point-set-topology

[high hill]

Is my axiom list correct, first of all?

oh ok

idm

but hmmmmmmmm I can't see what I've missed

[gaunt linden]

I'll admit I can't guess what more it is you want either.

[gritty widget]

You can't iterate disconnected -> disjoint union topology infinitely

[high hill]

can you not? 🤔

[gritty widget]

You can't

[gaunt linden]

You can go all the way to sets of the form { x \in X | x is topologically indistinguishable from a } for every a.

[gritty widget]

And then it's obvious

So in short, spaces of which Kolmogorov quotient is a discrete space

T_0

[high hill]

hmmmmmm I still do not get the infinite vs finite

$\bigsqcup_{i\in I} A_i$ where $A_i$ are all indiscrete spaces

[gaunt linden]

I get the impression that you're just trying to get Shuri to say what he has already said, but with particular magic words?

[high hill]

Does this not work for some cardinalities of I? Or are there some spaces I've missed

[gaunt linden]

A topology with a basis that is a partition of X?

[high hill]

ah ok

Well I can certainly see all the spaces I've listed are sigma algebras

as for the converse... that doesn't sound so easy

[gritty widget]

{1/n} u {0}

Does it work with this?

[gaunt linden]

With which topology?

[gritty widget]

You can't just iterate taking {1/n} here to take disjoint union

[gritty widget]

Euclidean

[gaunt linden]

The discrete topology is certainly a sigma-algebra.

With the Euclidean topology it's not a sigma-algebra. {0} is a closed set, but is not open. So not closed under complement.

[gritty widget]

My point is you can't just take finite amount of clopens, then more etc

And get a disjoint union

[gaunt linden]

Sorry, it's not clear to me what it is you're even arguing for.

Can you make a more self-contained statement of your thesis?

[high hill]

Well as for the proof --- all I can really say is that if we cannot write it as a union of indiscretes, we can still write it as a union of connected spaces, at least one of which isn't indiscrete. But a connected space cannot be a sigma algebra (we must have a non-trivial/proper open set whose complement isn't open)
(the open set in the connected space whose complement is not open can be found back in the union)

[gritty widget]

I agree

[gaunt linden]

You can identify your indiscrete summands exactly as the equivalence classes of the "topologically indistinguishable" relation.

[high hill]

I shall have to look up that term 👀

[gaunt linden]

indistinguishable?

[high hill]

yes (I uh... haven't really taken topology properly properly yet)

[gaunt linden]

Two points are topologically indistinguishable iff they are in exactly the same open sets.

[high hill]

Ah I should know this, but probably forgot

[gritty widget]

Nah

I cant think of one right now

but ill let you know later today if I can find one

[high hill]

no no - we got this as the answer

what was missing was my argument was finite ig

[gritty widget]

?

What was finite?

[high hill]

What I wrote as my argument about 'splitting' up disconnected spaces

The answer is indeed this

[gritty widget]

Oh I thought you said it could be an arbitrary disjoint union

I tried scrolling up but cant find easily.

[high hill]

By arbritrary disjoint union I was emphasizing a disjoint union of any cardinality

[gritty widget]

Yeah.

[gaunt linden]

So the topologies that are also sigma-algebras are exactly those that can be produced by picking an arbitrary equivalence relation and declare its equivalence classes to be the basis for a topology.

[high hill]

Ah I finally clocked it 👌

this was pretty cool

[high hill]

https://i.imgur.com/nxKnS85.png

If we have all of these, do we have a metrisable space? Or not quite

[gritty widget]

@high hill no

But if you assume T_3 and countable basis, then you do

This is called Urysohn metrization theorem

[gritty widget]

This isnt obvious

and ive seen the proof

Just embedd X in Hilbert cube and dab away

[cursive vigil]

Just started topology

What treatises do you human beings recommend? I just started using topology without tears

[marsh forge]

Hatchers notes

[rancid umbra]

munkres

[cursive vigil]

