point-set-topology

[dry jolt]

Idk if this is the same question, but you can characterize cohomologically trivial G-modules in terms of vanishing cohomology of the p-part when restricted to p-sylow subgroups

[zinc hearth]

If fixed points are functors then how do they act on morphisms hmmm?? They’re fixed!! You can’t do anything. Just give up 🆙….

[umbral panther]

There are projective modules over ZG that become free over QG, but aren’t free. The class group of Z[zeta_n] is about the same as for Z[Z/n]

[slate bane]

Is it true that the fundamental group of any simply connected open set of R^2, minus one of its points, is Z?

[jolly umbra]

yes because by the Riemann mapping theorem, it's homeomorphic to the open unit disk 😂

maybe overkill

[slate bane]

I didn't know there was such a theorem! Mm, maybe that consequence is not a trivial result and it's not overkill

[umbral panther]

Definitely overkill

[slate bane]

Oh, how could you do it then?

We saw it can't be trivial through a retraction on a small S1 around the puncture, that induces a surjection of the fundamental group onto Z, but I was wondering how to go further

[fallow ginkgo]

What is K_0[Z[S_3]]? In other words, the projective class group of the integer group ring of S_3. I need it for number theory reasons.

[tawdry valve]

Maybe a Van Kampen argument would be easiest? Call U the simply connected open set (ie before we remove the point). Then, U is the union of (U minus the point) + (a small disc centered at the point).

Then, Van Kampen tells you that pi1 of U is the pushout/amalgamated free product of the two fundamental groups. I.e,

pi1(U) = pi1(U - pt) * pi1(disc)/ <i(a) j(a)^-1>

where “a” ranges over hmtpy classes of loops of the intersection, and i, j are the maps sending “a” into both pi1(U - pt) and pi1(disc). Since pi1(disc) is trivial, j(a) is trivial, and we just get

pi1(U) = pi1(U - pt)/pi1(intersection),

We know the LHS is 0, so pi1(U - pt) is pi1(intersection), and the intersection is disc - pt, which has fundamental group Z (since it deformation retracts onto a circle).

[fair tusk]

Not sure I understand the big disjoint union notation for quotient spaces - how does a quotient space get formed with one? (For context, I’m reading about smash products and I don’t see how S^1 x S^1 / S^1 v S^1 is a sphere - the wedge sum is only one space, not two to be identified via quotient)

[tawdry valve]

The notation X/A means you collapse all of A to a single point—your equivalence relation is x~y is both x and y are in A, and as always x~x

and then X v Y can be thought of as a sub space of X x Y, where your copy of X is X x {y0} and your copy of Y is {x0} x Y, so the two spaces are glued together along x0 and y0

[fair tusk]

So there’s never a case where that refers to combining spaces larger than points? I guess the question comes from being able to identify opposite sides of a square to make a cylinder

[tawdry valve]

oh yeah that definitely happens too! In general quotient spaces are when you start with a top space X and an equivalence relation ~, and then you get a set X/~ together with the finest topology that makes the projection map continuous.

[slate bane]

That's pretty clever, thank you!

[tawdry valve]

yeah no problem! Hatcher does this sort of thing for applications of Van Kampen to cell complexes. It’s a little funny I think to do Van Kampen “backwards”—you compute pi1 of a piece by knowing the whole space, a different piece, and the intersection.

[fair tusk]

Ah okay, thanks - so then the smash product in that first case just has the wedge sum contract onto one point?

[tawdry valve]

so a common case of this is the X/A thing I described, but you’re totally right that you can get the cylinder by gluing two sides of [0,1]^2. In that case, you instead have the equivalence relation where

(0,y) ~ (1,y),

and (x,y) ~ (x,y) as always

[slate bane]

Yes this is the first time I see such a use of it, super cool

[tawdry valve]

yup

[fair tusk]

Awesome thanks

[slate bane]

I had a follow up question on this but I'll send it tomorrow I guess! Now I gotta go

[exotic rose]

very new to algebraic topology, so apologies if this is a silly question, but does homotopy equivalence and isomorphism as sets imply homeomorphism?

[jolly umbra]

by isomorphism as sets do you mean bijection?

[exotic rose]

yeah

[jolly umbra]

then no

consider [0,1] and (0,1)

[exotic rose]

ah beautiful

thanks

[obtuse meteor]

Does anyone know how to show that if M1, M2 are aspherical and have dim >= 3 then M1 # M2 is not aspherical

[boreal glacier]

Can anyone help me with a question that i wrote it in "real-complex-analysis"

[boreal glacier]

sry i dont know

[dry jolt]

Here's how I would prove it, assuming the connect sum itself is closed. Let G = pi_1(M1), H = pi_1(M2). By Van Kampen, pi_1(M1 # M2) = G * H (free product). I claim that a group of (co)homological dimension n >= 3 cannot split as a free product. There's a long exact sequence in cohomology for the splitting which gives 0 -> H^n(G * H) -> H^n(G) \oplus H^n(H) -> 0.

On the other hand, both G and H are infinite index subgroups of their respective free product. Let X be the cover of M1 # M2 corresponding to G. Then X is a non-compact aspherical manifold with fundamental group G, hence its top cohomology vanishes. That is, H^n(G) = 0. Similarly, H^n(H) = 0. Thus, H^n(G * H) = 0.

I can think a bit more if the connect sum is non-compact what might go wrong, but I would probably have to review dualizing module stuff (which I should really do anyway lol).

Also if something goes wrong in the above, it might be easier to take homology LES and/or take some suitable choice of coefficients

[unreal stratus]

goated

i had another idea but may be wrong

Ah I couldn't quite get it to work but found a nice answer by John Klein which used a similarish idea

Use the cofibration sequence S^{n-1} -> M#N -> M v N where n = dim M = dim N

the map M#N -> M v N induces an iso on fundamental groups and the latter is aspherical,so is both are aspherical then it is an equivalence

[unreal stratus]

contradicting this

[dry jolt]

Wait that's way better lol

[unreal stratus]

My idea was to study this sequence along with universal covers but not sure how that'd work

Like idk what uh M~ # N~ looks like

Hopefully it is (M # N)~ lmao

but i doubt that

Anyway the point would be that M~ and N~ are both contractible

So if M#N were aspherical then (M#N)~ would be contractible

and then the sphere would be contractible

Something like that feels like it should work since S^n-1 is aspherical for n >= 3

[night heron]

Why isn’t the set of odd natural numbers not open in the finite-closed topology of natural numbers

[unreal stratus]

Well, is its complement finite

[unreal stratus]

But then this would also work for n=2 so that can't work ig

[night heron]

Oh nvm. I forgot that the subsets need to be finite

Thx lol

[unreal stratus]

univeral cover of RP^3 # RP^3 is R x S^3 not S^3 # S^3

Maybe it works up to homotopy or smth

Apparently universal cover of connected sum of aspherical manifolds of dim n >= 3 splits as a wedge of (n-1)-spheres though (up to homotopy) which is curious

[dry jolt]

If it was aspherical, wouldn't the universal cover be contractible?

[unreal stratus]

exactly, so this implies that

wait sorry I mistyped, corrected now

[dry jolt]

Oh ok lol

That is quite interesting

[unreal stratus]

i guess like what is the easiest way to see this thing with fundamental groups lol

presumably you can argue using cellsor smth

[merry geode]

Wish I knew what cofibration seq is

[unreal stratus]

Roughly it is like a cokernel

[merry geode]

But a sequence is not like a cokernel, is it?

[unreal stratus]

Uh do you know about cokernels being certain pushouts

idk how much stuff u know lol

[merry geode]

I can just only identify cokernel as either a coequalizer, or an element of right-exact sequence

[unreal stratus]

Yeah coequaliser is fine for this lol

[merry geode]

Ah, so the coequaliser can be thought of as pushout as well. Interesting

[unreal stratus]

so one way t o put this isthat if you take a map f: X->Y of spaces and form the coequaliser with a constant map at a basepoint of Y, then you get the "cofibre" of f

[Really this is in oo-categories, but dw too much lol]

And like if f is a nice map, like an inclusion of a subcomplex of a CW complex X -> Y, then that is just Y/X

[merry geode]

How infty cat

[unreal stratus]

like literally just quotient out the little thing

[merry geode]

Ah, makes sense

[unreal stratus]

Yeah

[merry geode]

But of course, topology being topology, the category would not be nice in general

[unreal stratus]

Yeah this is why you need something more refined

Usually you require f:X -> Y to be nice (a "cofibration", hence the name cofibration sequence)

[merry geode]

Like equalizer
X -f -> Y -> Z
-0 ->
where Z being cofibre?

[unreal stratus]

Or you can do it in oo-categories and weaken the notions enough such that you can just make the definition work straight up like this

[unreal stratus]

yeah

but this is a diagram in an oo-category, so don't take it literally for topological spaces

it's just easier to describe that way lol

But yeah it's an analogue

You can get long exact sequences from these sequences, which is nice!

[merry geode]

Why is it diagram in infinite category?

[unreal stratus]

Oh I mean like for it to all work out nicely lol

[merry geode]

Ah, I see. Makes sense

[unreal stratus]

Like the notion I have in mind doesn't literally work in spaces since it's up to homotopy everywhere

But because the "up to homotopy" is baked in, it works well with homology and cohomology

like uh

if A is a subcomplex of X, we have a cobfiration sequece A -> X -> X/A

this gives rise to the usual exact sequence like

H(A) -> H(X) -> H(X/A) on homology

:)

[merry geode]

I recently learned a term "exact triangle" >.>

[unreal stratus]

oof yeah

[merry geode]

Nicely packs up the LES (with degree map)

[unreal stratus]

Btw so if you want to see how coequalisers are pushouts

if you have like this thing

X -> Y -> Z
->

where the two horizontal maps are f and 0

you can view this as the pushout of 0 <- X -> Y

[merry geode]

Y - -> Z
X -0-> Y

[unreal stratus]

where the right map is f and the left map is just 0

[merry geode]

Ah yeah

[unreal stratus]

it's nice to know heh

[merry geode]

Now, now I want to know why
S^{n-1} -> M#N -> M v N
is a cofibration sequence

Is there a natural map S^{n-1} -> M#N?

