hausdorff spaces are totally disconnected

how do i show this?

can someone give me a hint?

[tender halo]

uhh its obvious?

like do the obvious thing and you'll prove it

im unsure how to give a hint because there is really only one thing to try

[opaque scroll]

According to Wikipedia it's not true...

An extremally disconnected space that is also compact and Hausdorff is sometimes called a Stonean space. This is not the same as a Stone space, which is a totally disconnected compact Hausdorff space. Every Stonean space is a Stone space, but not vice versa.

https://en.m.wikipedia.org/wiki/Extremally_disconnected_space

[tender halo]

what? it is true

[final grotto]

its not trivial for me

[tender halo]

they are asking Stonean -> Stone, not the other way around

[tender halo]

take two points and their Hausdorff neighbourhoods

take the closure of one of the neighborhoods

what can you say about it

[final grotto]

the closure is clopen

and furthermore the closures of these are disjoint

because its extremally disconnected

[opaque scroll]

Oh I see. The naming threw me off. Stone being a generalization of stonean is unfortunate.

[final grotto]

and how does this help me by proving its totally disconnected?

[opaque scroll]

Consider a connected component with more than one point

[tender halo]

well now you have that any two points are in different clopen sets

thats like bad for connectedness

[final grotto]

but we have to show connectedness

[opaque scroll]

Didn't you want to show total disconnectedness?

[final grotto]

yes

[opaque scroll]

So the total opposite of connectedness...

[tender halo]

fun fact an antidiscrete space is both extremally disconnected and connected

[opaque scroll]

Like consider a connected component with more than one point, then you're argument says that it's not connected, hence contradiction

So totally disconnected

[worthy nebula]

Is this argument valid?

Found it on Geoghegan's book

I get that $\mathcal{T}$ contains a maximal element, but I don't get why it is maximal in the set of trees of $X$.

[gentle osprey]

qrno

[worthy nebula]

I only understand why it's maximal in $\mathcal{T}$ itself, I don't get why it spans through all vertices of $X$.

[gentle osprey]

qrno

[final grotto]

i gonna think about it ty^^

[tacit basin]

Is it true that $\partial S=S'\setminus S^{\circ}$? That is, the boundary points are all limit points which are not interior points?

[gentle osprey]

Sara

[tacit basin]

Ah wait 1 is a boundary pt of {1} but not a limit pt nvm

[craggy cedar]

Yes, an equivalent definition is $\partial S=\overline{S}\intersect\cap\overline{S^c}$

[gentle osprey]

خرشوف
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[craggy cedar]

Wait sorry, I didn't think about that too carefully

[craggy cedar]

Should be $\overline{S}\setminus S^{\circ}$

[gentle osprey]

خرشوف

[tacit basin]

ye ik this i just thought the above could be true for a sec (it isnt), ty tho 🙏

[eager vigil]

Could anyone help me understand the latter part of this.
We must define a bijection between k-fold covers and representations of the fundamental group acting on {1,...,k}.
Am I right in understanding the equivalence relation of the action as being: $\alpha, \beta \in \pi_1 (Y)$ are equivalent if there exists an automorphism $\sigma \in S_k$ such that $\alpha$ acts on ${1, \ldots, k}$ the same as $\beta$ acts on $\sigma (S_k)$?
Here I presume that represented $\pi_1 (Y)$ as a transitive permutation is done by essentially saying that $\alpha (1)$ is the number of the sheet of the endpoint of the path that $\alpha$ lifts to starting in sheet number 1?
Idk if all that made sense...

[gentle osprey]

CoffeeMan

[lusty trench]

Let $f : X \to Y$ be a covering map. Given a base point $y \in Y$, the fundamental group $G = \pi_1(Y, y)$ acts on the fiber $F = f^{-1}(y)$ as follows. Given $[g] \in G$ and $x \in F$, take a loop that represents $[g]$ and lift it to a path $h$ starting at $x$. We define $[g] \cdot x$ to be the final point of $h$.

Now assume $F$ is an abstract set on which $G$ acts transitively somehow. Let $X = \widetilde Y / H$, where $\widetilde Y$ is $Y$'s universal cover and $H \subset G$ is the stabilizer of an arbitrary element $x \in F$. Factor the universal covering map $\widetilde Y \to Y$ as $\widetilde Y \to X \xrightarrow f Y$, and then take just $f$.

Finally, if we had chosen another $x' \in F$, then the resulting covering space $X' = \widetilde Y / H'$ would be isomorphic to $X$. To see why, take $[g] \in G$ such that $[g] \cdot x = x'$, and notice that $[g]$ induces a deck transformation $\widetilde Y \to \widetilde Y$ that sends $H$-orbits to $H'$-orbits.

[gentle osprey]

Eduardo León

[swift fjord]

Suppose that T_X isn't maximal, so there is T' containing it. What must the intersection of T' with A be?

[manic seal]

This sounds like just plain ol graph theory

[ruby crown]

I don't understand what motivates this definition, any ideas?

[red yoke]

That just means this diagram commutes

[white oxide]

You might have read it off the commutative diagram, but you want to ensure that if you pass to the bottom row then taking the differential there shouldnt give you a different result than first applying the differential and then sending it over

More geometrically, you want that property to send cycles to cycles and boundaries to boundaries

[ruby crown]

I am trying to calculate H_0 here but I can't wrap my head around why Z^3 / Z^2 would be isomorphic to Z and how the free abelian group of {v_0, v_1, v_2} is isomorphic to Z^3. Regarding the latter confusion, doesn't this depend on the dimensionality of v_0? e.g. if they are all 1-dimensional and non-zero, the free abelian group is just Z.

[uneven scroll]

Hi, not sure if this is the correct channel for my problem:

I know that this task is asking for the lebesgue number but i dont quite understand how this is true

[plain quail]

$$
\coprod_{f:[n] \rightarrow[m]} X_m \times\left|\Delta^n\right|
$$ Can someone help me understand the notation here? What does it mean for the coproduct to be running over f:[n]->[m]? Could someone give me a simple example?

[gentle osprey]

Timanaku

[unreal stratus]

The coproduct will be over the set of all homomorphisms f:[n]->[m] in the simplex category

You could write this as the coproduct over the set Hom([n],[m]) but this is convenient

[white oxide]

It is the group generated over {a,b,c} mod the subgroup generated by a-b, b-c and a-c.
You can express a-c as the sum of the former two. Next up a+(a-b)=b+(a-b), so a+N=b+N. Similarly b+(b-c)=c+(b-c), so b+N = c + N. So you just a group expressed over a single formal symbol, a+N, since: (k1 a + k2 b + k3 c ) + N = (k1 + k2 + k3)a + N

[ruby crown]

Yep, makes a lot of sense why H_0 would be isomorphic to Z

[white oxide]

ultimately you just have to practice the formal yoga, trying to see what carries over from the linear algebra setting is a good practice as well

[ruby crown]

I still don't understand why the free abelian group of a,b,c would be isomorphic to $Z^3$. I see this happen a lot where people don't specify what n would be with $a,b,c \in F^n$. However, for some reason, it is taken for granted that the free abelian group of a,b,c is just the direct sum of Z with itself 3 times.

[gentle osprey]

FrankF

[white oxide]

I mean, you can write out the isomorphism explicitely?

[ruby crown]

but isn't the problem that a,b,c could be linearly dependent and the isomorphism wouldn't work

[white oxide]

no, the free abelian group of some set treats its elements as formal symbols

the a, b and c are just "basis vectors" of the group

[ruby crown]

huh, ok, that's weird.

Has anyone btw followed https://sites.google.com/view/introductiontotda/schedule-and-index. I wonder if there is a better source for learning this stuff

[white oxide]

Unless you care about the universal property of free objects, hence free abelian groups, the free abelian group over X is just Z^|X| with some prefered notation

[plain quail]

Ah I see, I have another question in context to geometric realization. In the definition, that the geometric realization is $$|X| = \operatorname{colim}\left(\coprod_{f:[n] \rightarrow [m]} X_m \times \Delta_n \underset{f^}{\stackrel{f_}{\rightrightarrows}} \coprod_{[n]} X_n \times \Delta_n\right)$$ Is $$f^{*}: X_m \times |\Delta^{n}| \rightarrow X_{n} \times |\Delta^n|$$ defined by $$(x,p) \mapsto (X(f)(x),p)?

[gentle osprey]

Timanaku
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[plain quail]

My end goal is to understand how the colimit definition is equivalent to the quotient space definition.