I tried that but idk i dont feel like im ready for that yet

[plain raven]

did you read the set theory chapter

or did you jump ahead to the second chapter

[cursive vigil]

The set theory

[marsh forge]

Skip set theory tbh

[gritty widget]

Ah yes. The Cantor way

And then you imagine the set of all sets and your head explodes

There's couple important identities in topology imo

$f(A\cap f^{-1}(B)) = f(A)\cap B$ is one of them

[gentle osprey]

Blitz

[uncut surge]

Isn't f(f^{-1}(B)) = B only true for uuuuh surjective f

Yeah surjective

[gritty widget]

This holds for all B iff f is surjective

My formula holds for all f

[uncut surge]

i see

that's a cool hack

[marsh forge]

I have very few such formulas memorized

[gritty widget]

There's not a lot of them

Another one is $f\left(\bigcup_i U_i\right) = \bigcup_i f(U_i)$

[gentle osprey]

Blitz

[gritty widget]

Also useful

[marsh forge]

I mean that one is just obvious

I wouldn’t even call that a formula

[gritty widget]

Hmm... sure, I have it memorized by heart though

[marsh forge]

?

[uncut surge]

well it's nice to remember that maps of any kind interact well with intersections and unions

the proofs are obvious after like a minute of inspection but it's nice to not to have to put that minute of work in

[marsh forge]

They don’t interact well with intersections

[uncut surge]

uh oh

[gritty widget]

Yeah, only an inclusion

[marsh forge]

It’s not even a proof in the union case tho lol

[uncut surge]

life is harder than i hoped

[marsh forge]

You’re applying f to the same elements either way

[gritty widget]

That's why there are ways to circumvent intersections, such as finite fibers

Then in some special cases you can still extract something like that

[uncut surge]

mathematics is horrible

[gritty widget]

I'm thinking of some compactness thing

Too specific probably

[gritty widget]

Are there a non-trivial continuum C such that there is a point x in C which makes C\{x} totally disconnected?

[gritty widget]

https://mathoverflow.net/a/16630/150060

Nope

[sleek thicket]

Set theory is trivial because you can just pick a particular model and check what happens in it

[patent gorge]

ayoooooooooo

[viral yoke]

Let X be a topological space, let A be compact in X, and let U be open in X. Prove/disprove A\U is compact.

The specific problem I have here is that X is a metric space but a topological space should be sufficient

[rancid umbra]

i don’t think this property always holds in a general setting. it holds if your space is hausdorff tho

i’d like to see a ce if anybody has one. can’t think of one immediately

[rancid umbra]

anyway, do you need help proving this in the case of a metric space?

[gritty widget]

Closed subspace of a compact space is always compact

[rancid umbra]

dang. compact subspaces of hausdorff spaces are closed. that’s what i was thinking

my bad

[gritty widget]

Yeah

@viral yoke pinging you just so you know

It's enough

A\U is closed in A

[sleek thicket]

no blitz is right

[gritty widget]

Nothing

[sleek thicket]

A\U is the intersection of X\U and A, so it's closed in the subspace top.

[gritty widget]

Proof still holds

[rancid umbra]

let C be an open cover of A\U. then C union {U} is an open cover of A. it admits a finite subcover {C1,…,Cn} which covers A, hence covers A\U

i was just being lazy ig and didn’t want to use open covers.

no need for any fancy theorems

i’m such a noob

[gritty widget]

No need for reproving fancy theorems either

[viral yoke]

Being a metric space is nothing special either

[viral yoke]

I forgot A\U is in A

That was the line I needed

Thank you

[kind depot]

hey yall i was trying to find all the two and three degree covers of RP2 v S1. for 2 sheeted i found two covers and for 3 sheeted i could not find any. i dont think any exist of degree 3 but am not sure how to prove it. any ideas?

My two sheeted covers

[kind depot]

?