[hollow tinsel]

Mostly natural. The usual way that you define M # N is by a) puncturing each of M and N, b) gluing together an annular neighborhood of the the punctures, right?

[merry geode]

Yeah.. Oh

Sorry, I am dumb so I failed to realize this myself*

[hollow tinsel]

The S^{n-1} is some separating embedded sphere in that annular neighborhood

That doesn't make you dumb !

The embedding should be unique up to a) homotopy and b) potentially composing with an orientation reversing map of S^{n-1}

[merry geode]

Thanks..

[merry geode]

Ahh, so truly natural (up to homotopy & reversing map)

Yeah the sequence makes sense now, thanks a lot Shelby!!!

Also Potato, thanks to you too!!

[unreal stratus]

I guess it as well defined as the connected sum is, basically

And the data used in making the connected sum can be used to pick this sphere

Ultimately it doesn't matter for these purposes as it is just a sign error essentially :)

[obtuse meteor]

sorry how does this show G*H has cohomological dimension >= 3? Or why is that true a priori?

[dry jolt]

Assuming M1 # M2 is aspherical, the cohomological dimension of G*H agrees with the dimension of M1#M2 if it is closed

[obtuse meteor]

why can't there be an equivalence in a cofibration?

I guess I don't know anything abt cofibrations lmao

Maybe you can get it from a (co)homology exact sequence that that would be bad

[hollow tinsel]

Oh, hi Faye!!! It's been a really long timeeeee

Hopefully you are doing well :)

[hollow tinsel]

This is the right idea! It's not a priori impossible for the cofiber of a map to be an equivalence, but if this one were, you could compute the reduced (co)homology of the S^{n-1} using the long exact sequence, and see that it would have to vanish in all degrees, which it doesn't !

Btw, I think the result you are looking for holds even if you don't ask the original manifolds to be aspherical.

The follows from that fact that if an embedded sphere S^{n-1} -> X separates a manifold X into two non-compact pieces, then that sphere has to represent a non-trivial element in H_(n-1)(X, Z/2). This follows by examination of the intersection pairing between homology and locally-finite homology (essentially take a path "to infinity" on both sides of the sphere, which intersects the sphere exactly once; that intersection number depends only on the homology class of the sphere, which must therefore not vanish)

You can use this fact when X is the universal cover of M1 # M2. What you want to show is that if neither M1 or M2 is simply connected, then any lift of the S^{n-1} from M1 # M2 to X separates X in that way. You can do this, for example, by explicitly constructing a model of X, as a infinite connected sum of the universal covers of M1 and M2, where the summands are indexed by cosets in pi_1(M1) * pi1_(M2).

The general theorems here are:
a) "a contractible manifold of dim >= 2 can have only one end" (by poincare duality)
b) "the ends of the universal cover of a compact manifold are the same as the ends of its fundamental group"
c) "any non-trivial free product has infinitely many ends"

[vast estuary]

Not sure if this is the best place to ask, but why is the collection of ellipsoids contained in A compact? and to make sure I understand the topology correctly, we're identifying all ellipsoids with the cartesian product of R^d with M_d(R), right?

A is a bounded open subset of R^d, by the way

[red yoke]

Is A supposed to be compact instead

Oh wait unit ball is open

If we replace unit ball by closed unit ball and A by its closure then it should be clearer why it is compact

@vast estuary

[merry geode]

What is set of all ellipsoids contained in A

[red yoke]

The subset of R^(n×n) × R^n carved out by L(B)+x ⊂ A

[merry geode]

Ah, similar to a moduli space
Yeah allowing degenerate ellipsoid is crucial, I guess

[red yoke]

||More precisely, this is a closed subset with compact image under projection to either component||

I guess this also characterizes compact subsets of the collection of affine transforms

[vast estuary]

Why can we do that? The author considers the open unit ball

also we're looking at the set of ellipsoids, so the ball shouldn't matter, right

[merry geode]

Maybe E \subset A iff Ebar \subset Abar.

[vast estuary]

still not clear 😦

[red yoke]

Oh wait I think this is false for any nonempty open A

You can consider a sequence of open balls that get smaller and "converge" to some boundary point of A

Compact A still works though

[red yoke]

Since 0 is in L(B), the image of projection onto the x-space will be A

And the image of projection onto the L-space is compact due to a bound on singular values of L

(or any other reasonable justification)

[radiant holly]

what is topological sum?

[red yoke]

Topological disjoint union probably

[unreal stratus]

yeah i've seen that terminology in older textbooks

[slate bane]

Hey, yesterday in the end I ended up with a doubt about this

Van Kampen actually tells us that pi1(U) = pi1(U - pt)/K, where K = N(b*(pi1(intersection)), that is, the normal closure in pi1(U - pt) of the image of pi1(intersection) through the map b* induced by the inclusion b: intersection -> (U - pt)

[slate bane]

Previously, for the point I made here about this retraction, I think we know that b*(pi1(intersection) is a subgroup of p1(U - pt) isomorphic to Z.
We concluded that p1(U - pt) = N(b*(pi1(intersection)), as pi1(U) was trivial
So basically this should only tell us that, if G=p1(U - pt) is the group we need to determine, there is an element γ in G with infinite order such that G is its normal closure

[radiant holly]

Alright what is topological disjoint union?

[rancid umbra]

@radiant holly
wikipedia page describes it concisely imo

https://en.wikipedia.org/wiki/Disjoint_union_(topology)?wprov=sfti1#

[slate bane]

It shouldn't be enough to say G is generated by γ, that is it's Z, unless I'm missing something of course

[tawdry valve]

hmm I think you're totally right, I'm not sure if this proof can easily be fixed... maybe the Riemann mapping theorem is the way to go lol

[jaunty summit]

is there a name for a space with the same definition as a topology except for the empty set being open

[night heron]

Do you mean not clopen? Cause the empty set is both open and closed

[upbeat glen]

oh no

[rancid umbra]

why?

[lime sable]

empty union is the empty set

[radiant holly]

How a topological space be compact space?

[jaunty summit]

i know but you can have spaces like the cofinite filter over a set which have the T1 seperation property but dont contain the empty set as an intersection

[jaunty summit]

im using the definition where you just say that the whole set and the empty set are open which implies the same thing

i guess im saying that i want to find topologies where T - {empty set} has the finite intersection property

so the empty set cannot be written as an intersection of two nonempty sets

[zinc hearth]

if every open cover has a finite subcover? By definition?

[night heron]

Isn’t that just the T0 space

[rigid gorge]

compactness implies this condition, but not sure how to show the other way around (may be false idk). the setting is in C but seems like it's a topological statement

[tender halo]

does not seem true to me

[radiant holly]

Thank you

[fair tusk]

having trouble visualizing the X/B case here - how does a circle form from contracting B?

[obtuse meteor]

Contracting B identifies the two endpoints of A, and so turns A into a circle. It's a bit less obvious why the other side remains a sphere

[night heron]

Can I prove that the interior of the subset of a topological space is open by saying it’s a subset of an open set by definition and therefore open?

[night heron]

I was thinking when the open sets are disjoint. I don’t think you can have a closed set disjoint from the other disjoint open subsets

[night heron]

Trying to show that the closure of a subset is closed by showing its compliment is open in the topology but my sleep deprived brain is failing me

[zinc hearth]

What definition of closure are you using? This is immediate if you take the closure to be the intersection of all closed sets containing your subset

[night heron]

Currently limiting myself to the closure being the disjoint union between the interior and the boundary

Other definitions would make the proof much easier but I’m trying to follow along with the material

[night heron]

Nvm got it

[naive trench]

On a topological space X if D is dense and DcScX then S is dense too?

[ebon galleon]

yes closure is monotone

[naive trench]

Yay, thanks

[pastel idol]

The classical way to do this would be to consider the Quillen model structure on the category of topological spaces (I shall call it QTop). Localise at the set of weak equivalences, namely, weak homotopy equivalences between spaces. This yields the homotopy category of QTop. It seems like this is known to you already; if so consider it a remark for lurkers.

Recall that local equivalences are precisely equivalences before localisation of a model structure. Hence a map is a categorical equivalence of simplicial sets (in particular Kan complexes) iff it is a Kan equivalence before adding in any inversions. Combine this fact with the facts that (i) isomorphism classes of objects are weakly homotopy equivalent spaces and (ii) killing homotopy groups by attaching cells yields a category of homotopy n-types out of Ho(QTop), which are the inclusions in your mapping space, and I believe this gets you what you want (if not see last line)

Note that boundary inclusions of the sphere into the disc of one dimension higher are cofibrations, so you would not be localising at weak equivalences and it's not obvious to me the idea you had would work, but I've thought about it for a sum total of 180 seconds so it's worth a try anyway!

[wind bronze]

Is a function continuous if and only if it preserves the closure operators of the two topological spaces ? (or, equivalently, acumulation points of an open set of the domain get sent to acumulation points of the image of said open set, in the codomain).

I am having trouble figuring this out. Any help would be appreciated!

[red yoke]

Yes since this is equivalent to preimage of closed sets being closed

[tender halo]

uhh is it

constant function has no accumulation points in the codomain

[red yoke]

I meant, continuous iff image of closure is in closure of image

For any subset of the domain

[tender halo]

i guess by accumulation points they meant points of the closure

instead of limit points

[red yoke]

Also I don't think this is true if one only quantifies over open sets

[tender halo]

its not

take a map from an indiscrete topology to discrete topology

also if by "preserves" you meant equality and not what Arki said then it will be the definition of a closed continuous map

closure of image equal image of closure

[wind bronze]

Aha

Thanks!

[radiant holly]

can you give me a example on about proper maps

[hollow vault]

Any book reccomendations for topology

[worthy olive]

I heard hatchers was good but I haven’t read it

[hollow vault]

Isn’t Hatcher’s algebraic

I’m looking for normal topology

[worthy olive]

Usually analysis books have some topology

But idk about more advanced stuff

[hollow vault]

I know there’s one by Dover, the one by Munkres, and another that’s open access

[plush folio]

I've read much of both Lee's intro to topological manifolds and Munkres, and I slightly prefer Lee. The explanations are a bit better, and it emphasizes the characteristic properties of the various topological constructions

Topology without Tears is not worth reading imo

[hollow vault]

Why?