[final grotto]

bruh i get it it was really obvious i havent got connected in mind but yeah makes sense 😂

[lusty trench]

Assuming that X is a finite set.

Otherwise the free Abelian group generated by X is a strict subgroup of Z^|X|.

[quick bough]

could anyone help me understand this example https://mathoverflow.net/questions/59938/examples-for-non-naturality-of-universal-coefficients-theorem

it says that X is a co-H-space, why does it follow that we can add maps in its homotopy class [X,X]?

[unreal stratus]

In fact you can add maps in [X,Y] where Y is any other space. The co-H-space structure gives you a map f: X v X -> X. Given two elements of [X,Y], you glue representatives together to get a map X v X -> Y. Compose this with f to get a map X -> Y.

In fact any suspension is a co-H-space, in particular spheres, and then this reproduces homotopy groups.

[quick bough]

could you expand on the glueing part

oh, nvm

[unreal stratus]

Well XvX is the coproduct in pointed spaces basically

[quick bough]

yeah

i realized that

[unreal stratus]

also dually for H-spaces you can add stuff in [-,X]

[quick bough]

yup

[unreal stratus]

I guess another thing is that like

[quick bough]

i’ve got another question

regarding that post

[unreal stratus]

Hm nvm doesn't quite work ll

[quick bough]

why is the map in the post zero in H(X,Z), but in H_3(X; Z) it’s nonzero

[unreal stratus]

Well it's non-zero in Z/2 coefficients not Z

[quick bough]

in Z/2, not Z

sorry

why is it nonzero in H_3(X; Z/2)

and why zero in H(X,Z)

[unreal stratus]

I think this is by thinking abuot what the maps mean in terms of homology since we are working with RP^2

[quick bough]

hm, i don’t quite see it

[unreal stratus]

Okay I'm reading the post now lool

So the first thing is why f is 0 on Z-homology

[quick bough]

yup

[unreal stratus]

this is just cause it factors through S^2

like

reduced Z-homology of M is concentrated in degree 1 and so f_* is 0 in all other degrees automatically

but it's the zero map on H_1 as H_1(S^2)=0

[quick bough]

that makes sense, yes

[unreal stratus]

now the next bit hmm

Well first thing is that H_*(M;Z/2) is Z/2 in degrees 0,1,2

well this is just what is dne at the start of that answer

as in why f_* is non-zero on H_2(-,Z/2)

follows from LES

[white oxide]

Huh? But the free abelian group should be left adjoint, so the corresponding free abelian group of an infinite X is just the coproduct of free abelian groups over single points of X i.e. over copies of Z?

[quick bough]

there are given two LES

or do you combine them to get that f_* is nonzero on H_2

or well

[unreal stratus]

Eh so the first just says M->S^2 is an iso on H_2

[quick bough]

yes

[unreal stratus]

and the secnod says the same

[unreal stratus]

so yeah this

[quick bough]

oh, i get it, so you just have M -> S^2 and S^2 -> \Sigma M are isos on H_2

got it

[unreal stratus]

yes

uwu

[quick bough]

that makes sense, thank you

[unreal stratus]

np

you also want to check the two addition in homology (from co-H-space structure and the usual thing) agree

which should follow from the Eckmann-Hilton argument

[lusty trench]

Yes, but Z^|X| denotes the product, not coproduct, of |X| copies of Z.

[quick bough]

wdym precisely?

[unreal stratus]

Well like

You can add two maps f,g:X->X to get a map f+g: X -> X using the co-H-space ting

they then use the fact that like

[white oxide]

whoopsie

[unreal stratus]

(f+g)_* = f_* + g_*

where RHS is the addition in H_*

[quick bough]

oh, right

what's the eckmann hilton argument?

[unreal stratus]

This i guess is basically what happens when you want to say the Hurewicz map pi_n(X) -> H_n(X) is a homoomorphism

Well it's like

A theorem that if you have some set with two "compatible" additions then they coincide

[quick bough]

i see

[unreal stratus]

https://en.wikipedia.org/wiki/Eckmann–Hilton_argument

Like uh

[quick bough]

yup, just looked it up

[unreal stratus]

A neat example is there are two ways to add maps S^1 -> S^1

Like pointwise multiplication and also the co-H-space structure on S^1 (which induces the addition in the fundamental group of S^1)

you can check these coincide

then uh if you know the identity S^1->S^1 generates pi_1(S^1), you know that like

z |-> z^n is n times that generator

[quick bough]

yes

[white oxide]

It is what you namedrop when justifying why some functor assigning groups to stuff, assigns an abelian group to this thing in particular

Although any time that has been happened writing out the homotopies explicitely was honestly easier

[worthy nebula]

Oh, okay lol

Don't know how I didn't think of that

Thank you!

[hollow geyser]

@unreal stratus I finally solved this problem. I basically had to rewrite the whole thing from scratch.

I was wondering if I could get an easier proof by showing that the images of g and h make an open partition of R, contradicting its connectedness.

[jaunty stratus]

can someone verify my proof for this problem?

[red yoke]

👍

[jaunty stratus]

alright tysm

[granite mesa]

I’m confused how we know such U and V exist

Edit: I realized it was a consequence of a previous exercise

[ruby crown]

Hi guys, I am trying to understand the proof for interval persistence module being indecomposable. Are the main observations that make this proof work the following?

1. Choose V_k = 0 if $k \not\in [i,j]$ and $V_k = F$ otherwise. Choose $W_k$ to be always 0.
2. Direct sums between linear maps are linear maps.

What I don't understand however, is why a cannot be 0 in the second picture.

[gentle osprey]

FrankF
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[earnest ibex]

The Hopf map from $S^3$ to $S^2$ is not nullhomotopic, if I further compose it with the inclusion $S^2 \to S^1 \vee S^2$, is it still not nullhomotopic?

[merry geode]

You have retraction S^1 \/ S^2 -> S^2, right

[gentle osprey]

ImHackingXD

[earnest ibex]

Sorry, I meant the wedge sum lol

[merry geode]

If it were nullhomotopic, you can compose with the retraction to get nullhomotopic Hopf map

[merry geode]

Yeah and I understood it as wedge sum

You still have retraction, no?

[earnest ibex]

You are absolutely right

Thanks

[merry geode]

Np!

[earnest ibex]

Another question of a similar nature: are there maps $S^n \to S^{n-2}$ which are not nullhomotopic?

[gentle osprey]

ImHackingXD

[red yoke]

This is just asking whether πn(S^n - 2) = 0, and S^n - 2 is homotopy equivalent to S^(n-1)

So that's a no

[unreal stratus]

Wait are you using S^n - 2 to mean S^n minus two points

Ouchies

[red yoke]

(I'm assuming that's the case)

[unreal stratus]

I assumed S^(n-2) was meant

[red yoke]

Oh

The same point holds in that case

[dusk slate]

What does zero differential mean in homology?, here by differential I mean boundary homomorphism

[hidden crag]

some differential sending everything to zero i assume

needs more context though

[unreal stratus]

Yeah

[earnest ibex]

I meant this, my bad

[earnest ibex]

Wait, so for example, if we have that pi_4(S^2) is not the trivial group, doesn't that mean that there is a map S^{4} to S^{2} which is not nullhomotopic?

[red yoke]

Yea

[empty grove]

They should make homotopy groups which are not as fake

[opaque scroll]

I guess the more famous example would be that the Hopf fibration is a map S^3 -> S^2 that is not nullhomotopic

[tacit basin]

for x = (x_1, ... x_n), y = (y_1, ..., y_n), x, y in R^n with the euclidean topology, is it true that ||x_i - y_i|| <= ||x-y|| for all i?

[umbral panther]

Higher Homotopy Groups Are Spooky
Enter the world of 3-dimensional holes in 2-dimensional objects.

[naive trench]

Yes

[tacit basin]

Ty

[fair idol]

Can someone give me a sanity check... Homology is only necessarily a module it's not always a ring right?

[tacit basin]

Is the issue with this that my sequence might not converge to p? Like if say epsilon_n converged to 1/2

[obtuse meteor]

homology lol

[opaque scroll]

I don't think it's every a ring in any meaningful/natural way, but it's sometimes a coalgebra.