[hasty pasture]

I have a space that is k circles all attached at opposite ends, so think of it as a figure 8 but with k circles instead of 2. Is this the same as the Wedge product of k copies of S^1?

the wedge product usually identifies all spaces at a single point, but in this case we have a long chain of them

is it possible to create a homeomorphism between these spaces?

[sleek thicket]

It's homotopy equivalent but they're not homeomorphic

[hasty pasture]

ok, but then they would have the same fundamental group?

[sleek thicket]

Yep

[hasty pasture]

ok cool

[sleek thicket]

I'm about to explain why they're not homeomorphic, sorry

The wedge sum has a point p such that X \ p has 8 connected components

Right?

[hasty pasture]

yeah

[kind depot]

any answers to covering spaces of RP2 v S1 btw? i cant find any of degree 3

[sleek thicket]

But this isn't true for the other

[hasty pasture]

im trying to visualize why, is it because its all path connected?

[sleek thicket]

No

You can only get two components at most

There's two types of points. Points where adjacent points meet and points where they don't

The first type of points will disconnect the figure eight into two components

Right?

[hasty pasture]

ok

[sleek thicket]

If you remove the point connecting the ith and (i+1)th circles, you'll get the component {jth circle where j <= i} and {jth circle where j > i}

And the second type of points will still be path connected if you remove them

So this shows they're not homeomorphic

[hasty pasture]

ooh i think i get it, so if you remove it in the "flower-like" configuration, you get separate components (petals), but if you remove one point in the straight line configuration it only divides it into two

is that it?

[sleek thicket]

Yep, exactly!

And a homeomorphism would preserve this property

[hasty pasture]

ok gotcha

what kinda homotopy should i use btw? like would i just stretch the connection points on each circle to eachother inductively?

[sleek thicket]

Yeah

It's kind of ass to write down

[hasty pasture]

sounds like it, might just handwave and hope its okay 😬

[sleek thicket]

One way to think of it is as contracting the interval between the two endpoints to 0

(inductively)

[hasty pasture]

ah yes that makes sense

[sleek thicket]

That might be okay to write down

[hasty pasture]

cause the arcs are contractible, on their own

so it should be ok

thanks for the help!

[sleek thicket]

Np!

[kind depot]

for classifying covering spaces can any of yall expalin how they are equal up to conjuugacy? im a little confused on that point

[hasty pasture]

do you mean with regards to the galois correspondence for subgroups of $\pi_1$ and covering spaces?

[gentle osprey]

llspacebarll

[kind depot]

ye

im trying to prove there are no covers of degreee 3

and its to do with order of a subgorup of gthe freegorup

[empty grove]

There are 3 sheeted covers

[kind depot]

uhh

oh

[empty grove]

Take the 3 sheeted cover of S^1

[kind depot]

mm oh wait

duh idk why that didnt occur to me

[empty grove]

And at each of the 3 evenly spaced points put a copy of RP2

[kind depot]

wait

uhhh

oh ic

[empty grove]

What did you have in mind

[kind depot]

and i can build any n sheeted covering using that

[empty grove]

Yes

[kind depot]

well i couldnt figure anything out but i was looking at covers of S1 v S1 for inspiration

like the inscirbed triangle in a circle

[empty grove]

Ah

[kind depot]

are there any else tho? like i guess this makes sense as the subgroup (ab)^n of Z cross Z/2

[empty grove]

I think you might be able to say that the restriction of the cover of the wedge to either of the 2 spaces that are being wedged should be a covering map on its own?

[kind depot]

uhhhh elbaorate

not sur what u mena resticitin of the cover to the wedge

[empty grove]

So you have S^1 sitting inside this wedge

Suppose you have a covering map p to this wedge

[kind depot]

mmm

[empty grove]

Then p: p^-1(S^1) → S^1, the restriction of the covering map, is this guaranteed to be a covering map?

[kind depot]

uhh on the wedge $$S^1 \mathbb{R}P^2$$? Uhhh i think so

[gentle osprey]

Optimism

[empty grove]

If it is, then we know what it must be because S^1 has exactly 1 3 sheeted cover, and then we just need to look at how RP2 is attached

[kind depot]

ohhh

ok

so in terms of subgorups

the circles + projective concatenations is like (ab)^n whereas this is id x b or a x id right?