Also, Do you have any views on the Dover book

[plush folio]

I just didn't find it very illuminating. The concept of continuity is introduced way too late, it doesn't convey the intuition very well, and it states theorems and definitions in the most confusing way. For example, they would always write "there exist a set U such that x is in U and U is open" instead of just writing "x has a neighborhood U"

I haven't read the Dover book, so I don't know about that one, sorry

[hollow vault]

Gotcha so between Lee and Munkres

[plush folio]

Yep, those are both good, but I'm sure there are other good ones

Topology: A Categorical Approach is free online, but it might be a bit difficult for a first introduction

[chilly lynx]

Can someone help me with this question? ,How can I prove that two bases generate two different topologies?

more specifically these bases

I already tried that they are bases but I don't know how I can prove that they generate different topoligas

[hollow vault]

I’m as stuck as you are, but I would first take a guess as to what exactly these bases are generating.

[opaque scroll]

So the main aproach would just be to show that one basis element in one is not open in the topology generated by the other.

For example you might notice that no M(f, epsi) is ever contained in a U(f', epsi')

(I'm assuming the set in question is continuous functions on [0, 1])

[rancid umbra]

what does hatcher mean when he uses D_+ ?
I couldn't find it referenced anywhere else in the text before this

oh, he str8 up says what it is :/

reading is hard

to verify that i am understanding correctly,
for n = 1, D_{+}^{2} is essentially the upper half of S^2 with the equator, correct?

hence the notation?

[lusty trench]

That interpretation works even for n > 1, me thinks. It's just the upper half of S^{2n}, considered as the unit sphere in C^n x R, this in turn considered as a subspace of C^n x C.

[rancid umbra]

alright, cool. thank you!

another question, how do you show that a space can be realized as a CW complex? are there any techniques other than explicitly constructing the maps?

[lusty trench]

I don't know.

[night heron]

I don’t quite understand the justification for the second statement

Can anyone help me?

[rancid umbra]

can you send more of the page?

[uncut surge]

Since $B_1 \cap B_2$ is open, it can be represented as the union of elements of the basis, so there are sets $B_i \in \mathcal{B}$ with $\cup_i B_i = B_1 \cap B_2$. Since $x \in B_1 \cap B_2$, this means that there must be an element $B_i$ in the union with $x \in B_i$. That's your $B_3$

[gentle osprey]

Lartomato

[night heron]

Oh I see. Thank you

[finite stratus]

If G is an open set, with x \in G then does there always exist a closed set F containing x which is a subset of G?

[quartz horizon]

Not necessarily

[zinc hearth]

with no other information on your topology I'll just say "what if {x} is open but not closed?"

[finite stratus]

oh thanks; is this false for hausdorff spaces too?

[quartz horizon]

It’s true for Hausdorff spaces

[finite stratus]

can you give a hint as to how to prove this?

[quartz horizon]

Think of the simplest possible closed set you could try

Turns out it works

Figure out why

(hehe that rhymed)

[finite stratus]

ohh yeah because every singleton is closed. thanks

[quartz horizon]

Mhm

[bright acorn]

If you have an infinite dimensional real inner product space $(X,\langle \cdot, \cdot \rangle)$, is it true that the inclusion $O(X) \hookrightarrow \mathcal{B}(X)$ of orthogonal operators on $X$ into the bounded invertible linear operators of $X$ a deformation retraction?

in the finite dimensional case this is just Gram-Schmidt

but this is not so clear to me in the infinite dimensional case

btw, I am considering both spaces with the usual tolology induced by the operator norm

(but i am also curious about what happens wrt other topologies too)

[gentle osprey]

MisterSystem

[night heron]

What’s a good introductory Algebraic Topology book

[coarse night]

Hatcher

[winged viper]

Or Lee topological manifolds if you want something more gentle

[lusty trench]

Are non-smoothable, non-triangulable topological manifolds nicer enough than arbitrary CW complexes to justify studying them separately?

[red yoke]

Does this work for separable Hilbert spaces

If you do the nth step during time [1/n, 1/(n+1)]

[fading vale]

if you replace isometries by partial isometries this should be true

by the polar decomposition

Unsure about this general case

maybe for bounded invertible operators the partial isometry is an actual isometry and you win? but i kinda doubt

[night heron]

Thx

[night heron]

That’s what I started with

[grave solstice]

how do you show that every perfect set in a complete metric space contains a compact perfect set?

this is 30B.1 in Willard btw

[drifting gorge]

facts which may be helpful to recall:
-a closed set in a complete metric space is a complete subspace
-every neighborhood of a point x in the perfect set P contains a point in P other than x

[night heron]

Doesn’t completeness of a metric space imply closure of the set

[drifting gorge]

the set is already closed since it is perfect

[night heron]

Oh I see. I’m not familiar with perfect sets yet

[inland thistle]

Hi, I want hint for this problem, and this is from my topology class. I think I can understand what the distance injective subsets of R are like, but I dont know how one would prove it is uncountable. I think I know the definition of uncountable sets. It is the set such that there exists no surjection from N to it, or there is no injective function from the set to N.

one idea was that we could refer to each subset with number of points the set contains, and then we could always find some other distance-injective subset of R

but I dont think thats how its done

[lusty trench]

What would happen if you take a basis of R as a Q-vector space? (I'm not claiming that this is an answer.)

Let B be such a basis and take four elements x,y,z,w in B such that d(x,y) = d(z,w) > 0. We may assume x < y and z < w. Then y - x = w - z, so x,y,z,w can't be pairwise distinct, i.e., at least two of them are equal. By symmetry, you can reduce to two cases: x = z and y = z.

If x = z, then y = w and you're done.

I'm still thinking about the y = z case.

Oh, if y = z, then 2y = x + w is a linear dependence relation, so again x,y,w aren't pairwise distinct.

Since we can't have x = y, then either x = w or y = w.

If x = w, then x < y = z < w = x, contradiction.

If y = w, then z < w = y = z, contradiction.

[lusty trench]

Does the above solve your problem?

[inland thistle]

I’m not very sure what you meant above, but how would your argument say there exists a uncountable subset of R with that property

[lusty trench]

A Q-basis of R is necessarily uncountable.

Because a Q-vector space with countable basis is again countable. But R is known to be uncountable.

[inland thistle]

I see, that makes sense

[lusty trench]

Then I proved that a Q-basis of R is distance-injective in the sense of your original question.

[inland thistle]

I see, I guess I will check your answer later again

thanks for your help

[lusty trench]

np

[opaque scroll]

A somewhat 'less constructive answer':
Consider the poset of distance injective sets. Note that the union of a chain of distance injective sets is again distance injective, so we can use Zorn's lemma.

Now assume the maximal set is countable. Then you have a countable number of points and a countable number of distances. If you consider all points that are one of those distances away from one of your points, or are a midpoint between two of your points, that's still countably many. Since R is uncountable there must be possible to append a new point to the set and remain distance injective.

This also shows the stronger statement that any distance injective set is contained in an uncountable distance injective set.

[lusty trench]

I also used Zorn's lemma implicitly, when I said “let B be a Q-basis of R”.

But I guess your proof is nicer, because it doesn't invoke any unneded structure.

[umbral panther]

Did you make any progress on this? All I can think of is to prove the stronger statement that it contains a cantor set

[rancid umbra]

one idea would be to try to find a totally bounded subset of the initial perfect set

[tender halo]

I can't imagine you can get an easier proof than just saying it contains the cantor set

[umbral panther]

Partly there’s the evidence from it being an exercise. Why not assign the stronger statement? Is the point of the exercise to make the jump to the cantor set? or is there some technique I’m missing?

[tender halo]

but like, every compact perfect set will contain a copy of the cantor set, so its not a stronger statement

also i think if you say "prove that it contains a copy of a cantor set" it makes the exercise easier

[grave solstice]

the section this exercise is from is titled "The Cantor set"

but I did not find anything that is particularly useful for this problem on the section, or not that I noticed

[grave solstice]

You can just assume that your metric space X is complete and has no isolated points

maybe you can pick p,q in X such that q not in B_1(p) (the balls here are closed). Then, by using completeness, maybe you can find x in B_1(p) and y in B_1(q) such that d(x,y) is minimal among all such pairs. Then repeat the construction with p,x and y,q but look at the balls of some small enough radius, etc

well I'm not very sure that gives you what you want, even in R. You would have to take the closure

but I think then it's clear that what you get is perfect (closure of a thing without isolated points is perfect)

[grave solstice]

Once you have p,x,y,q, for constructing p,z,w,x you want w,z to be inside some ball centered at p (the smallest such that it contains x) etc.

[grave solstice]

let X be a metric space, $A=\bigsqcup_{n=1}^\infty A_n$ be the union of disjoint finite subsets $A_n\subseteq X$. Suppose $r_n$ is a sequence of positive real numbers with $r_n\to 0$. Suppose that for each $n$ the closed balls centered at the points of $A_n$ cover $A$. Is then $\ol A$ necessarily compact?

[gentle osprey]

croqueta3385

[grave solstice]

I can further assume that the balls are disjoint and that each ball of radius r_(n+1) centered at some point of A_(n+1) is contained in some ball of A_n ...

I think this is true by sequential compactness criterion. Like if B_1, B_2,... is a decreasing sequence of closed sets with diameter -->0 then the intersection should just be a point

[grave solstice]

except maybe this is not true lol. In Rudin it is required that the B_i be compact too

[red yoke]

It's true

In a complete metric space

[opaque scroll]

Assuming X is complete, then by Heine Borel a set is compact if it is closed an totally bounded. And it's not hard to see that your set is totally bounded

If X is not complete, then it shouldn't be true

[grave solstice]

yes I am assuming complete

is the intersection of a decreasing sequence of closed subsets of a complete metric space necessarily nonempty?

uhh

[opaque scroll]

[n, infinity) in R

[grave solstice]

yeah, this is relaed to spherical completeness

If X-->Y is a continuous bijection of separable complete metric spaces, is it necessarily an homeomorphism?

I went to Kechris book. Why is f "therefore an embedding"?

in case more context is needed, a Cantor scheme is this

[opaque scroll]

There's a bijection between [0, infinity) and the circle.