[umbral panther]

A lot of questions have good sanity checks with degree zero homology and cohomology

For a space with finitely many points, homology and cohomology seem the same. But for the integers, cohomology is functions on the space, which form a ring under pointwise multiplication. Whereas homology are functions with finite support. These do not form a ring. There is no candidate for the unit. Actually, you should think of them not as functions, but as measures

[unreal stratus]

I am now wondering how to prove they cannot form a ring lol

One thing that is clear though is that if homology were a functor into rings then it'd clearly not work with units since e.g. no element is always in the image besides 0 in H_0

[opaque scroll]

I mean, they can form a ring.

You have a surjective map H0 -> Z, then picking any section you get that Z is a direct summand. Then just define multiplication of things in the other summand to be zero, and thats a ring.

This is a totally arbitrary ring structure though

[unreal stratus]

Sure yeah lol

[eager vigil]

Why do we need the abelianization of pi_1 for it to be well-defined here?

[red yoke]

The path from x0 to f(0) is arbitrarily chosen

[eager vigil]

Ah, that makes sense

Thanks!

[surreal island]

How to prove that binormal absolute neighborhood retract is locally contractible?

[fierce lily]

I am confused about under what condition will this equality not hold?

[red yoke]

Equality holds when a has no successor and b has no predecessor

[fierce lily]

is there any examples of successor of a as we have interval (a,b)?

[red yoke]

Any interval on Z

[fierce lily]

I mean if I say a=0, b=2, and I can have a+1 as successor of a, so it should exist at anytime when there is an interal (a,b)

[red yoke]

Successor as in immediate successor

The minimum among all elements greater than a

In general the closure of (a, b) is [inf of elements greater than a, sup of elements less than b]

[fierce lily]

so you mean no successor means no specific inf greater than a right?

[red yoke]

It means there is no minimum element greater than a

i.e. inf of elements greater than a is a itself

[fierce lily]

Suppose A and B are subsets of X. if x is in closure(A), can I argue that x is also in closure(A union B)?

[quartz horizon]

Yes because the closure of A is a subset of the closure of A union B

If you’re not sure why then you should try proving it

[fierce lily]

I think if x is in closure A, then there must have open balls U containing x intersects A, since it intersects A, it must intersects A union B

[quartz horizon]

That makes sense

[sly geyser]

again, why the fixation on neighbourhoods?

[red yoke]

Point-set topology is about generalizing stuff to the context of topological spaces

And a topology is just a way to talk about neighbourhoods

[inland laurel]

any serre fibrations should from a topological space go to a topological space?

[inland laurel]

and every serre fibrations a homeomorphism this is correct?

@umbral panther then every homeomorphism is a serre fibrations this is correct?

thank you @umbral panther

[eager vigil]

Hm, now that I'm looking at it again. How does modding out by the commutator group fix this?

[umbral panther]

We made an arbitrary choice of lambda. Different choices differ by an element of the fundamental group. Changing by such an element changes both lambda and its inverse, so the result is conjugation by that element

[gentle osprey]

CoffeeMan

[eager vigil]

Sorry, I must be missing something obvious here...

[umbral panther]

We want to prove that it is necessary and sufficient to mod out by conjugation. To show it is sufficient you should consider arbitrary lambda and lambda prime. But to prove it is necessary, choose lambda prime. Choose it to be the concatenation of a closed loop with lambda.
lambda.f.lambda^-1
vs
g.lambda.f.lambda^-1.g^-1
exactly conjugation

[eager vigil]

Sorry, not sure I understand. In the above, f is not necessarily a loop, so lambda.f.lambda^-1 is not well-defined

[eager vigil]

I.e., in my construction, alpha and beta are not necessarily the same, so they don't cancel

[umbral panther]

If f isn’t a loop, I have no idea what the claim is. How is it a homomorphism? How can you use the same lambda on both sides?

[eager vigil]

Yes, this is also what I'm really confused about. It just says that f is a path in the singular p-chain group

Oh... I think the author made a typo and for some reason ran with it throughout the whole page

Hm... no, throughout the next two lemmas, he also keeps using the map on general singular 1-simplices

[umbral panther]

What is pi tilde?

What is lambda?

[gentle osprey]

CoffeeMan

[umbral panther]

Lambda is a choice of path to the base point for every point of X? A continuous choice would require the space to be contractable

I guess you can define these things, but whether they are well defined is not important. We don’t care about the homomorphism on 1-chains, only on the restriction to cycles, where it does break into loops

[eager vigil]

Yes, I suppose so. I will try to work through the rest of the section with that assumption to see if things work out. Thanks for the help!

[red yoke]

From what I can infer, their construction is to choose an arbitrary path γp from x0 to p for each p, then to construct the linear map by sending the basis element f to γf(0) f γf(1) right

And then to induce a map on H¹

[eager vigil]

Yeah, except the gamma f(1) is taken inverse, but yea

[red yoke]

So you just need to verify that 1-boundaries are indeed sent to 0 and that this map is independent of choice of γp

[eager vigil]

I can't seem to show that it should be independent of the choice of path though

[red yoke]

A 1-cycle consists of paths f1, f2, …, fn such that fi(1) = fi+1(0)

(or some linear combination of these forms)

So each γ should appear twice in a way that "cancels out"

[eager vigil]

Oh god

That's the thing...

The lambda paths are presumably by construction not arbitrary

For each point we choose a predefined lambda

[red yoke]

Oh hm perhaps there are multiple senses of arbitrary

[eager vigil]

Then the reason it maps into the abelianization is just because one needs the maps to be a homomorphism and because the domain is abelian

Yeah, I guess

But this way, it all checks out

Thanks a bunch

[rancid umbra]

is there a space with fundamental group S_n?

[opaque scroll]

For any group there is a space with fundamental group that group

[rancid umbra]

is it just like, the cayley graph? what is the general construction?

[opaque scroll]

Take a presentation of your group. For each generator wedge together that many circles. For each relation glue a disk to the circles in accordance with the relation

[rancid umbra]

so for S_n i just need a wedge of two circles

[unreal stratus]

There is also the notion of a classifying space BG of a group G. (G can have a topology but ill assune discrete rn) BG is a path connected space whose fundamental group is G and whose higher homotopy groups vanish. This can be constructed by taking some contractible space EG on which G acts freely and then quotienting out by the G action

That may sound abstract but for S_n you can consider the subspace of R^oo x ... x R^oo of pairwise distinct points (a_1,...,a_n)

That is contractible (I hope this is believable) and S_n acts freely

Taking the quotient you can view this as the "space of subsets of cardinality n of R^oo", sometimes denoted B_n(R^oo) or various things

Idk if that is hekpful actually lol but the classifying space is very important so it's important to see once ig

[opaque scroll]

Well, a wedge of two circles with a bunch of disks glued to it in interesting ways, sure

[unreal stratus]

Like you can kinda imagine what a path looks like in that space above (take your subset and physically move around some points lol)

[rancid umbra]

this sounds kinda familiar.

i have a follow up question:

Given a group G, is there always a Hausdorff space X for which G acts freely and properly discontinuously on it? Then pi_1(X/G) would be isomorphic to G

[unreal stratus]

Yes, indeed the EG I mentioned can be taken to be a CW complex

So more than just Hausdorff

[lusty trench]

Where can I find a proof of this theorem of Leray?

Seems like a generalized Leray-Hirsch theorem.

[umbral panther]

I think it preceded Leray Hirsch. The first spectral sequence. Proved in a POW camp

Oh, this is more special than the full Leray, although more general than Leray Hirsch

This is often called the Serre spectral sequence

[lusty trench]

Ah...

[umbral panther]

The filtration is the preimage of the cellular filtration on the base. The main issue is understanding what is a spectral sequence and why a filtration gives one

[lusty trench]

And then also why it converges to something useful.

Although that's probably outside of the theorem's scope.

[umbral panther]

It’s a filtration on the cochains of the total space, so it converges to the cohomology of the total space

[lusty trench]

The cohomology of the total space is the total complex of the limit of the spectral sequence, right?

[rancid umbra]

if X is homotopy equivalent to a CW complex, it should be true that X has a CW complex structure, correct?

[umbral panther]

The cone on any space is homotopy equivalent to a point. The cone on the cantor set has no cw structure

[rancid umbra]

hmm. right

with respect to this problem in hatcher chapter 0, I can show part (a), and I can give the mobius band and the annulus the correct CW complex structures to form a CW complex which has them both as (strong) deformation retracts, but how do you use part (a) to get to part (b) with mapping cylinders?