[empty grove]

Hmm yeah I think so

[kind depot]

actually i dont think sp

does it not have ot be up to conjugacy

cause i was trying this but like the parity forces an even number of both

unless u have an explicite image in mind

[empty grove]

Yeah I'm not sure lol

[kind depot]

ye alg top makes my brain hurt

lol reading hatcher is like a nightmare

wait can u draw out ur cover

lol i dont see how that works

it still requires even parity of cricles

which becomes degree 4

nvm

i just got it

[gritty widget]

people usually don't like Hatcher, but some consider it a good first exposition

[rancid umbra]

what’s the knock on hatcher?

[gritty widget]

knock on hatcher?

what does that mean

[rancid umbra]

why don’t people usually like hatcher?

[gritty widget]

because his proofs is geometry and intuition

[rancid umbra]

oh. so he just leaves it to the reader to fill in the blanks?

[gritty widget]

and no normal human being is going to fill in that much details, or so I heard

[kind depot]

100% agree

the books is like impossible to read i basically learn from online lectures tbh

[gritty widget]

yeah, but he leaves too much for the reader and intuition and it feels wrong

[kind depot]

the issue is that by leaving out most details he goes really fast so its been an issue considering my prof thinks thats a good pace

what are homotopy classes of spaces btw? we literally never talked abt them and i cant really find it in hatcher prior to homology

[gritty widget]

of spaces?

I usually hear this with regard to maps

[kind depot]

maps

yup

ik waht it is

but i acnt find any way to compute it

[gritty widget]

I think there is not much you can do unless it's something more specific

[kind depot]

mm ok im specifically looking at [RP2, S1] which i think is H1 (RP2) x Z =~ Z/2 x Z but not sure how to show its the same as homology classes

[jolly garden]

https://www.youtube.com/watch?v=o1kDNbnX-ps&ab_channel=VisualTopology%26Geometry

Yeah, what can you say about it? Does anyone know? (It's in the video I just posted)

[golden gust]

one homotopy class?

[hot night]

What is a fiber bundle intuitively?

[empty grove]

It is a topological space E along with a continuous map E → B, such that the inverse image of each point of B looks like a fixed space F

The point is that if we understand B and F then we can use that to understand E

[gritty widget]

intuitively a fiber bundle looks like a product

you have some base space

and at each point you have a fibre attached

R^2 is a an R1 bundle over R1

a cyclinder set is a line bundle over a circle

it can also be viewed as a circle bundle over a line

a torus is a circle bundle over a circle

these examples so far are all “trivial” fiber bundles

because they really are globally product

a non trivial example is a möbius band

if you imagine a möbius band

if you looked at a small portion of it

it looks like a cyclinder

but globally

there is a twist

[golden gust]

so a mobius band is a line bundle over a circle, but not a circle bundle over a line?

[gritty widget]

yes a möbius band is a line bundle over a circle

but it is not a circle bundle over a line

so you can see that fiber bundles are like a twisted product

but the twisting only happens in the fibres

[gritty widget]

oh sorry

i think i might be wrong

i am wrong

i get confused with this example because

[empty grove]

I can't see a continuous projection onto a line though

[gritty widget]

as a circle bundle over the interval

it’s not a *vector * bundle

i think it’s

picture the möbius band as rectangle before glueing

the fibre over the point x on the interval

is the line from x to 1-x

[empty grove]

Aren't these intersecting

[gritty widget]

i think? this is okay

[empty grove]

But fibres can't intersect right

Because then the intersection is mapping to 2 different points

[cursive vigil]

Is a möbius tetrahedron a tetrahedron that has its cohomology unaffected by its cofibers and kernels?