[unreal stratus]

Compact Hausdorff trick ig

Assuming C is the cantor set or smth lol

[grave solstice]

yes, C is the Cantor set

[unreal stratus]

Yeah

[grave solstice]

what is this trick?

[unreal stratus]

Then it's a map from compact into Hausdorff right

So it's a closed map

[grave solstice]

ah, right

[unreal stratus]

I'm not sure what a polish space is but they're at least lch right

[grave solstice]

Polish= completely metrizable + separable

[unreal stratus]

Ah lol

Then sure

[grave solstice]

but I mean, I don't think they used separable then lol

uh lol Kechris also discusses topo groups

[grave solstice]

by the way, Polish spaces needn't be locally compact

actually, C_p is a Polish space

[unreal stratus]

Ah okay sure

thank

[rancid umbra]

https://math.stackexchange.com/a/3707008/803927

perfect subsets of complete metric spaces are closed and uncountable. this may help

[heady skiff]

here A is a subset of Y right?

[umbral panther]

Yes, A and B have to be subsets of Y. Otherwise A could be all of Y and B a point far from Y and that would be a silly separation

[night heron]

Well yeah since Y is the union of A and B

[night heron]

Tried to teach myself point set topology from a 50 page set of notes

Horrible idea

[hidden crag]

Hatchers point set notes are doable

[night heron]

Yeah it’s straightforward until I try to do the proofs for the propositions and lemmas on my own before reading them

[timid verge]

if you have some topological space and you're considering fundamental group

why is concatenating a loop with the same loop but travelling in the opposite direction homotopic to the constant map at some point

It's not obvious to me what the continuous deformation is

[opaque scroll]

Imagine thinking about this as walking along the loop then turning around and walking back. Now imagine turning back just a little bit sooner, and then a little bit sooner until you never left.

[timid verge]

I was thinking about that but I still have one concern

if I do that sort of deformation

originally the loop going to the p and coming back in the opposite direction back to p encloses some area

but if we turn back earlier, this area isn't enclosed anymore

should I be worried about this happening during the continuous deformation? and if not, why

[opaque scroll]

Seems maybe you're conflating the function with its image. Like the shape of the image will change, but you're not deforming the image, you're deforming the path

[timid verge]

by the function are we referring to the homotopy or something else

[opaque scroll]

I meant the path, i.e. the loop

[timid verge]

can you explain the idea that the shape of the image will change, but we are not deforming the image but we are deforming the path

When i think of the deformation, it is true that I am thinking of it in terms of how the curve looks in the topological space

ok I guess there is no restriction on having to visit "p" three times when you deform

[opaque scroll]

Like think about a piece of string on the table. You can move the string around curl it up into a ball, make a loop or a figure 8.

The shape of the string changes, but you're only continuously deforming how the string is placed on the table

The same thing is happening here, you're changing the path continuously. Where will you be at time t never abroubtly changes, but the overall shape you trace out might

[timid verge]

I see what you are saying

I think where I was getting confused before is

if we are considering R^2 as our space for instance

are all curves homotopic to each other?

vs if I puncture the plane at the origin

[timid verge]

thx for your help btw

[opaque scroll]

More formally if f is your loop then
f(2t) for t < 1/2
f(2(1 - t)) for t > 1/2

is exactly the loop that first goes f, then f in reverse.

Then you can define a homotopy h(s, t) =
f(2ts) for t < 1/2
f(2s(1-t)) for t > 1/2

[rapid olive]

what ways are there to extend a metric from a field to its field of rational functions?

[grave solstice]

what makes you think this is possible?

discrete spaces are metrizable, you could take Q with the discrete metric, then the induced metric on Z is the same as the usual one

but with the usual metric, Q is not discrete

well you would need to be a little bit more careful if you want both metrics restricted to Z to be literally the same, but it is possible

maybe you just wanted the existence of an extension?

[rapid olive]

sorry, within the question "what ways are there?" i also meant implicitly "is there a way?"

[grave solstice]

lol for some reason I misread and thought you were asking about ring --> field of fractions. But I think still in your case K--> K(x) extensions, if possible, are not necessarily unique, which is what I tried to say

[grave solstice]

I think this is difficult

[rapid olive]

i kind of assumed if they existed they wouldn't be unique

[grave solstice]

A usual situation is when the metric on K comes from a valuation. (real) Valuations always induce metrics, and valuations can always be extended in field extensions. Given a fixed valuation on K, there are many valuations on K(x) that extend it

[rapid olive]

i see

[grave solstice]

If you have an absolute value on K, you can define an absolute value on K[x] by taking the max of the coefficients fo the polynomial, for example (absolute values on a domain extend to the field of fractions by multiplicativity)

[rapid olive]

so simple for like
the p adics

[rapid olive]

oh ok

i think this might actually work in general

[grave solstice]

but here are other examples

[rapid olive]

interesting

[grave solstice]

I also wondered about the classification of all such extensions, given that the situation for algebraic extensions is quite nice for non-archimedean valuations, you would need to be able to control the situation for purely transcendental extensions if you hope to say something about arbitrary extensions. But even in that book I just quoted, I don't think this is pursued

[rapid olive]

i see

and of course not all abs values come from valuations

[grave solstice]

I don't have an example of the top off my head, but I'm pretty sure these are not all, in case you are wondering. There might be a lot of freedom

[grave solstice]

Only non-archimedean ones (the absolute value on R certainly doesn't come from a valuation)

[grave solstice]

You may be able to define something like v(...)=min { v(a_i) +f(i)} where f is some function of i, or other variations. But didn't work this out myself

in any case, if you just have some arbitrary metric on K it is not clear to me that you can extend it to K(x)

well, if X is a metric space , and Y is some set with X subseteq Y maybe you can define a metric on Y by $d_Y(x,y)=\begin{cases} d_X(x, y),,\text{if},, x,y\in X\ 1,,\text{else if},, x\neq y\ 0,,\text{otherwise}\end{cases}$

[gentle osprey]

croqueta3385

[grave solstice]

wait this doesn't work haha

if the metric on X is bounded, then you can change the 1 by something big enough

[umbral panther]

You should make the valuation bounded by extending the ring in which the valuation takes values by adding an infinitely large element

[rapid olive]

well not arbitrary

i think by metric i really mean absolute value

[grave solstice]

yeah I suppose that if you want your metric to respect the field structure you end up with just absolute values

but you could also ask for metrics whose induced topology respects the field structure (like making the field operations continuous), but this is probably way more annoying. Idk if metrics will extend in this setting. Or if you have a topologial field structure on K and L/K idk if there is a topological field structure on L that extends that of K

[lusty trench]

I think the polynomial ring K[x] has a natural topology as the countably dimensional K-vector space. Then K[[x]] can be topologized as the inverse limit of the various K[x] / (x^m). And then K((x)) can be topologized as the union of the K[[x]]-modules generated by 1, 1/x, 1/x^2, 1/x^3, etc.

Finally, K(x) has the subspace topology given by the inclusion in K((x)).

[mild meteor]

chat, I am having a lot of trouble understanding mapping cylinders and just quotients in general I think. so here's the example I want to do. X=S1. Y=S1. f:X->Y and I want to quotient out the space (X x [0,1])UY/~ where (x,1)~f(x) and I picked f(x) to be the constant map. sending all points in X to a single point in Y.

but I am having trouble actually understanding what this looks like. or anything about it really

so X x [0,1] is just the ordinary cylinder. the union is the cylinder + just a unit circle somewhere. and (x,1) would be the top of the cylinder.

I have a suspicion that this would be a cone

[umbral panther]

My vague recollection is that this process destroys the topology on K

[lusty trench]

Ah, too bad, then.

[umbral panther]

This topic is higher local fields

[grave solstice]

I wondered how one could generalize local fields. I suppose that's a nice one

[unreal stratus]

so $((S^1 \times I) \coprod S^1})/((x,1) \sim )$ for some $ \in S^1$? You've taken a cylinder and a circle and crushed the top of the cylinder to a point of $S^1$. So you'll get something like a cone with a loop on the end

[gentle osprey]

Elliptic Potato
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unreal stratus]

up to homeomorphism

Often if you want to do this rigorously you can do stuff bit by bit. Like there's a map (S^1 x I) u S^1 -> cone u S^1 which crushes part of it, then map that to (cone u S^1)/ (tip ~ one point in S^1)

you can show this induces a map from your quotient to that last bit which is a homeomorphism

[mild meteor]

Where does the loop at the end come from?

Wouldn't it all collapse into the point?

I think my issue is I don't know which is in which space

[unreal stratus]

Well you've got an S^1

And you've only glued one point of it to the cylinder

[unreal stratus]

the pairs (x,1) are in S^1 x I

[mild meteor]

I'm starting to get a picture

[heady skiff]

in the equivalence class generated by these relations, why do we not have all four sides identified to a single point? because for instance, (1/2, 1) ~ (1, 1/2), and (1/2, 1) belongs to the top edge whereas (1, 1/2) belongs to the right edge

[opaque scroll]

Why do you say (1/2, 1) ~ (1, 1/2)?

[heady skiff]

Nvm I forgot the definition of an equivalence relation ig

[night heron]

Is it possible for an open ball to be a closed set?

[thorny agate]

In certain topologies yes

You could be dumb and take the discrete topology where every subset is open (and thus every subset is closed also)

In metric spaces I don't think so unless you have a dumb base set

[night heron]

Ahh I see

[unreal stratus]

yeah you can just have discrete metric spaces for example

[night heron]

Thx guys

[unreal stratus]

or, for example, bounded spaces where the whole thing is a ball

[thorny agate]

Oh yea bounded spaces are a better example

[night heron]

I see

[umbral panther]

In the cantor set everything that is at least 1/3 away is at least 2/3 away, so the open ball of radius 1/2 is closed

[thorny agate]

💀 atrocious

topological spaces were a mistake

[night heron]

So are closed balls always closed sets?

[umbral panther]

Yes. That sounds like a good exercise

[tranquil bloom]

this term doesn't appear to be explained in some of my notes

I'm guessing it just means "is non-zero in degree k"?

[opaque scroll]

That is exactly what it means.

Or rather "0 in degrees not k"

[tranquil bloom]

tyty

[alpine nest]

The Cantor set is cool and good and my friend

[rancid umbra]

is there an example of a connected top space whose homeomorphism group does not act transitively?