[umbral panther]

I don’t know what is the relation between a and b

[rancid umbra]

was that a question or are you saying that you don't know

[shadow charm]

glue two mapping cylinders

the trivial map and z->z^2

[rancid umbra]

z^2 gives you a twisted cylinder

[shadow charm]

it's a mobius band

a little hard to visualize

but imagine that the bottom S^1 is the equator of your band

[rancid umbra]

i’m imagining something like this

just a full twist around tho

[shadow charm]

mm that's not quite right

[rancid umbra]

and then a regular cylinder on the bottom

oh?

[shadow charm]

since z^2 has two solutions you get identifications at the bottom

like it's twisted but there's overlap

[rancid umbra]

wait why would you get identifications at the bottom?

[shadow charm]

yeah we might be thinking bottom and top differently here

okay so

[rancid umbra]

this would be for conj(z) then?

[shadow charm]

youve got S1xI/~ where ~ identifies (z,0) with (z^2,0)

[shadow charm]

a simple twist would just be any rotation of S^1 but at the end you just get S1x I again

[rancid umbra]

i think hatcher defines mapping cylinder as a quotient of the disjoint union of X x I and Y, where you glue Y to X x {1} via the map f : X —> Y

[rancid umbra]

with this, i can see how it’s a mobius band, but not with hatcher’s notion of a mapping cylinder

[shadow charm]

yeah sorry im getting confused lol

[shadow charm]

this is obviosuly much more gross

[shadow charm]

so this

with X= S^1, Y=S^1, f:z->z^2.

[rancid umbra]

yea

[shadow charm]

so (z,1) goes to z^2 in Y

[rancid umbra]

right

[shadow charm]

So a point w in Y is identified with both square roots of w in X x 1

yeah

[rancid umbra]

yes two

okay

[shadow charm]

These square roots are just polar opposites right

[rancid umbra]

yes

[shadow charm]

So you can kind of see it as if we're gluing X x {1} to itself by an antipodal map

[rancid umbra]

right

[shadow charm]

You cna visualize this as taking S^1 and adding a twist to it, then projecting back down on itself

[rancid umbra]

okay

[shadow charm]

The twisted S^1 is exactly what the boundary of the mobius band looks like

[rancid umbra]

that’s a mobius band

with the two cell of the cylinder attached

[shadow charm]

to be clear, what actually become the boundary of the mobius band here is X x 0 not X x 1

the equator is X x 1

[merry geode]

Hmm what was mapping cylinder

Is it that you attach cylinder to a space using a map?

[rancid umbra]

this

[shadow charm]

Its as if you started with two twisted S^1 s, joined them into a sort of wrapped cylinder, then collapsed the X x 1 S^1 so it has no twist

Idk if you see what i mean

[rancid umbra]

hmm. lemme draw it in a few minutes

gotta go do something

[merry geode]

Like I × S^1 \cup S^1 where attaching map is given by some S^1 -> S^1

[shadow charm]

you can also refer to this https://math.stackexchange.com/questions/474466/mapping-cylinder-of-z-rightarrow-z2

[shadow charm]

in our case it's this, in general the cylinder need not be based on S^1

[merry geode]

Indeed

It sounds fun rotating the cylinder twice!

[merry geode]

Do we get Mobius band in this case?

[shadow charm]

As I said above, the mapping cylinder for the map z->z^2 is a mobius band yes

[merry geode]

Hmm, z -> z^3 seems a bit difficult; you can think of going over S^1, rotating 2pi/3 on each circling

Ah, it's not a manifold, I guess.

[rancid umbra]

the auxiliary cut method is really neat and visually pleasing. would u happen to know any references on auxiliary cuts? or is it more just a thing one begins to recognize.

why are you allowed to cut and reglue in different orders?

it seems so counter compared to anything i’ve done with continuous constructions, but maybe it’s just the quotient topology working out nicely?

[shadow charm]

not sure what you mean by auxiliary cut here? Do you mean doing your quotients in different orders?

[merry geode]

Did we cut something somewhere?

[rancid umbra]

in the stack exchange post

sorry, that wasn’t clear

[shadow charm]

Yeah okay it’s what I thought

You just learn to recognize I guess but it is pretty useful

I wouldn’t say there’s any particular resource on this that I know of

[rancid umbra]

any quick justification?

my intuition is that quotient maps don’t need to be visualized/realized via an animation/homotopy. just that the right pieces get glued together

[quartz horizon]

Ooooh do you have an example

This is a classic case where category theory is useful

[rancid umbra]

the stack exchange post

[quartz horizon]

I see i see

Yeah i definitely think category theory is useful here

[rancid umbra]

do enlighten us

[quartz horizon]

Have you heard of the universal property of the quotient?

[rancid umbra]

ye

[quartz horizon]

What is it?

[rancid umbra]

any map out of the base space which respects the quotient map factors uniquely through the quotient map

[quartz horizon]

Hmm

I mean that’s one way to think about it

There’s another way i prefer though

Would you like to hear it?

[rancid umbra]

sure!

[quartz horizon]

Ok so

Let’s go back to sets

So you’ve got a set X, and some equivalence relation R on it

And you can form the quotient set X/R

[rancid umbra]

mhm

[quartz horizon]

And you get told what the set X/R "is" - its elements are equivalence classes of elements of X

For the categorical perspective, you care more about what X/R "does" - how it relates to other sets

And this is how - functions out of X/R naturally correspond to functions out of X respecting R

Meaning - functions out of X such that whenever xRy, then f(x) = f(y)

[rancid umbra]

right

[quartz horizon]

note, by the way, that nothing in the categorical definition actually requires R to be an equivalence relation

You can make sense of it for R being some arbitrary relation on X

Where you want to define a set "X/R" by what it does - functions out of X/R are the same as functions out of X respecting R

It turns out such a set exists - you take the quotient by the smallest equivalence relation generated by R

But I find the categorical definition a little cleaner - you don't need to work with the generated equivalence relation directly, you just use "functions respecting R"

[tender halo]

just say "coequalizer in Set"

[quartz horizon]

I could but I don't really need to

Does this make sense @rancid umbra ?

[rancid umbra]

yea

[quartz horizon]

Cool

The same diagrams actually work for topological spaces

So, you have a topological space X, and some relation R on it

And you want to define a quotient space "X/R"

You can define it by just telling me what it does

Namely, that continuous functions out of X/R correspond to continuous functions out of X respecting R (so whenever xRy, f(x) = f(y) )

Again, it turns out such a space exists - you quotient X by the smallest equivalence relation generated by R, and then use the quotient topology

[rancid umbra]

right. makes sense

[quartz horizon]

Cool

Now, let's go to that post

We want to see why this sort of thing works, right?

So, let's think about that top space

We can think about what the top space "is", by saying which particular topology is has on which particular set

But again, the categorical perspective is to focus on what it does

And what it does is the following - continuous functions out of that space correspond to continuous functions out of the rectangle which respect the relation

Where the relation on the rectangle is

[rancid umbra]

the obvious one

[quartz horizon]

pRq iff p is on the left edge, q is on the right edge (blue), or p is on the first half, q is on the second half of the bottom edge (green)

What I've described isn't actually an equivalence relation, but again it doesn't matter

Because the categorical definition supports arbitrary relations

Now, let's look at the bottom space

Can you tell me what it does?

[rancid umbra]

maps out of it correspond to maps on the two rectangles respecting the equivalence relation. but that’s just the same as maps out of the original rectangle respecting the original equivalence relation, since the red cut identifies points down the middle

[quartz horizon]

Exactly! This comes from something called the "gluing lemma"

[rancid umbra]

ik that guy

[quartz horizon]

Where if you have two closed subsets of a topological space C1 and C2

Such that C1 U C2 = X

Then maps out of X correspond to maps out of C1 and C2 that agree on their intersection

[rancid umbra]

oh that’s super clear

[quartz horizon]

Mhm!

This is what you might call a "pushout" categorically, but you don't even have to use that terminology

The point of the categorical perspective is to focus on what the thing does

And if two things do the same thing, then they're isomorphic

[quartz horizon]

This comes from Yoneda

Basically the fundamental theorem of category theory

[rancid umbra]

okay neat. thank you for the detailed explanation!

[quartz horizon]

Np!

I hope it's clear how the rest of the manipulations in that post work

[rancid umbra]

so, i know what equalizers are, but i don’t know what they do

would u mind explaining what they do?

[quartz horizon]

Sure, though it might be better to move to #category-theory ?

[dim perch]

hello people, I want to crowdsource a reference for an easy-to-prove topological fact

have any of you seen the following statement appear in a textbook or paper?