[gaunt linden]

Google just seems to find this concept, which is projective geometry rather than topology: https://mathworld.wolfram.com/MoebiusTetrahedra.html

[gritty widget]

this is true for vector bundles

but it’s it true for fiber bundles

https://en.m.wikipedia.org/wiki/I-bundle

i thought the möbius band was not a circle bundle over the interval

but this seems to suggest otherwise

oh shit

never mind

[gritty widget]

damn you Moldi

you shook my confidence

[cursive vigil]

I literally made up that sentence on the spot I never knew that a möbius tetrahedron actually existed lol

[fading vale]

[modest nexus]

Hi! I'm kinda new in alg top, and I have the following problem: if X and Y are connected CW complexes, then homotopy type of X wedge Y does not depend on the point where we're gluing the spaces. Can someone give me a hint?

[glossy pine]

lmao legend

[cursive vigil]

Lol

[lunar yoke]

Note that you can view a path as a homotopy

[modest nexus]

do you mean I can take a homotopy of paths?

[lunar yoke]

no i mean a path itself is a homotopy

[modest nexus]

how?

a path is a map from I to X

where I = [0, 1]

[lunar yoke]

yes

so if it starts at x you can see it as a homotopy {x} x I -> X

with homotopy starting at inclusion map

and ending at map x -> endpoint of path

[modest nexus]

hmm, I guess I got it, but correct me if I'm wrong: so I handle my gamma: I -> X path as H : {x0} x I -> X homotopy, where x0 is the starting point gamma(0), H_0 (x) = x0, H_1(x) = endpoint = gamma(1)

[cursive vigil]

lol

[lunar yoke]

yes, you just define H(x,t) = gamma(t)

but im kind of sceptical now if you have the tools to solve this question

maybe there is some easier proof idk

but do you know about homotopy extension

[modest nexus]

yes

[lunar yoke]

oh alright

good luck then

[modest nexus]

thanks 😄

[empty grove]

Wait so was it or was it not?

[hasty pasture]

Hey, im trying to apply the seifert van kampen theorem to a space constructed from chains of of circles all connected to a larger circle in the centre. I decomposed it into the individual "arms" of the space where the interesection is just the circle in the centre. My question is, is the kernel generated by the inclusion mappings trivial?

the inclusion of a map on the inner circle to any of the arms seems to just give a path in the circle like usual regardless of which arm it is included into, so i am thinking that any of these generators will always give you the constant loop.

i can share a picture if needed

Wait, I just realized my decomposition is probably wrong since the sets are not open

[fading vale]

@hasty pasture If you send a picture I can try to help

Also if you're working with circles and by each "arm" you mean the union of these circles or whatever you can generally thicken them a little bit to make them open and then deformation retract onto the arms so that you can still describe it in terms of pi_1(S^1)

The same thing you do to compute pi_1(S^1 wedge S^1)

[hasty pasture]

This is the space for k=3

the arms are at the k-th roots of unity

and there are k copies of S^1 in each arm

as well as k arms

so my thought was to decompose the space as each arm union the circle in the middle, now im thinking about how to get that open

and furthermore is the kernel trivial?

@fading vale

[fading vale]

Oh i see

Okay yes your idea is correct, what you can do is basically "thicken" the center circle and the arms a bit

[hasty pasture]

ok, and then use a deformation retract like you said?

[fading vale]

Its exactly how you compute pi_1 of the wedge like i said before

Yes

And then the inclusion will be trivial for the same reason that the inclusion of each copy of S^1 into the wedge is too

[hasty pasture]

ok, and i assume thats because this space is very similar to a wedge product?

it almost seems like it should be homotopic to one

since we can contract each arc to a point

[fading vale]

Pretty much yes, if I write the arms as like A1, A2, A3, and the center circle as B

then the existence of nbhds around A1 and B in A1 wedge B for example that deformation retract imply that pi_1 of their wedge sum is their free product

This is a general fact for spaces X and Y with this kind of retraction property

[hasty pasture]

ok

[fading vale]

So technically speaking what you do is apply this three times with A1 wedge (A2 wedge (A3 wedge B)))

And turn all the wedges into free products

[hasty pasture]

ah makes sense yeah

[fading vale]

and then u already know pi_1(A_i) and pi_1(B)

So ur done

[hasty pasture]

ok, so it ends up being something like $ZZ...Z$ $k^2+1$ times?