[red yoke]

[0, 1]

[rancid umbra]

which two points?
1-x takes zero to 1 and 1 to 0

oh shoot

like, no point on the interior can get mapped to 0 or 1

what about without boundary?

[alpine nest]

How about a lemniscate?

The "crossing" point has to be a fixed point of any homeomorphism

[rancid umbra]

i see. points near the crossing point must also get mapped near the crossing point. something something cut points as well seems to give a proof

[rancid umbra]

i think i have the intuition, but is there an easy way to see this?

further pushing this, what about an example where the top space is a connected, embedded submanifold without boundary?

[plucky veldt]

Let (X, 𝜏) be a discrete topological space. Prove that (X, 𝜏) is homeomorphic to a subspace of R if an only if X is countable

obviously Z is a discrete countable space

and in a discrete space every singleton has to be open, so in R every point has to have an open interval around it that doesnt contain another point of the subspace

how to prove that such subspace has to be countable?

[red yoke]

Isn't that just a connected manifold

[plucky veldt]

in other words, that every uncountable subset of R has to contain a convergent sequence

not sure if this is more topological, set theoretic or analytic

[rancid umbra]

lemniscate (figure eight) is connected and has a manifold structure but it is not an embedded submanifold

[red yoke]

You mean every uncountable subset of R has non-isolated points

You can use the fact that only countably many disjoint intervals can fit in R

[red yoke]

Yea but asking for embedded submanifold is just asking for any manifold right

[alpine nest]

Not sure about easy, I can't remember the proofs, but probably something about removing the point making the space disconnected

[red yoke]

Yea that works

[rancid umbra]

i dont see what you mean

[red yoke]

You're asking "is there a manifold X and an embedded submanifold Y ⊂ X such that Y is ..." right

What's the difference from just asking "is there a manifold Y such that …"

[rancid umbra]

ig ur right

[alpine nest]

Is the lemniscate a manifold? Doesn't the crossing point cause issues?

I don't see how it would have a neighborhood homeomorphic to R^n

[rancid umbra]

it has a manifold structure, but it is not a manifold when thought of as a subspace of the plane (with the subspace topology), so not an embedded submanifold of the plane

[alpine nest]

Wait, so what is "a manifold structure"?

[rancid umbra]

like, if you have a map from the open unit interval to the lemniscate, you can define a smooth structure on the graph of f since it is homeomorphic to the open unit interval

[alpine nest]

I've always assumed that a manifold is a space that's locally homeomorphic to an open subset of R^n

[alpine nest]

It's not, not if the graph passes through the intersection point twice, which it might

[rancid umbra]

if what is not what?

oh misread

[alpine nest]

The graph won't be homeomorphic to the unit interval if it passes through the intersection point twice

[rancid umbra]

right, but you can choose a map so that it doesn’t

[alpine nest]

I don't see what it has to do with being a manifold though

What's your definition of "having a manifold structure"?

[rancid umbra]

it is locally euc, second countable, and hausdorff

but that is irrelevant

the topology you place on the set matters

when you place this topology on the lemniscate, then you get a smooth manifold

but when you place the subspace topology on it, inherited from R^2, it is not even a manifold

[alpine nest]

I mean, yes, of course I was talking about the inherited topology on the lemniscate, what would be the point otherwise?

[alpine nest]

This just turns the lemniscate into an interval, which fair enough, but for what?

[rancid umbra]

well, because i wanted a manifold without boundary whose homeomorphism group acted transitively

but uh

i guess that example fails

oh wait

did i only ask for top space

ah, i did

okay okay

my fault

i see what arki is saying now

thank you Outsider

oh but the automorphism group for a connected topological manifold without boundary acts transitively

[plucky veldt]

yes, or that every uncountable subset of R has to contain a cluster point

[plucky veldt]

right, how is that proven?

[rancid umbra]

Q is dense and countable

[plucky veldt]

ok, I guess you can use the fact that every interval contains a different rational number, which means their number is countable

https://proofwiki.org/wiki/Set_of_Pairwise_Disjoint_Intervals_is_Countable

[rancid umbra]

the axiom of choice is convenient here but it is not needed

for bounded intervals, there are only finitely many rationals with bounded denominator, so you can find the first one that is non-empty and then choose the smallest one

for an interval not bounded above, use the ceiling function.
for an interval not bounded below, use the floor function.
for R, choose 0

[rancid umbra]

in the definition of chain connectedness, why do we need finite chains? why is it not enough to use any length chain?

i guess what i am asking for is an example of a disconnected space which is chain connected (where the chains can be of arbitrary length)

[opaque scroll]

What would it mean for two points to be connected by an infinite chain? Like an infinite sequence doesn't end, so which points are they connecting?

[rancid umbra]

for example, what if the shortest chain between two points is omega+1, so that it does end at omega

or maybe you want to define notion of length as cardinal of the chain, so that a chain is of length kappa (a cardinal number) if there is a map f from kappa + 1 into the open cover such that f(0) = x, f(kappa) = y, and f(alpha) intersect f(alpha+1) is non empty for each alpha in kappa

[opaque scroll]

Okay, but what's the relationship between the omega first parts of the chain and the next?

Like, I have a sequence
x1, x2, ..., x_omega. How is x_omega related to the previous points?

[red yoke]

A total order with successors and predecessors is necessarily a concatenation of copies of Z

If endpoints are allowed, then the copies of Z at the endpoints are truncated

So the only way two endpoints are meaningfully "connected" is via a finite chain

[worthy olive]

Hi im reading Vick’s homology theory, and I’m having trouble understanding what it means to add sets $X_{a_{i}}$

[gentle osprey]

CJ_:)

[zinc hearth]

It looks like they’re using it to denote a coproduct, which is beyond bizzare

[worthy olive]

Can you do a coproduct with more than 2 objects?

[unreal stratus]

Yes, you can do it over any set

(Well, depending on context, but the notion makes sense)

[worthy olive]

But if it’s the coproduct, what does it mean for $\sum_{i=1}^{n} f_{a_{i}}^{b}=0$? In the definition of R?

[gentle osprey]

CJ_:)

[zinc hearth]

it never says that, it says that the "sum" of those functions evaluated at the x_a is zero, aka is just the identity element in the coproduct since we're dealing with abelian groups

[nova fjord]

Is there a simple way to see that you get a klein bottle if you act on each copy of the circle in S^1 x S^1 via Z_2 and take the resulting orbit space?

[lusty trench]

The easiest way would be to use a parametrized embedding of the Klein bottle in R^4.

Or rather, the parametrization would be a map R^2 -> R^4 whose image is the Klein bottle. So, strictly speaking, the map is thr universal cover of the Klein bottle.

[nova fjord]

Is there an easier way via the usual presentation of the torus as a square with opposite edges identified?

[umbral panther]

Is that true?
How are you acting? If it has fixed points, the quotient isn’t a manifold. If you act on the two factors separately without fixed points, you preserve orientation, so the quotient is orientable, the torus, not the Klein bottle

If you have an unorientable surface covered by the torus, it is the Klein bottle. This is a good criterion, even if you can’t use it as a definition without proving uniqueness

[nova fjord]

(x,y)g = (xg, g^{-1}y) is how I'm acting

[umbral panther]

How does Z/2 act on the circle?

[nova fjord]

A rotation by pi, right?

[umbral panther]

Ok, then the quotient is a manifold, but an orientable manifold, the torus, not the Klein bottle

[nova fjord]

Is what I'm describing not the same as what this problem is saying? (This is from Homotopical Topology by Fomenko and Fuchs)

[umbral panther]

Is this one of those books with fake exercises to keep you on your toes?

[nova fjord]

There are typos in this book here and there

Is the actual relation I need to factorize by (phi, psi) ~ (phi, pi - psi)?

[umbral panther]

That has fixed points
You can’t treat the two factors completely separately

(x,y) ~ (x+pi,-y)

[nova fjord]

Isn't y --> -y the same as adding by pi (in terms of angles)?

[umbral panther]

Adding pi preserves orientation. Negating reverses it

Also, adding pi has no fixed points. Negation has two fixed points, which you need the x coordinate to avoid

[nova fjord]

Ah gotcha. So in terms of the unit square, this corresponds to flipping the direction of one of the identifications

Instead of identifying opposite edges, one of the pairs is identified with a twist

Thanks!

[viral robin]

Very basic quetion but whta is the dimension of the reunion of 2 manifolds (for example the complex plane added to the unit sphere of R³) ? Is it defined locally ?

[rancid umbra]

the reunion?

[viral robin]

Yes

[umbral panther]

The disjoint union of two manifold is not usually allowed as a manifold. But if it is, then dimension is a locally constant function

[viral robin]

Thank you !

[umbral panther]

I imagine the Lebesgue dimension is upper semi continuous

[mild meteor]

I'm trying to construct a mobius strip via a mapping cylinder, and I am not quite sure how to go about it.

how do we talk about specific sides of the box

which is kind of a stupid question ngl

maybe I should wait until I reach the cw complex part

[robust drum]

in the source I'm reading, taking a solid cylinder (D x [0, 1]) and removing n tubes gives a handlebody with 2n-1 handles. But for some reason all the pictures I'm drawing make me think it should be just n handles. ex in the 2 tube case.

like i feel like if we have n tubes we join all of the tubes on one side to get n tubes on the top and 1 tube on the bottom, slide it up to the top so we have a sort of network of n+1 tube openings on top all connected, then between each pair of openings there is a filled tube that we can poke up into a handle to get rid of the tube there - because we have n+1 openings we have n+1-1 = n total handles

Can someone explain what's wrong with my picture/logic?

[robust drum]

maybe I'm interpreting the original source wrong

[merry geode]

What are you doing at the third stage?

[robust drum]

I bring the hole on the bottom of the cyllinder up to the top of the cyllinder

or do you mean the stage after that

the stage after that I am basically taking solid black tubes sitting on top of the empty green tunnels and poking them up out of the manifold - after doing that, we go from having the green stuff be just a tunnel network to being only an indentation in the top of the manifold

which we can just deform away, and we're left with a handlebody with 2 handles

[merry geode]

Hm, I can see 2 handles arising, but you still have "tube" going below which needs to be taken care of.