  If M is a (d-2)-connected d-manifold and x is a point in M, then the complement M - {x} is (d-2)-connected

(I understand that this is easy to prove, but I'm trying to unclutter a topology paper, and a reference would be much cleaner)

[unreal stratus]

||"It is easy to see that [...]"||

[dim perch]

nonono

i will not become meter pay i will not become hike mill

[final grotto]

hey can someone explain me Homotopy? why is it important that we look at the [0,1] and not for example [-2,2]?

[plain raven]

it's not important. these two intervals are homeomorphic topological spaces.

they are functionally equivalent in all contexts dealing with topology

[final grotto]

then why we use the unit intervall in the definition? and not for any intervalls?

[plain raven]

the unit interval is convenient to use as a coordinate system

[final grotto]

ah ok

thanks

[ebon galleon]

Dwyer Kan paper

[final grotto]

can u explain me why these intervalls are homeomorph?

[opaque scroll]

f(x) = 4x-2 is a homeomorphism [0, 1] -> [-2, 2]

[final grotto]

ah

ty

[dim perch]

bump

the bounty is that i will put a footnote with "[name] pointed the author to this reference" in the paper if you want lmao

[umbral panther]

Why not call it Blakers-Massey?

[dim perch]

ok yeah that totally works, ty

i forgor because im not a real topologist

[opaque zodiac]

No see the real pro move is doing what I recently did in a paper which is play the 'it's easy to see/clearly/recall that' card on a fact that is folklore known by like five people

[pastel idol]

All you really need to say fits in one line, so you could write something compact like “note that [statement]: a connected space is path connected and any two paths can be pushed off of {x}” to avoid clutter. I don’t know if there is a reference for this in published literature but I know it appears as an exercise somewhere in Lee’s Intro to Top Manifolds book

[dim perch]

I will say that I don't believe k-connectivity to be so simple

[umbral panther]

Citing an exercise sounds fun

[dim perch]

you essentially need to say that you can perturb a non-embedded k-sphere so that its "interior" misses the point always (for some "interior," which will itself be a positive-codimension disk)

which seems easier said than done from the purely topological pov

[umbral panther]

General position
Better: transversality

[dim perch]

you can't just like

say general position without elaborating lmao

[umbral panther]

The proof is left as an exercise to the reader. Specifically Lee x.x.x

Transversality is a black box that says exactly that

[dim perch]

this is reminding me why I decided not to become a real topologist

[opaque zodiac]

wait what are you then

[steel wing]

Is anybody familiar with this result involving cardinal functions:
https://dantopology.wordpress.com/2010/06/12/the-cardinality-of-first-countable-compact-spaces-i/

[pastel idol]

Yes, if you want to be completely detailed you need to argue you can go to a closed ball around the point and modify any paths through it by pushing them off the boundary (which you can in not much more than a few lines). But if you’re going that route you may as well just use the fact that no k-connected n-manifold with n>1 and k<=n has cut loci of dimension k or lesser because Euclidean space in dimension n>1 has no cut loci of dimension n, which might be easier to find a reference for

[opaque zodiac]

I have seen the proof before

I don't know if I could replicate it off the top of my head

it's a surprisingly hard thing to prove

[steel wing]

I was lucky enough to have as a professor the person who originally proved this result

[opaque zodiac]

ah nice

[dim perch]

I don't do enough geometry to parse this... what is a cut locus in a smooth manifold, and how does this help you null-homotope maps from k-spheres?

[dim perch]

an algebraic topologist lol

i am lost without the h-cobordism theorem

[opaque zodiac]

pants

[dim perch]

pants

[tender halo]

we were just discussing a related result about compact (instead of lindelof) spaces

i even wrote down the argument, you can search it up

[steel wing]

Feel free to show me whatever you're referring to

[tender halo]

#foundations message

[gentle osprey]

benjazul

[pastel idol]

You don’t need to think about it geometrically if you don’t want to (in particular I don’t care about the cut locus in the tangent space or anything like that here); you can just call it the minimal set of points in a space for which the removal of those points results in a disconnected space. If you want to say a little more you can say the cut locus of x is homotopy equivalent to M - x. Then the point is no local piece of the manifold has such a point and there is no way to k-disconnect the space. If you want to argue combinatorially you can say consider a triangulation of M by polyhedra and then recall a graph is k-connected by definition if its cut loci is of dimension greater than k. This can be found in most texts on graph theory so that’s a candidate for a reference

@dim perch

That probably isn’t quite what you’re looking for but I think it’s a decent fallback plan if you don’t find anything else suitable

[unreal stratus]

Can't you also use Hurewicz and (homology) excision

Like M \ x -> M induces isos on homology in every degree we care about and then fits in nicely with Hurewicz which we can apply since we have such strong connectivity hypotheses

[pastel idol]

One can use many things; it’s not difficult to prove the claim. The context is Nat wanted to refer the reader elsewhere instead of writing a proof (or asserting its immediacy or its recollectability)

[unreal stratus]

Yup exactly

I meant that saying Hurewicz + excision makes it entirely standard

But idk what sort of detail is allowed besides just a reference lol

[pastel idol]

This is definitely a good argument (but I don’t know that anyone’s bothered to write it down in published literature!)

[unreal stratus]

(Thanks)

Yours is nice and not smth I've seen before

[pastel idol]

Cheers

[umbral panther]

This is using Hilton Milnor to prove Blakers Massey. It is practically circular

[unreal stratus]

I don't see why this is circular, and I don't know why your need hilton milnor

But sure you may use BM or smth similar in part of the proof of Hurewicz, so you could just apply BM directly

[dim perch]

yeah this is what i meant by like

the easy proof

[unreal stratus]

oh lol

[dim perch]

it's just slightly frustrating because like

I need this for equivariant little cubes shenanigans

and the original configuration space reference (fadell-neuwirth) doesn't reeeeally argue this, but everyone cites it as though it does

and i detest gaps in the literature

[unreal stratus]

Ah okay lol yeah they use that a ton right lol

[fierce lily]

is it adequate to prove that box product(A) subset of box product(X) in order to prove the subspace relation here? I think it is because they are all topological space

[paper wedge]

i mean you would have to prove that they would both have the same basis

it is just set algebra ig

you would nwat to prove that the product topology on A_is

is the same as the subspace topology coming from the product of the X_is

@fierce lily

[fair tusk]

random thought: if homeomorphisms are the isomorphisms of the Top category, what are the automorphisms?

[umbral panther]

Self-homemorphisms

A great topic is the homotopy type of the group of self homeomorphisms of QxS^n, where Q is the Hilbert cube

[fair tusk]

oh damn that sounds cool

[steel wing]

I was actually surprised how simple the proof was for a problem that had been open for fifty years

[opaque zodiac]

which version of the proof did you see?

(Also sometimes there just aren't that many people thinking about something. This is a pretty solidly 'set-theoretic' topology question and a lot of topology research was moving to algebraic topology)

[steel wing]

The version that I linked. I don't think it's that hard to generalize to the full argument though.

[opaque zodiac]

ah right

[steel wing]

Yeah Arhangelskii is known for his work in General (set theoretic) Topology

[scenic ridge]

If I have a hilbert space H, what is the normal topology of the space of all self-adjoint linear operators: H-> H, bounded or unbounded. So one of the operators might only be defined in a dense subspace of H.

someone said something about norm-resolvent or graph

but im not quite sure what they mean

[fierce lily]

then suppose U_ is open basis element in X_. and it is clear that U_ is open in A_. Hence we know product(U_) is basis element of product(A). subspace topology on A has basis element A_ intersects U_=U_, hence subspace topology on product A_ is product U_

[umbral panther]

Given a reasonable operator, you can use the spectral theorem to define another operator called the resolvent, dependent on a parameter. Then you can ask that the resolvents converge in the norm or strong topology to define norm resolvent or strong resolvent convergence

[scenic ridge]

ok but how does this define the topology on the space? As the norm and strong topology are only defined for bounded operators right

[umbral panther]

If the operator is normal and closed, the resolvent is bounded for lambda in C-R

[umbral panther]

(0,1) is homeomorphic to R, which is complete

Metrizable and completely metrizable are about the existence of a metric, not about a specific metric

[ocean narwhal]

Yes; such a space is open in its completion (from local compactness) hence Gdelta in its completion (hence completely metrizable)

[final grotto]

how do I show that a Homotopy class exists?

be constructing?

by*

nvm

[silver ridge]

The first part of this proposition is completely obvious from drawing a picture, but I'm not exactly sure how to go about writing out the proof. I would appreciate a starting point. Also, I don't understand the part from "thus". I would appreciate any help (:

[quartz horizon]

You probably want to use the universal property of the closure

[silver ridge]

what do you mean by this

[quartz horizon]

So, how have you defined the closure?