[gentle osprey]

llspacebarll

[fading vale]

Is it k circles on each arm?

[hasty pasture]

yeah

[fading vale]

Ok yea

That should be right

[hasty pasture]

cool, thanks so much

i feel like i could just make an argument with the descending wedges instead of all this deformation retract stuff

[fading vale]

Also I think you're right that its homotopy equivalent to the standard wedge of k^2 + 1

But showing that formally would be much more annoying

Lol

[hasty pasture]

indeed

[marsh forge]

You can think of this space as a graph and find a maximal spanning tree for it, then contract the tree to get this equivalence

@hasty pasture

[hasty pasture]

oh i think i remember something like that in hatcher

right after van kampen in fact

[lunar yoke]

yes its used to show that the fundamental group of a connected graph is always free

[fading vale]

model categories

[fading vale]

also dont do this

Wasting peoples time trying to explain things to you

[kind depot]

cna yall esxplain how this implies that this is isomorphic to Finfty. Im trying to show $$[F_2, F_2] \approx F_\infty$$

[gentle osprey]

Optimism

[cursive vigil]

Well I learned something from it, so was it really a waste

[fading vale]

Its at minimum dishonest

[cursive vigil]

Dishonest if I said I knew what it was

[surreal ocean]

I just learned the definition of the support of a function. At first i thought it was a fancy term for the points which get mapped to a non-zero value but then i noticed the support is the closure of all points mapped to nonzero values. If Im not mistaken doesnt this allow for the possibility of there being a point in the support of a function which gets mapped to zero? If so, what idea are we trying to capture by defining the support of a function in this way?

[flint cove]

that's just a technicality, and the one justification I know is that you need to go to the closure anyway if you want to talk about compact support

[flint cove]

but you are correct in your analysis, eg a bump function ℝ→ℝ with supp = [-1, 1] is actually zero in -1 and 1

although I do wish there was a name for just f⁻¹((0, ∞))

[cobalt sonnet]

I guess it is useful that the complement of the support is open

So that, for example, $\text{supp}(f') \subset \text{supp}(f)$.

[gentle osprey]

IlIIllIIIlllIIIIllll

[cobalt sonnet]

Also it allows extension of the definition of support to tempered distributions as the complement of the largest open set on which the distribution vanishes

[surreal ocean]

ah i see that makes sense thanks

[jolly garden]

What's the idea behind long exact sequences? I suppose you somehow encode information in it, splitting the computation into easier pieces. I just encountered them when learning about relative homotopy groups. Given my interpretation is correct, how do they encode information?

[gritty widget]

You might also want to ask in #groups-rings-fields

[jolly garden]

Thank you for the advice. Fortunately, I've already come to terms with it.

[gritty widget]

the möbius band is not a circle bundle over the interval

[sleek thicket]

related: any bundle over the interval is actually just a product space

it's not too hard to prove

Wait, you may need more structure I'm not thinking of

Like the transition maps should be given by the action by a topological group or something

[marsh forge]

The idea is generally to break up the computation into an LES where you understand or already know a bunch of the groups in the sequence, as well as hopefully some of the maps

This allows you to better leverage computations you already know to do new ones

[gritty widget]

nice to see you pfp is still going

[sleek thicket]

I still cherish it <3

[tawny juniper]

This isn't so much of a desire for help as it is sheer curiosity.

I have just been introduced the fact that for n-dimensional euclidean space where n ≥ 5, there are only 3 regular convex polytopes.

WTF is going on that makes this true?

What's the culprit?