[robust drum]

I think that after you poke up both handles the tube is now just an indent

because you no longer have anything above it except for the handles

[merry geode]

The point is that it isn't. Idk how to explain this but

[robust drum]

the way I think abt it is if you pick a point in the green tube and look up either it's below an entrance or it's below a handle

[merry geode]

Basically you have another "loop" around it

[robust drum]

so if you poke up all the handles then there shouldn't be anything else above it

but maybe im visualizing badly

im also not sure because I feel like if we start instead with the cylinder with 3 holes in it and instead of even doing anything we just shorten the cyllinder a lot it looks to me a lot like a 2 holed torus

which would be genus 2

so I think maybe im just misinterpreting the original source

because this was like a side comment in a paper abt something else and also using some formula that I might be misreading

[merry geode]

I'm sure it should be of genus 3.

Ah wait

Sorry, you are right. It should be of genus 2

You can collapse it height-wise from the beginning and it literally is 2-torus..

So difficult

The very first stage looks iffy ime as well

Consider homotopically "making a skeleton", the first diagram is 2 S1-s while in the middle it is 3

[robust drum]

hmm. I think because this is a filled cyllinder the middle should be 2 s_1 also. The only iffy diagram is the second one, I'm quite sure the third only has 2 nontrivial loops in it up to homotopy

[merry geode]

Which one is the third?

[robust drum]

regardless of the skeleton talk all you need to do to get from the first to second drawing is to compress material vertically

[robust drum]

the one where all 3 are on the ceiling. Actually now I

hmm

I do agree the first diagram is homotopic to a wedge of S_1 with itself. I think that maybe the second one is too?

[merry geode]

Hmm, yeah maybe all are just homotopic to wedge of 2 S_1's

[robust drum]

I think they are. tho I can't quite picture the second one. But I am pretty sure that just "compressing material to join the openings" would be a homeomorphism

[plucky lake]

I'm trying to understand how to calculate the first simplicial homology group of the torus. If we let the 1-simplices of the torus be represented by a, b, and c, the kernel of the first boundary map will form the free abelian group whose set of generators is <a,b,c> and the image of the 2nd boundary map will form the free abelian group <a+b-c>. I know that for the purposes of computation, we can calculate H1=Ker(del1)/Im(del2) by establishing the relation a+b-c=0 and observing that Ker(del1) reduces to <a,b>. This means that H1 is isomorphic to Z2. However, I don't know how the cosets of H1 form a group that is isomorphic to Z2 from an algebraic perspective.

Can someone help me out please?

[gentle girder]

use the first isomorphism theorem, find an onto homormorphism Z^3 -> Z^2 with kernel <(1,1,-1)> - here is the hom: || (x,y,z) -> (x - y, x + z) ||

[plucky lake]

Thanks. My knowledge of algebra is very weak compared to my knowledge of general topology so I've been having issues understanding what happens "behind the scenes" during homology calculations. I'll study the first isomorphism theorem- thanks for pointing me in the right direction

[tacit basin]

is this ok to say idk why it feels iffy manipulating them like this

I can prove the second equality (what I need) directly in like 2 lines, but we did the left in class so its cleaner if I can just do this

[gentle girder]

in a general topological space the boundary is defined by $\bar A \cap \overline{X\setminus A}$, which if you take the complement gives us $\mathrm{int} A \cup \mathrm{int}(X\setminus A)$, so we are good even not in R^n

[gentle osprey]

smay
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gentle girder]

why is texit giving me a compile error

but like anyways yes you said a true thing

and it's true just set-theoretically

[tacit basin]

ok cool ty

Ah wait im silly, obv if $X=A\mathbin{\dot{\cup}}B\mathbin{\dot{\cup}}C$ then $A=X\setminus(B\mathbin{\dot{\cup}}C)=(B\mathbin{\dot{\cup}}C)^{;\complement}$

Tyty

[gentle osprey]

Sara

[grave solstice]

Suppose $X$ is connected and that for each $p\in X$ we have an open nhood $U_p$ of $p$. Given $p,q\in X$ is it true that we can always find finitely many points $p_1,\dots, p_n$ such that $p\in U_{p_1}$, $q\in U_{p_n}$ and $U_{p_i}\cap U_{p_{i+1}}\neq\emptyset$ for each $1\leq i<n$?

[gentle osprey]

croqueta3385

[grave solstice]

Given $p\in X$ let $V_p$ be the set of points $q$ such that the above is impossible. If $q\in V_p$ then $U_q\subseteq V_p$, obviously, so $V_p$ is open. If $q\not\in V_p$ then $U_q$ doesn't intersect $V_p$ and so $V_p$ is also closed. So $V_p$ is clopen, hence is empty or the entire space

[gentle osprey]

croqueta3385

[umbral panther]

I think the long line gives a counter example

[grave solstice]

why doesn't this work then?

[red yoke]

Does it

[umbral panther]

The long line is not a counterexample
But I suspect that there is a longer line that is a counterexample, while not being a manifold

[alpine nest]

The very long line

[red yoke]

Anything based on ordinals won't be a counterexample

[opaque scroll]

Any decreasing sequence of ordinals is finite, so I don't think there will be any problem

[opaque scroll]

Argument seems good to me. I think I actually had this question on a point set topology exam long ago.

Don't know if anyone managed to solve it though 😛

[night heron]

How do I do this

Nvm got it

[fair tusk]

When considering homotopy classes of loops, can the basepoint be passed to form a different loop? For context, I’m looking at the figure 8 - Wikipedia says its fundamental group is the free group of 2 words, but if we allow a loop to pass the basepoint and continue onto the other side, this seems to form a new homotopy class.

[night heron]

Nvm thought I did

[night heron]

Can someone help me with this

[fair tusk]

Shoutout to the freeform app, I can finally draw on my phone 🥳

[red yoke]

The fundamental group depends on a choice of fixed basepoint

Any path from x to y will induce an isomorphism π1(X, x) → π1(X, y)

And if the path is a loop at x then you're basically doing an inner automorphism

Which is not necessarily the identity map

[fair tusk]

Oh, I should’ve clarified, the basepoint is that middle point where the loops meet

[red yoke]

Oh sorry I misread your question

Yea it's allowed

[fair tusk]

Oh neat

[red yoke]

A loop is any continuous map from a circle to the space

[fair tusk]

So Wikipedia botched that one lol

That seemed a little strange

[red yoke]

It is generated by the two smaller loops though

[fair tusk]

True

[red yoke]

It is indeed the free group with two generators

[fair tusk]

Can the basepoint be skipped though?

[night heron]

[fair tusk]

Like if the loop passed through it and didn’t form two loops

[night heron]

Do I use the Archimedean property

[red yoke]

Not sure what that means

[fair tusk]

Like can a path pass over the basepoint and not be considered as a loop

Wait that’s stupid never mind

Thanks lol

[red yoke]

Just to check you know what the free group with two generators is right

[red yoke]

This would be the loop ab^-1

[fair tusk]

Pretty sure, its the group of words aba^1b^1 etc right

[red yoke]

Yea

[fair tusk]

Oh ok, for some reason it wasn’t clicking that a loop is formed at that basepoint

[wind bronze]

Is a topological space Hausdorff if and only if all convergent sequences have only one limit ? If yes, why ?

[rancid umbra]

what do you think?

[naive trench]

You can try to prove it, its a good exercise

[wind bronze]

I don't know if being Hausdorff only implies uniqueness of limits, or if it is equivalent to it.

I'll try proving equivalence, I guess, and will let you know if I get stuck.

[opaque scroll]

They're not equivalent, so I guess your time is better spent looking for counter examples

I think it's true if your space is first countable though, so you might try to prove that

[wind bronze]

Oh, okay

[radiant holly]

what is simplicial set?

[lusty trench]

“Simplicial X” means contravariant functor from \Delta to the category of X's.

Where \Delta is the category whose objects are [n] = {0, ..., n}, for each n \in N, and its objects are the monotone maps.

[radiant holly]

thanks

[round oyster]

where is Ranalysis channel?

[tender halo]

two stops down

[slate bane]

I'm sorry but is it true that a continuous functions f: A -> R where A is a subset of the reals, is always closed in its image f(A) ?

I mean it is topologically closed when viewed as a map f: A -> f(A), taking induced topologies by R

[languid patrol]

no

take the function $f: {-1, 0, 1, 2, \dots} \to \mathbb{R}$ given by $f(-1) = 0$ and $f(i) = 1/(i + 1)$ for $i > -1$

[gentle osprey]

Math_Discord_Final_Girl

[languid patrol]

then every subset of the source is closed

but the map is not closed

since the image of the set $\mathbb{N}$ within the source is not closed

[gentle osprey]

Math_Discord_Final_Girl

[slate bane]

Mm, right, and if we add that A is an interval?

[languid patrol]

well if A is compact then its continuous image is always compact

and within $\mathbb{R}$ compact sets are closed

[gentle osprey]

Math_Discord_Final_Girl

[slate bane]

A any interval I mean

I.e. a connected subspace of R

[languid patrol]

well then no

if A = (0, 1) then the identity map Id: A \to \mathbb{R} is already not closed

if A is a closed interval then it's true

[slate bane]

But id is closed in its image, which is A

[slate bane]

I am still considering this, adding that A is taken connected/an interval

[languid patrol]

are you asking if the map to the image is a closed map?

[slate bane]

Yes

[languid patrol]

okay then consider the map f(x) = x for x < .8, f(x) = 1.6 - x for x \geq .8

[unreal stratus]

Interesting choice of numbers

[languid patrol]

this gives a map from (0, 1) \to \mathbb{R} which is not closed (even in its image which is (0, .8] )

[languid patrol]

I wanted it to look nice

[unreal stratus]

Ye noice

[slate bane]

Alright thank you!

[hearty condor]

I am making an operator

Yay

A smoothness operator

And a unsmoothness operator

On the powerset of paths

[worthy fjord]

Hey I was just wondering if the fact that two topological spaces have isomorphic fundamental groups guarrantees that they have the same homotopy type?

[umbral panther]

No
The spheres S^2 and S^3 both have trivial fundamental group but are not homotopy equivalent

[worthy fjord]

ok thanks

yeah idk how i didn't immediately notice all of the counterexamples with trivial fundamental groups

[lusty trench]

Even having all homotopy groups isomorphic isn't enough in general.