[silver ridge]

the smallest closed set containing A, or the intersection of all closed sets containing A

[quartz horizon]

Mhm, that is one way to define what the closure is

The idea of a universal property is to instead focus on what the closure does, how it relates to other subsets

What I mean is the following

Let $A$ be a subset of a topological space $X$. The closure of $A$, denoted $\bar A$, is the unique closed subset of $X$ such that, whenever $C$ is closed in $X$, $A \subseteq C \iff \bar A \subseteq C$

[gentle osprey]

Pseudonium

[quartz horizon]

Do you see how this is equivalent to the definition you gave?

[silver ridge]

Yep

[quartz horizon]

This gives us a strategy to show that $\bar A^Y = \bar A^X \cap Y$

[gentle osprey]

Pseudonium

[quartz horizon]

Namely, we show that $\bar A^X \cap Y$ satisfies the universal property

[gentle osprey]

Pseudonium

[quartz horizon]

And by uniqueness, we'll be forced to have $\bar A^X \cap Y = \bar A^Y$

[gentle osprey]

Pseudonium

[quartz horizon]

Does that make sense?

[silver ridge]

Makes sense

Say C is a closed set in Y containing A.

Then there exists a closed set in X, B such that C = B \cap Y, by definition of the subspace topology

[quartz horizon]

Mhm

[silver ridge]

So, $\bar{A}^{Y} \subset B \cap Y$ for any closed set B in X containing A

[gentle osprey]

swifteeee

[quartz horizon]

This is true, but not what you want to focus on

You want to show that $\bar A^X \cap Y$ satisfies the universal property

[gentle osprey]

Pseudonium

[silver ridge]

hm. The one hump is that \bar{A}^X is open in X no?

[quartz horizon]

No

Or at least, it's not guaranteed to be open

[silver ridge]

I thought \bar{A} is closed by definition, so its complement is open?

[quartz horizon]

Perhaps I should spell this out a little more - what this means is verifying that, if $C$ is a closed set in $Y$, then $A \subseteq C \iff \bar A^X \cap Y \subseteq C$

[gentle osprey]

Pseudonium

[quartz horizon]

We're not taking any complements?

[silver ridge]

oh one second, I misread. I thought \bar{A}^X meant "closure then complement in X" but it means "closure in X"

[quartz horizon]

mhm

[silver ridge]

gimme 2 minutes

Given any closed $C$ in $Y$ containing $A$, we have $\bar{A}^{Y} \subset C = B \cap Y$ for some closed $B$ in $X$. As $A \subset Y$, we have $A \subset B$. So, considering the closure of A in X, we have $\bar{A}^{X} \cap Y \subset B \cap Y = C$. By the uniqueness of the closure, we have $\bar{A}^{X} \cap Y = \bar{A}^{Y}$, as required.

[gentle osprey]

swifteeee

[plucky veldt]

I don't think you need to introduce category theory here

[quartz horizon]

I don't quite follow how you get A is a subset of B?

You also haven't actually needed $\bar A^Y$ here as far as I can tell

[gentle osprey]

Pseudonium

[quartz horizon]

And you would need to show the reverse too - that if $\bar A^X \cap Y \subset C$ then $A \subset C$

[gentle osprey]

Pseudonium

[silver ridge]

$A \subset B \cap Y$. ?

[gentle osprey]

swifteeee

[quartz horizon]

Oh right yes

(universal property of intersection)

Ok so I can believe that you've shown $A \subset C \implies \bar A^X \cap Y \subset C$

[gentle osprey]

Pseudonium

[quartz horizon]

You'd just need to show the reverse implication

[quartz horizon]

The way I would do this is:

$A \subset C$. By definition of subspace topology, $C = B \cap Y$ for some $B \subset X$ closed.

So $A \subset B \cap Y$. This means in particular that $A \subset B$.

Then, using the closure in $X$, we have $\bar A^X \subset B$.

Then intersect with $Y$ to get $\bar A^X \cap Y \subset B \cap Y = C$.

[gentle osprey]

Pseudonium

[quartz horizon]

I think this is basically what you've done except I didn't need to mention $\bar A^Y$ anywhere

[gentle osprey]

Pseudonium

[plucky veldt]

if Y is closed in X then $\bar A^X \subseteq Y$, so $\bar A^Y = \bar A^X \cap Y = \bar A^X$

[gentle osprey]

mzdunek

[silver ridge]

Yes, but the point i want to make is that becasue this holds for any closed C in Y containing A, and we have \bar{A}^Y \subset C and \bar{A}^X \cap Y \subset C, both sets satisfy the property of being the "smallest closed set containing A" and hence must be equal

[silver ridge]

I appreciate this, I'll read it in a second, thanks

[quartz horizon]

Yes, but you don't actually need to mention $\bar A^Y$ to do this

[gentle osprey]

Pseudonium

[quartz horizon]

Also, this isn't a complete proof yet

You have to show the reverse implication - that if $\bar A^X \cap Y \subset C$ then $A \subset C$

[gentle osprey]

Pseudonium

[silver ridge]

Oh I see what you're doing

2min

[silver ridge]

$C $closed in $Y$, so there exists a $B$ s.t. $B \cap Y = C$. Then $\bar{A}^X \cap Y \subset B \cap Y$. This implies that $\bar{A}^X \subset B$, so $A \subset B$. Then $A = A \cap Y \subset B \cap Y = C$

[gentle osprey]

swifteeee

[quartz horizon]

There’s some implications here I’m suspicious of

It’s not true in general that if $X \cap Z \subset Y \cap Z$ then $X \subset Y$

[gentle osprey]

Pseudonium

[quartz horizon]

You know that $\bar A^X \subset B$ anyway though because of the definition of $\bar A^X$

[gentle osprey]

Pseudonium

[silver ridge]

Right, that's good enough

[quartz horizon]

Here's another way to think about it - $A \subset \bar A^X$ so $A \cap Y \subset \bar A^X \cap Y$, so $A \subset \bar A^X \cap Y$

[gentle osprey]

Pseudonium

[quartz horizon]

So if $\bar A^X \cap Y \subset C$, then $A \subset \bar A^X \cap Y \subset C$, so $A \subset C$

[gentle osprey]

Pseudonium

[silver ridge]

can you spoiler or something please

[quartz horizon]

This is basically an application of the Yoneda Lemma

[plucky veldt]

since $\bar A^X$ is closed in X then $\bar A^X \cap Y$ is closed in Y, and if B is a closed set in Y with $A \subset B$ then $B = Y \cap C$ for some closed set C in X, therefore $A \subset B \subset C$, which means that $\bar A^X \subset C$, so $\bar A^X \cap Y \subset C \cap Y = B$, so $\bar A^X \cap Y$ is the smallest closed set containing A in Y ergo it's the closure

[gentle osprey]

mzdunek

[silver ridge]

please stop

we're working through the answer

[quartz horizon]

Anyway, we've now successfully shown that $\bar A^X \cap Y = \bar A^Y$, right?

[gentle osprey]

Pseudonium

[plucky veldt]

wdym by spoiler

[silver ridge]

Yeah I'll tentatively agree lol. The strategy being an appeal to the definition of closure. The broad idea as I understand it is: show that \bar{A} \cap Y \subset C for any subset of C containing A (it satisfies the universal property of the closure of A in Y) and we need the second part in order to show that \bar{A} \cap Y is a closed set containing A without assuming so to begin with. So the first part shows that it is the smallest closed set and the second shows that it is a closed set containing A at all, right?