I should preface by saying that I am just a lowly analyst, so don't go too hard on me. xD

[cedar pebble]

one way to view this is in terms of finite Coxeter groups, and then the classification follows from the classification of these groups

such groups are classified by their Coxeter diagrams, and the point is that there are a few exceptional diagrams (H_3, H_4, and F_4) that give you the exceptional examples in dimensions 3 and 4

(from Vogan's slides: https://math.mit.edu/~dav/regpolyHO.pdf)

[tawny juniper]

Skimming, it reminds me of the argument about possible tessellations of R^2.

[cedar pebble]

"most of linear algebra" and "rest of linear algebra"

based

yeah exactly

the classification of regular polyhedra in R^3 is just like

[tawny juniper]

So, the issue is that there are too many possible symmetries in higher dimensions.

[cedar pebble]

you try all the regular faces you can have and you get everything

yeah

the reason why Coxeter groups are a good way to look at this is like

if you have some regular polyhedron in R^n, it has to be awfully symmetric

[tawny juniper]

This has to do with root systems and all that jazz.

[cedar pebble]

and not just by any group action, by like rotations and reflections across hyperplanes

yeah

[tawny juniper]

What happens if you remove the convexity requirement?

[cedar pebble]

then the stuff with Coxeter groups lets you classify these kinds of reflection groups in Euclidean space in a nice combinatorial way

oh hmmm

definitely becomes harder

I feel like I've seen the classification of non-convex regular polyhedra (or some other weakening of the definition) in R^3 and it's a huge mess

certainly you lose being able to attack the problem really cleanly with Coxeter groups's there's a lot of other junk you have to deal with

[tawny juniper]

Would this subject area be "geometry"?

(like, without any modifying adjectives xD)

I've always felt that pure geometric questions like these are really, really beautiful. I love it when (initially) unexpected phenomena like these emerge from simple objects.

[cedar pebble]

yeah this is combinatorial geometry

[bronze lake]

I feel like I'm losing my mind. Anyone familiar with knot theory, someone told me:

According to Fox's Quick trip through knot theory, one correct presentation for the 2-twist spun trefoil is <x,y | x*y*y*x^-1*y^-1, y*y*x*y^-1*x^-1 >

But I've spent like a half hour scrolling through Fox's 'A Quick Trip through Knot Theory', and I can't for the life of me find that presentation

and the only other presentation I've found in another source is different

[sonic hill]

What space do you get when you attach the boundary of D^2 to S^1 by a triple covering map?

[marsh forge]

its not like a specifically named space

I would just call this the cofiber of 3

[sonic hill]

Damn, so it's not something interesting?

[marsh forge]

i.e. the cofiber of the map $S^1\xrightarrow{3} S^1$

[gentle osprey]

ShamFan1999

[marsh forge]

I mean

this space is interesting

but this requires some explanation

it plays a "height 0" role of chromatic homotopy theory at the prime 3

[sonic hill]

Damn interesting

What's the mapping cone of the 4-fold cover? Since 4 isn't prime...

[marsh forge]

less interesting, i guess

[sonic hill]

Aw

[marsh forge]

This is effectively just like analogous to asking what the outcome of taking Z and killing 3Z is

it's Z/3Z. You can think that is intrinsically interesting or not, depending

[sonic hill]

Z/3Z is a basic fundamental object, so is Z/4Z

but is the cofiber of n-fold coverings useful?

[marsh forge]

I'm not sure what useful would mean, really

[sonic hill]

Maybe like, used in some important construction, an important example of something, an important counterexample of something, etc

[marsh forge]

it's a very elementary example of killing a generator in homotopy, but its only really used in constructions when you kill a prime.

[sonic hill]

I see

[mint sage]

How do we use that expression

[rancid umbra]

@mint sage lim f(x) --> f(a) as x --> a^+ means that for each e > 0 there exists a d > 0 such that if 0 <= x - a < d, then |f(x) - f(a)| < e

[mint sage]

But how to prove

[rancid umbra]

let U be an open subset of R. you want to show that V = f^{-1}[U] is open.
if V is empty, then we are done. otherwise, let a in V. then f(a) is in U. U is open, so there is an e > 0 such that
(f(a) - e, f(a) + e) is a subset of U

[rancid umbra]

apply this definition now and see if you can conclude that V is open

[mint sage]

oh

thx

[cursive flume]

does there exist any principal S^n,n even bundle?