[gentle osprey]

Kapa

[white oxide]

Math just isn't advanced enough to tackle such problems

[worthy olive]

Idk but you could try making specific examples

[alpine nest]

Aren't you going to have a cardinality problem for starters? S_n has cardinality n!, and the set of integers mod n has cardinality n

An isomorphism between finite groups of different cardinality might be challenging to construct

[white oxide]

such cowardice smh my head

[tribal palm]

yes, for n=1,2

[unreal stratus]

Unfortunately n=0 reverses the pattern

[alpine nest]

integers mod 0 is one of my favourite groups

[umbral panther]

What about n= -1?

[gaunt linden]

What about n=-1? Just compute the automorphism group of F_{-1} and you will see the answer.

[umbral panther]

F_{-1} is R, which is rigid
But what is F_{-2} ?

[languid patrol]

F_1 pour tout le monde, mais -2 ce n’est pas bien, il a besoin de plus

[dim perch]

this is the best argument for why we should only care about spaces whose homotopy type is that of a CW complex (e.g. CGWH)

[plain raven]

I doubt it is the case that every CGWH space is homotopy equivalent to a CW complex.

I also think that "having all homotopy groups isomorphic" is strictly weaker than "weak homotopy equivalent" and you seem to be conflating them.

https://math.stackexchange.com/a/283200/427832

[unreal stratus]

it is strictly weaker ye

[unreal stratus]

Hm is this a common viewpoint

[dim perch]

iirc CW approximation holds for CGWH topological spaces (i.e. they're weakly equivalent) and the distinction between a conservative and essentially injective functor is somewhat pedantic for an imprecise discussion

[unreal stratus]

Or do you mean homotopy type in the sense of weak things lol

[dim perch]

this yea

[unreal stratus]

but then it sort of means anything

hm i thought CW approximation holds even for non CGWH things

[plain raven]

Every topological space is weakly equivalent to a CW complex.

[unreal stratus]

yeah

[dim perch]

huh

[unreal stratus]

e.g. hatcher does cw approximation without having mentioned cgwh stuff at all

[dim perch]

cards on the table, I'm a homotopy theorist, so ig I mix up my reasons

[plain raven]

https://math.stackexchange.com/a/1537965/427832

[dim perch]

is it just for mapping spaces that one asserts CGWH?

[unreal stratus]

it's just i've heard it sometimes stated only for cgwh things in like homotopy things

[unreal stratus]

as in, is this the main reason to care about cgwh?

well exponentiability is one thing but another thing is that quotients and (finite?) products distribute as one would like

and product of cw complexes is a cw complex in full generality if you use cgwh topology i think

not as sure about that last point but i think it is true, at least for finite products (but with possibly infinite cw complexes)

[dim perch]

hm

[unreal stratus]

idk, i also identify as a homotopy theorist and mostly care up to weak equivalence but i would phrase the situation slightly differently aha

to me it is that you can model spaces up to (w)he in various different ways, and often it is convenient to take CW models, but other times that is not possible (for example if you use a construction that involves extra structure, even if you only care about the output up to some weaker equivalence)

hm

[merry geode]

Wait whh

[dim perch]

I think I've never actually had to use a topological model for an object which wasn't CGWH, with the possible exception of accidentally touching the topological space underlying a scheme

[unreal stratus]

tbh same, i think cgwh is very nice

[dim perch]

this fact about quotients not playing nicely with products is somewhat disturbing to be lol

do you have an example or a reference?

[unreal stratus]

Well not got an idea off the top of my head sorry

hm

usually it works if one of the spaces involved is LCH, so the example would have to be relatively hairy

like stuff with Q can given counterexamples lol

[lusty trench]

If you have an arbitrary space X, can't you construct its Postnikov tower ... -> X_(n+1) -> X_n -> X_(n-1) -> ... -> X_0, and then X will be weakly homotopy equivalent to the inverse limit of the X_n's anyway?

[unreal stratus]

I think there is a weird counterexample with smash product not being associative up to homeomorphism for example like involving smash of Q with Z and stuffs

[umbral panther]

Lots of enumerative combinatorics gives functions of q, the cardinality of the field. If you substitute q=-1, you get the Euler characteristic of the corresponding real space

[dim perch]

yea this is true

[dim perch]

horrifying

[heady skiff]

how important are the countability axioms for algebraic topology? i'm reading through munkres (point-set) as a prereq for studying some algebraic topology, but honestly point set is very boring for me

so I'm wondering if it's sufficient to just cover topological spaces, continuous maps, connectedness, compactness

[dim perch]

ok I'm understanding this a bit better now, and learning that I don't care too much about the distinction--a relevant theorem is that the inclusion of e.g. CG topological spaces into topological spaces forms the right adjoint in a quillen equialence between the quillen model structures on each

[unreal stratus]

well, i have never worried about them lol

[dim perch]

I can only speak for myself (a homotopy theorist who pretends to do algebraic topology), but I have found that I've cared about point-set facts outside of the fundamentals of connectedness and compactness an extremely small number of times while doing algebraic topology

they are mostly technical mid-proof inputs in classical alg top ime

[unreal stratus]

sometimes it can be useful to know like, idk a manifold is second countable or smth is compact to use nice open covers

but not sure beyond that lol

[heady skiff]

i see

in that case, I'll probably just study up to compactness then and leave out some other topics

thank you

[dim perch]

seems like a good move, as long as you're comfy coming back to spot-study holes that may come along, like e.g. you at some point may feel you need to know about paracompactness for the duration of a technical proof

[white oxide]

If you had your relevant share of analysis classes you should have a decent enough familiarity of things like compactness and connectedness

[heady skiff]

oh yeah definitely

also despite having taken two algebra courses i'm sure there's some topics that will need refreshing as well

[heady skiff]

yeah i've taken real and complex so i'm familiar with compactness

although connectedness is a bit weak

[white oxide]

Most important prerequisuites are algebra (drawing diagrams) and topology (drawing filled squares) ofc

[winged viper]

I would recommend reading Lee rather than Munkres since he covers basically all (and only) the point set needed to get started in alg top

[heady skiff]

although I'm not entirely sure how the notion of compactness in euclidean space/R generalizes to arbitrary topological spaces

well I can guess lmfao

but i want to rigorize that understanding

i'm planning on doing rotman

i heard hatcher is very hand wavy and I hate hand-waviness

[white oxide]

Well, for AT you really only care about things that are locally path connected. You might get the connectedness stuff warmed up in a weekend day tbh

[heady skiff]

i see

so it would be okay to just start then

i know rotman also has a brief cover of category theory since i need a refresher on that too

[white oxide]

well, a good number of AT authors expect that this is is your first time where category is the work horse and will teach you whats necessary

[heady skiff]

ah okay

[lusty trench]

I don't have much issue with Hatcher's handwaviness, but he yaps for a really long time (or rather, a large amount of space, sometimes several pages) about geometric intuition.

[grave solstice]

What do you guys think about Dieck's book on AT?

[lusty trench]

Haven't used it. Maybe I should.

[white oxide]

wdym

[heady skiff]

nvm lol

oops shouldn't have deleted

i can't read

[white oxide]

the reading comprehension devil takes no prisoners

[unreal stratus]

Dieck's book is nicely organised but don't think it is good for a beginner

[white oxide]

Its good reference and a place absorb some homotopy cat terminology before moving on to more dedicated books

[radiant holly]

X={1,2,3}
T|X={X,{},{1},{2},{1,2}}
Open covers={X,{},{1},{2},{1,2}}
{X,{}}
{X,{},{1}}
{X,{},{2}}
{X,{},{1,2}}
{X,{},{1}}
{X,{1}}
{X,{2}}
{X,{1,2}}
Mean (X,T|X) compact space

Is it correct?

[lusty trench]

Any finite topological space is compact.

[radiant holly]

But my exam say show open cover

[unreal stratus]

wdym

show every open cover has a finite subcover?

[radiant holly]

Yes

[unreal stratus]

then you can just say that every open cover is already finite so you can just take that cover

[lusty trench]

Any finite topological space is compact for dumb reasons.

[white oxide]

how are you even getting a non-finite open cover here

[lusty trench]

An open cover of X is, first and foremost, a subset of its topology, which is in turn a subset of the powerset 2^X. But, if X is finite, then 2^X is finite as well.

[white oxide]

(I'm wondering, when would you ever want duplicates of the same open set in an open cover?)

[lusty trench]

If you treat an open cover as an indexed family of open subsets, then you might want to assign the same open to two different indices.

[grave solstice]

sometimes points are counted with multiplicities, so maybe in some context this would makes sense, but idk.

[lusty trench]

Yeah, that's a good example.

[white oxide]

Okay yeah, that makes sense

[radiant holly]

X=R
T|X={R,{}}
And open covers={R,{}},R mean (X,T|X) compact space A subset X and A={1,2} and T|A={A,{}} mean (A,T|A) subspace (X,T|X) and (A,T|A) compact because A finite. And A compact subset of X did I get right? @lusty trench

[heady skiff]

what would be the vertices of the 2-simplex?

[lusty trench]

The standard basis vectors.

[heady skiff]

would they just be (1, 0, 0), (0, 1, 0), (0, 0, 1)

ah ok

how would one show that a 3-simplex is a tetrahedron if it is a subset of R^4?

don't we only have geometric intuition up to 3 dimensions

[unreal stratus]

the usual constructions of n-simplices take place in R^{n+1} but can be embedded into R^n

i mean, for example the one simplex is (often taken to be) the line between (0,1) and (1,0) in R^2

[lusty trench]

The equation \sum_i x_i = 1 defines an affine subspace of R^(n+1) isomorphic to R^n. You could take, for example, the first n standard basis vectors of R^(n+1) to the standard basis of R^n, and the last standard basis vector of R^(n+1) to the origin.

Then your question becomes “why is the convex hull of the origin and the standard basis of R^3 a tetrahedron?”

But the presentation in your book is a bit nicer, because it treats all the vertices equally.