[quartz horizon]

Yeah, that's the idea

We've defined $\bar A^Y$ by saying how it relates to other sets

[gentle osprey]

Pseudonium

[silver ridge]

with the TeX bot you can cover the output, which can be uncovered by each user by clicking on it, if you really want to post an answer.

|| this is a spoiler in discord ||

[quartz horizon]

It satisfies the property that whenever $C$ is closed in $Y$, $A \subset C \iff \bar A^Y \subset C$

[gentle osprey]

Pseudonium

[quartz horizon]

Suppose we had two sets which satisfied this, $\bar A^Y_1$ and $\bar A^Y_2$

[gentle osprey]

Pseudonium

[plucky veldt]

ah, sorry

[quartz horizon]

Since $\bar A^Y \subset \bar A^Y$, we can go backwards to deduce that $A \subset \bar A^Y$

[gentle osprey]

Pseudonium

[quartz horizon]

So, $A \subset \bar A^Y_1$, and going to the right gives $\bar A^Y_2 \subset \bar A^Y_1$

[gentle osprey]

Pseudonium

[plucky veldt]

my first message showed how the first part of the proposition implied the second, my second message showed how to prove the first part

[quartz horizon]

By symmetry, we also get $\bar A^Y_1 \subset \bar A^Y_2$

[gentle osprey]

Pseudonium

[quartz horizon]

So they have to be equal

[silver ridge]

yep

[quartz horizon]

This is why the universal property def works

We can use $\bar A^Y_1 = \bar A^Y, \bar A^Y_2 = \bar A^X \cap Y$

[gentle osprey]

Pseudonium

[quartz horizon]

Do you need more help with the Q?

[silver ridge]

No I see how this works now

I appreciate your time (:

[quartz horizon]

Ok cool

Can I also ask you something?

[silver ridge]

Go ahead

[quartz horizon]

I'll admit that this was kind of an experiment on my part

I haven't actually worked with the universal property of the closure before myself

So... did it seem weird or

Like I'm trying to see if I can gently introduce categorical concepts

To people without any background in category theory

[silver ridge]

It didn't seem too weird because I've read from other books who use "universal property" without introducing any category theory

[quartz horizon]

mhm mhm

[silver ridge]

I vaguely remember something about the universal property of R[x1, .... xn] that I need to brush up on

but no, it felt fine

[quartz horizon]

That's great!

In general, universal properties are about saying "arrows into X" or "arrows out of X" have a nice description

So here - $\bar A^X$ is a closed set, such that saying $\bar A^X \subset C$ is the same as saying $A \subset C$

With the latter being a "nice description", and $\bar A^X \subset C$ being an arrow out of $\bar A^X$

[gentle osprey]

Pseudonium

[plucky veldt]

$\bar A$ is a closed set containing A by definition, so $\bar A \cap Y$ will necessarily be a closed set in Y containing A, since $A \subset Y$

[gentle osprey]

mzdunek

Pseudonium

[silver ridge]

gotcha

[quartz horizon]

So all this will say is that "arrows out of R[x1, ..., xn]" have a nice description

[silver ridge]

yes, something about unique homomorphisms to overrings of R

iirc

[quartz horizon]

mhm

[silver ridge]

okay, back to studying. I appreciate your time, certainly cleared things up

[quartz horizon]

np!

[plucky veldt]

do we assume here that we have a category P(X) of the space X with the morphisms being inclusions?

or just the closed sets, or open sets

[quartz horizon]

So if you know category theory, the closure is an example of an adjunction

[plucky veldt]

I'm not sure how you translate this into purely categorical language

[quartz horizon]

Given a topological space X

We can form two posets

The first is the poset of all subsets of X, where an arrow from $A \to B$ just means $A \subset B$

[gentle osprey]

Pseudonium

[quartz horizon]

The second is the poset of all closed subsets of X, with the same arrow def

The closure is defined such that $A \subseteq C \iff \bar A^X \subseteq C$

[gentle osprey]

Pseudonium

[quartz horizon]

So it's the left adjoint to the forgetful functor from closed subsets to subsets

[plucky veldt]

I don't really know adjoints, but I know that inclusion relation is a partial order on P(X)

looks to me that you could also have the poset of all open subsets, but it seems like that is not necessary

[quartz horizon]

This would get you the interior

[quartz horizon]

Yes, and any partial order makes a category

[quartz horizon]

With suitable reversing of arrows

[plucky veldt]

seems like the poset of all open subsets should be isomorphic to the opposite category of the poset of closed subsets

[quartz horizon]

Yes, complement does the trick

[glossy fossil]

TeXit

[rancid umbra]

1. What does Hatcher mean here when he says that one could always unlink the two loops if they intersect?

2. separately, what does he mean when he says that C can be unlinked from A and B? i'm not seeing how that can be done

[rancid umbra]

for 2, i think adding some perspective helped me out

[merry geode]

This one looks awfully close to trefoil knot

[rancid umbra]

yea, it does

[red yoke]

If we allow the loops to intersect A, then obviously we can unlink it

[opaque scroll]

Like if you just stretch C bigger, then it encompasses everything.

Or you can think of it as pulling A through C I guess

[merry geode]

Gotta look at those intersections closely, it seems so tangled

[opaque scroll]

Well, there's really only one intersection that matters. The one right in the center of C

Everything else can just be moved past each other

[red yoke]

^ just pull A through C

[rancid umbra]

are intersecting loops considered to be unlinked? what he’s saying isn’t obvious to me

[merry geode]

If you allow intersecting A, then you can just pull circle through it.

[dim perch]

you do not need to intersect A

just make C a bit bigger

[rancid umbra]

thanks, pulling A through C helped for me

[merry geode]

Ah I was commenting on 1.

[rancid umbra]

i’m not seeing how

they’re stuck together

[merry geode]

About 1. ?

If you allow intersection, they are not stuck because they can pass through each other.

[dim perch]

fun fact: the fact that there is no geometry-preserving isotopy unlinking 2 figure when A,B,C are realized as linearly embedded circles of a shared radius is known to the chainmail community, as it appears as a foundation of e.g. "orbital chain"

i.e. we use this to make jewelry

[rancid umbra]

?

how

[red yoke]

I mean, if the loop are allowed to intersect each other

[merry geode]

Yeah

[red yoke]

Hatcher just wants to say "the loops can cross through themselves, but not each other"

[dim perch]

true

[red yoke]

Yes

[rancid umbra]

nah fam

how

[red yoke]

You drew it

[rancid umbra]

no, i have the beginning and end stages

i have no idea what goes on in between

[merry geode]

Ah, so this one is not homotopic (and not about homotopy)

[red yoke]

"the loop must not be allowed to intersect A, since otherwise we could just separate them like this"

[rancid umbra]

why can u separate them? they’re joined at a point

[red yoke]

So he just says they're not allowed to pass through

[merry geode]

I think "intersection is allowed" means this process is allowed

[red yoke]

As in, they can't intersect at some point in time when you're moving them

[rancid umbra]

ohhhh

that makes sense

[merry geode]

I guess one issue of presenting this is that.. it is quite an obvious restriction

[dim perch]

ah I found my favorite instance of this in nature (i.e. the luxury jewelry market)--Lorraine's turkish orbital earrings would not be possible without this geometric phenomenon

[red yoke]

Pretty circles

[dim perch]

its one of the best chainmail earring designs we've come up with using homogeneous ring geometries

[rancid umbra]

thanks guys

[bright acorn]

What are u guys favorite introductory references on orbifolds?

[steel wing]

Is anybody familiar with this result involving cardinal functions (I posted this yesterday above):
https://dantopology.wordpress.com/2010/06/12/the-cardinality-of-first-countable-compact-spaces-i/

[eager vigil]

Quick question: does he mean it is exact and splits here? Not just exact? Cause from what I can see, the exactness has nothing to do with the splitting

[rancid umbra]

is every group homomorphism between fundamental groups the induced map of some continuous map between ambient spaces?

[merry geode]

You cannot take tensor product of exact sea and expect it to be still exact.

[paper wedge]

probably no

[rancid umbra]

why not?

[paper wedge]

probably things outside my scope

but i would just say no 😄

i googled it check this out

https://mathoverflow.net/questions/166153/realizing-homomorphisms-between-fundamental-groups

idk wtf is "obstruction theory"

but ig the answer is no

[merry geode]

I heard of obstruction whenever we can measure how something is impossible

[rancid umbra]

appeartently its in hatcher

section 4.3

thanks moamen

[merry geode]

Like, you cannot always construct X -> R from (Path of X) -> R by addition - this obstruction is H^1, and one may say it is an

[paper wedge]

yw

[paper wedge]

yeah ig

sounds cool

[merry geode]

I guess in this instance, (realizing pi1 hom) is prevented by H^3

[eager vigil]

Could you walk me through why it might not be? My experience with tensors is minimal, sorry

I'm following what the author says here though

[paper wedge]

ig he meant that the tensor product is not a left-exact functor

cuz u just consider the usual counterexample of 0 --> Z ---> Z -->Z/2--> 0

where Z-->Z is multiplication by 2

[rancid umbra]

if two paths are homotopic, then their change of base point maps are identical. is the converse true as well?

i can't seem to prove it if it is

coming up with a c.e. seems a bit tricky

example like this seem to suggest that the converse is true i suppose

[umbral panther]

The converse is not true

[rancid umbra]

right, taking X to be the circle, x0 = 1, x1 = -1,
p0 being the top semi-circle from -1 to 1,
p1 being the bottom semi-circle from -1 to 1

p0 and p1 are not path homotopic, but their change of basis map agrees on the generator based at 1

does that seem correct?

[merry geode]

Yeah so injective map may not be injective after tensoring.

Consider doubling map 2: Z -> Z.
After tensoring with Z/2Z, it becomes:
0: Z/2Z -> Z/2Z.
Since it sends 1 to 2 = 0, it is a zero map.

To understand in detail, you do want to study the tensor product.

[red yoke]

If n ∈ N and m ∈ Z/2, then n⊗m is sent to 2n⊗m = n⊗2m = n⊗0 = 0

And all the finite linear combinations similarly go to 0

[merry geode]

Yeah, also Z \otimes Z/2Z is just Z/2Z, right

[red yoke]

That too

[merry geode]

I am too immersed in the base change approach

[red yoke]

Since Abelian groups are Z-modules, this is called a tensor product over Z, as demonstrated by the fact that you can move the integer 2 to the other side of ⊗

In general a tensor product of R-modules lets you do rm⊗n = m⊗rn

[merry geode]

Mostly defining property of tensor product (that my dumb ass forgot )

[rancid umbra]

i think this picture:

[merry geode]

What is addition here?

[rancid umbra]

path concatenation and then homotope

[merry geode]

Ah, it is a bit confusing imo, + is often commutative.

[red yoke]

In particular all base point changes are the same for an Abelian group

[rancid umbra]

hm. but i didn't use the fact that Z is abelian here?

[red yoke]

Since there are no nontrivial conjugations

[rancid umbra]

did i?

[merry geode]

Yeah, you did not

[round sage]

do my chats work

[merry geode]

In case of circle, you have pi1 = Z, abelian

[round sage]

sorry my chats werent working nd this was the active one i saw im sorry to bother!!

[merry geode]

You need to try one with two holes yea

[rancid umbra]

okay

and then go around different holes

[merry geode]

Yea

[rancid umbra]

got it

really enjoying this stuff; i love drawing pictures like this

[merry geode]

Oh seems like you are going for algtop?

[red yoke]

It's time to homological algebra

[rancid umbra]

reading it over the summer. hoping to take two more classes next fall to get some letters of rec and then yea, apply for grad school. it would be cool to study this i think

[merry geode]

Localization of the Top w.r.t. homotopy

[rancid umbra]

wut

have i not met the bane of AT yet?

[merry geode]

This is nowhere close to bane of anything, but there is this concept called "localization" where you get "true" homotopy category

In the sense that the Top category quotiented by homotopy map is "naive".

[merry geode]

This is a categorical one btw

[rancid umbra]

i have heard of the naive homotopy category, but i didn't quite understand why they were making the distinction

what's the issue with just quotienting by homotpy/what does it fail to capture?

[ebon galleon]

(wrt weak equivalences)

[merry geode]

Oof, yeah

Idk what's the real differences, weak equiv is homotopy equov on my smol CW

[merry geode]

Sorry I do not know, as I am not studying algtop in focus. @ebon galleon do you know?

[ebon galleon]

i mean i don't really do that much alg top so i can't really speak too too much
https://math.stackexchange.com/a/2876582 this is a decent answer i think. In short, I think it's that in practice a lot of homotopical considerations turn out in practice to be invariant under weak equivalence, not just homotopy equivalence (e.g. singular (co)homology). There's also the formal aspect of "It's good to be able to work in a nice category of (potentially bad) spaces" as opposed to restricting to a "bad category of good nice objects (like manifolds or CW/cell complexes)". In passing to the more general setting for categorical reasons, it works better to use weak equivalences rather than homotopy equivalences, although Whitehead's theorem tells you that they are "essentially the same" for CW complexes (i.e. produce the same homotopy category)

idk i've thought more about general model categories and simplicial sets than classical alg top so i don't have the best motivation

[merry geode]

I mean, why is the naive homotopy category bad as a category?

[rancid umbra]

hmm. this isn’t working out with OO. did you have anything else in mind?

i need them to agree on generators, but its kinda not working out

[merry geode]

Strange, my intuition says there shall be counterexample there

[merry geode]

Seems like this works as a counterexample

[rancid umbra]

no, i need them to agree on generators

beta_{p0} needs to be equal to beta_{p1} for every loop based at x0

where p0 = a1 + a2, p1 = b1 + b2

and beta is the change of base point map

[merry geode]

Why would it be equal?

[rancid umbra]

im trying to find an example of a space X with non-abelian fundamental group where two change of base point maps agree everywhere, but the two paths that the change of base point maps arise from are not path homtopic

[red yoke]

Two base point change maps are the same iff the two elements induce the same inner automorphism right

[rancid umbra]

i have no idea :/

[red yoke]

i.e. the same element in G/Z(G)

[rancid umbra]

yes, yea that makes sense

well, only if the elements are loops, right?

not more general paths

[red yoke]

Yes

But the same argument holds

[rancid umbra]

wait, for paths?

[red yoke]

Two paths induce the same isomorphism iff the loop they form is in the center

[rancid umbra]

alr, i buy that

[red yoke]

The center of π1(S¹ ∨ S¹) is trivial, so that's bad

[rancid umbra]

hmm. i don’t know any other spaces with non-abelian fundamental group

[red yoke]

Compact surfaces?

[rancid umbra]

should i postpone this until i learn van kampen’s theorem?

cus uh. i know the fundamental group of like four spaces lol

[red yoke]

Perhaps

[rancid umbra]

alr. are any of the surfaces arising from the fundamental square good to look at?

like a mobius strip? or a klein bottle?

well

mobius strip is bad

cus that’s just a circle

so i guess that just leaves the klein bottle

[red yoke]

Yea that has nontrivial center

[rancid umbra]

should this be true of any space whose fundamental group has non-trivial center?

[red yoke]

Well this is always true

[rancid umbra]

i meant, can you always find an example like this if the fundamental group has non-trivial center?

[red yoke]

Yea cuz of this

[rancid umbra]

oh, that was a hint

oh okay cool

converse?

if you have non-trivial center then can you find paths like this?

i suppose you just split the loops

[red yoke]

Yes

[rancid umbra]

okay. cool. ill keep this in mind and try to prove it later.

[radiant pollen]

I am trying to show that this https://pub.ista.ac.at/~edels/hexasphere/ (old april fools joke) doesn't exist. I am to assume it's a regular CW complex. I have an idea on how to go about it and am curious if anyone can give input if it's a right direction

Basically, the vertices of the hexagons are my 0-cells, the edges my 1-cells and the hexagons themselves my 2-cells. Then I was aiming to compute the cellular homology groups and was hoping that the result would be different from the known Homology groups of a 2-sphere.

However.. So far it seems that at least for H_2 I get the expected result (it's a group of integers). Before proceeding to calculation of H_1, could someone provide some input on this approach?

[red yoke]

If one assumes the hexagons are all regular then this is impossible cuz three hexagons sharing a vertex must lie on a flat plane

Alg top approach also works

I think the easiest method is to look at the ||Euler characteristic||

[radiant pollen]

I think my approach should end up in a similar argument (using the formula for Euler characteristic expressed in terms of betti numbers) and seeing something go wrong

[red yoke]

Perhaps

But I'm not sure how you're calculating H^n from this

[radiant pollen]

I might have made a mistake so let me clarify what I was planning to do:
So in this hexasphere we would have $h$ 2-cells (hexagons), $3h$ 1-cells (edges) and $2h$ 0-cells (vertices).
So, cellular homology gives a complex
$\dots \to 0 \to \mathbb{Z}^h \to \mathbb{Z}^{3h} \to \mathbb{Z}^{2h} \to 0$

[gentle osprey]

Faputa

[radiant pollen]

Then I proceeded as follows, using the known homology groups for the sphere and the rank-nullity theorem

(so I didn't really use the fact on how the boundary map looks like, I thuoght perhaps a contradiction can be reached just from these arguments).

Does this look correct? (I should say that I have barely done any alg. top. before)

[red yoke]

Yea

[radiant pollen]

But considering I've basically been subtracting/adding number of faces/vertices/edges this is basically the euler characteristic approach that you suggested?

[red yoke]

You can also shove this argument under the rug using formulas for Euler characteristic