[unreal stratus]

i guess another way to view it is that like

oh i guess what you are saying would be equivalent lol

[sly geyser]

can a topology be equivalently defined as a family of subsets that don't have the empty set and X, are closed under (possibly infinite) intersections and finite unions (closed sets)?

[alpine nest]

That wouldn't be an equivalent definition of topology, but you can indeed define a topological space in terms of specifying the closed sets.

You still need to request that 0 and X be closed though

[sly geyser]

why would 0 and X be closed?

[alpine nest]

Because their complements need to be open

[sly geyser]

wait I just noticed

are 0 and X both closed and open at the same time?

[alpine nest]

Yep

They're clopen sets

(no joke, that's genuinely the term)

[sly geyser]

now that I think about it

what actually is a "closed" set anyway?

(intuitively)

[alpine nest]

There's probably many intuititons, but mine is "one where taking limits won't take you out of the set"

[lusty trench]

For me, it's “conceivably the zero set of a function”.

[tender halo]

metric moment

[lusty trench]

Metric? No.

More like locally ringed moment.

[tender halo]

thats considreably worse

[lusty trench]

Why? Locally ringed spaces are beautiful.

[tender halo]

closed sets are angular and jagged and opens sets are puffy and soft

closed sets are definetly the Kiki of the two

[alpine nest]

Extremely, but outside metric spaces there's not much intuition to speak of

[lusty trench]

No idea what you mean by “jagged”. My only functions are polynomials and, occasionally, power series.

[alpine nest]

Some people have all the luck

[tender halo]

not all of us are born with a ringed spoon in our mouths

[tender halo]

there are like a million ways to define a topology

[alpine nest]

Well, a topological space

[tender halo]

in terms of open sets, of closed sets, of closure operators, interior operators, boundary operators, limit operators

[lusty trench]

If you're an analyst, then you probably use neighborhood filters.

[tender halo]

neighbourhood system

[alpine nest]

Metric

[tender halo]

weak and strong topologies generated by a family of functions

[grave solstice]

some people in analysis do use locally ringed spaces

[lusty trench]

Do you mean it like “use spaces that happen to be locally ringed?” or like “actually use the structure sheaf and sheaves of modules over it?”

The former is obviously true - who doesn't use manifolds? The latter, I've never seen it, at least...

[grave solstice]

I meant the latter. I am thinking about people like Cartan and Oka. But this is very close to AG anyway

[lusty trench]

Oh, complex geometers.

In my mind, “analysis” means “real analysis”, with all its horrible pathologies, like functions that are everywhere smooth and nowhere analytic.

[fading vale]

If you just want a sheaf theoretic perspective a lot of real people use this too

Sometimes people just call this algebraic analysis, it's a fairly developed theory

[lusty trench]

Ah, cool. I just happen not to know any people who do that.

[grave solstice]

that seems a limited and uncommon (I think) view of analysis, although I suppose it is harmless

[fading vale]

I think it's pretty big in France

Culturally

I was looking at faculty at sorbonne literally today and thinking "why tf do they have so many algebraic analysts "

[grave solstice]

mmh interesting

[white oxide]

project mbappe project groethendieck

[merry geode]

Wow, interesting

Maybe R is a residue field?

[languid patrol]

A very common function space in pde is L^1_loc which is the sheafification of L^1

The product of L^1 loc functions is not L^1 though so it’s not a sheaf of rings

[naive trench]

My friend L^1_loc

[lusty trench]

Now that I read a bit about what algebraic analysis is, well... The professors I took analysis with would simply say that's algebra, period.

[obtuse meteor]

One note about this: these functions, while pathological, do come up in many natural contexts: e.g in dynamics, or in hyperbolic geometry

[fading vale]

Uh that is their prerogative but it exists to solve problems in systems of PDEs

So I think it meets the criteria of the question

[honest shore]

Does not the 2nd condition imply the first one

[winged viper]

is there a reason why you think it does

[honest shore]

It's on wikipedia

😭

[winged viper]

it does not imply the first one

you're asking if second countable implies hausdorff right?

[honest shore]

Yea

[winged viper]

an example is the two point space with the indiscrete topology

like X = two points and the topology is {emptyset, X}

there are also examples which satisfy properties (2) and (3) but not (1)

i think if you keep reading he should mention them

[honest shore]

Ohh its a language issue, Hausdorff space is called a separated space and secondly countable imply that the space is separable

Not the same thing

Thanks I'll see that

[winged viper]

ah i see

yeah it's a bit confusing lol

[honest shore]

In French I mean

Yea

[merry geode]

I hate the notion of "separable space"

[gentle girder]

lol that is interesting

[unreal stratus]

Yeah there are a few differences in terminology in french topology like this oop

a famous thing (ig not just limited to french) is that the definition of compact differs from the English one

[gentle girder]

woah

[radiant holly]

(X,T|X) and (Y,T|Y) two compact space mean f:X--->Y a proper map did ı get right?

[merry geode]

Uh so I cannot dislike it?

[unreal stratus]

I don't think this is true in general unless you assume Y to be Hausdorff

[slate bane]

I don't know if this fits here, as it's not a striclty formal question, but could someone help me with this intuition/visualization?
I think the pinched sphere and pinched torus, as subspaces of R^3, are homeomorphic, as they should both be identifiable with the same quotient of the sphere (two poles identified), but I can't imagine the explicit homeomorphism/deformation that brings one into the other

[umbral panther]

If you pinch a torus by contracting a circle to a point, it is the same as if you pinch a sphere by identifying 2 points. These are two different kinds of pinching

[slate bane]

Yes, so the two subspaces of R^3 I mention are homeomorhpic right?

Maybe, is it possible the map is only possible to be "seen" in R^4 as when I say that two nested circumferences that are tangent in a point in R^2 are homeomorphic to the wedge of two S1, but visually one needs to "lift" the inner one above the outer in the space and then lay it back in the plane?

[wispy veldt]

why would you

[radiant holly]

Let X={1,2} T|X={X,{},{1}, {2},{1,2}} and Y={a,b,c} T|Y={Y,{},{a},{b},{c},{a,b},{a,c},{b,c}}
f(1)=a
f(2)=b be (X,T|X) compact and (Y,T|Y) hausdorff mean f:X---Y proper map. true?

[merry geode]

What does “separated” mean for a space with dense countable subset?

[unreal stratus]

Sure but the sets are finite so it's automatically proper

[radiant holly]

thank you

[final grotto]

hey when is a set closed in the cartesian product X x Y?

[radiant holly]

ı think XxY new a set mean we need to create a new topology

but this is my idea

[gaunt linden]

But there's only one topology that it is "reasonable" to put on a binary product.

Namely the one generated by products of an open subset in X with an open subset in Y.

So a subset of X×Y is closed iff its complement is a (possibly infinite) union of such open rectangles.

[plush folio]

I need some help interpreting the solution to an exercise:

It seems they first show that D^n \ {0} is homotopy equivalent to S^(n-1) and therefore has trivial fundamental group. Then they assert without proof that the fundamental group of D^n is trivial. Therefore the induced homomorphism is an isomorphism, since every homomorphism between trivial groups is an isomorphism

Is this correct?

[quartz horizon]

“a subset is closed iff its complement is open” lol

i mean i get what you’re saying

but it’s also such a non-answer

[gaunt linden]

I did try to see if I could describe it more directly based on closed sets, but didn't quickly find anything.

[quartz horizon]

i see

[gaunt linden]

It's unclear whether the OP wanted a definition or a neater characterization, so I could at least come up with the former.

[radiant holly]

let a function be f:X--->Y, f:X---->Y homeomorphism and at the same time proper map, is it possible?

[unreal stratus]

Every homeomorphism is proper

[radiant holly]

ah ok thanks

[lusty trench]

How many maps between trivial groups do you know that aren't isomorphisms?

[unreal stratus]

yes your reasoning is correct sheddow

It's just a bit ofa weird thing lol, like showing they are isomorphic is a bit of a silly thing when both are 0

D^n is contractible, indeed it is convex and contains origin so you can do H(x,t)=tx

[plush folio]

None, but that's not the confusing part. Like Potato said, it just seemed like an odd approach (or odd exercise?) when both are the trivial group

is it possible to find a homotopy inverse (I don't know the correct terminology, but like an inverse in the category of homotopy equivalences) to the inclusion map i? For example, f : D^n -> D^n \ {0} where f maps 0 to (1, 0, 0, ...) or something?

[lusty trench]

The inclusion D^n \ {0} -> D^n isn't a homotopy equivalence.

But having isomorphic fundamental groups is a waaay weaker condition.

[plush folio]

Aha, that was my suspicion

I see, thanks 👍

[fierce lily]

I have a question about the proof of topologies of Rk is strictly finer than topologies of R. For the second paragraph, we let (a,b) be the basis element of T, and we can let x=1/3. In this case, how can we choose basis element of T'' that contains x=1/3 here?

[lusty trench]

From your proof, I guess R_\ell is the Sorgenfrey line, but what is is R_K?

[white oxide]

given the yapping in the second half, K being the set 1, 1/2 , ... would seem fitting

[unreal stratus]

yes

[fierce lily]

R_k means the set with basis elements of the form (a,b)-K and denote K to be set of all numbers in the form of 1/n while n belongs to Z+

so I can clearly see that (a,b)-K definitely be the subset of (a,b). However, what if x in (a,b) is of the form 1/n like 1/3?

[white oxide]

well, you probably have your old open sets in there as well

otherwise you'd be correct in observing the the sets (a,b) - K fail as a basis for a topology, they never cover any element in K

[fierce lily]

old open sets refer to?

[white oxide]

of the form (a,b)

take the old topology, throw in those (a,b)-K sets and consider the topology they generate

[fierce lily]

Do you mean just consider the set as of the form (a,b)-K?

i mean let old sets be of the form (a,b)-K

[white oxide]

the topology of R_k is generated by open sets (a,b) and (a,b)-K

[fierce lily]

then for each x, it must be in both (a,b) and (a,b)-K, and we can conclude that x is in (a,b)-K subset of (a,b), is that correct?

[white oxide]

I meant and in the natural language sense

sets looking like (a,b) are open in R_k, and sets looking like (a,b)-K are also open in R_k. Due to the former sets this gives you a stricter finer topology (there is no open set of the form (c,d) containing 0 that fits in any (a,b)-K )