[sinful seal]

Let $X$ be a topological space and $f, g: X \to X$. Do $f\circ g$ and $g\circ f$ being homotopic to the identity map implies $f$ homotopic to $g$?

[gentle osprey]

Tₑ𝘖(n)

[marsh forge]

no

consider S^1->S^1 given by the +1 and -1 map

In particular if [X,X] has a nontrivial group structure then you can always just take any map and its inverse

[sinful seal]

But in this case the composition of these maps wouldn't be -1 that isn't homotopic to id?

[marsh forge]

we are talking about the additive group not the multiplicative structure

Oh wait I am the one being dumb here.

Yes okay use nobody's example. My bad.

[gilded crane]

say i have topolgical spaces (X,tau) , (Y,pi), with A being a subset of X and B being a subset of Y. the interior of A and B, int(A) and int(B) are both open in X and Y respectively, but why is int(A) x int(B) open in X x Y (with product topology)? i dont fully understand the product topology definition, i feel

[plain raven]

Maybe you should post the definition of the product topology

from the source you're working with

[gilded crane]

its similar to munkres definition

[marsh forge]

Hint: try writing these interiors as unions of base sets and then apply the above stuff

[rancid umbra]

saw somebody state this on stack exchange without proof or source of proof. is this true or false:
if X is a compact metric space and f is a continuous map on X, then there is a set A such that f(A) = A

[empty grove]

Should be true by starting with A_0 = X, A_{i+1} = f(A_i), and A = intersection of A_i's

[rancid umbra]

i see. didn’t think of this. thanks

[gritty widget]

From compactness, if X is non-empty then so is A

[coarse night]

I like leaf's answer better

[wise walrus]

I'm doing intro algebraic topology. What does it mean to "provide an explicit homotopy"? I presume it's wanting me to give a function on the unit square to the top. space defining a continuous interpolation between two (presumably homotopic) loops?

am i right with that?

[plain raven]

Yes.

[plain raven]

Did you solve this? It follows like, immediately from the definition, you just have to understand what the definition says

Take A to be {0,1}, with X_0 = (X,tau), X_1 = (Y,pi)

so the basis for the topology on X x Y is given by all U x V where U is open in X and V is open in Y

taking U = int A, V = int B gives the result

[quartz edge]

I'm looking at may's explanation of the fundamental groupoid and skeletons of categories, etc

he defines a groupoid to be a category in which every morphism is an isomorphism

he then says a group can be thought of as a groupoid with only one object

this doesn't make sense to me. i don't really see what only having automorphisms is useful for here

because the homset on a group to itself is the endomorphism set

not aut(G)

in what sense exactly is a group a groupoid with one object?

[plain raven]

He's saying that you think of the elements of the group as being automorphisms of some fixed abstract object

[plain raven]

right, so here's your miscommunication. he's not saying that given a group G we consider the full subcategory of Grp consisting of the object G, which would just be the endomorphisms of G. What he's saying is to any group G we can associate a category C where Obj(C) = { * }, a one-object set, and Hom(*,*) = G, with composition given in the obvious way

conversely for any one-object groupoid, Hom(*,*) is a group under composition

[gilded crane]

Yep, this is what I got to in the end, it was just a case of me not understanding the definition at all

[quartz edge]

huh. ok, like group actions on it

this actually makes sense considering a group action of G on A is induced by a map G -> Aut(A)

[plain raven]

Yeah, right. And like, pretend that this is more specifically an isomorphism

[quartz edge]

ye

wait, do we need to?

the elements of Aut(A) seem to be a superset of the elements of the homset for the groupoid

if i'm thinking about this right

may not be, just woke up

[plain raven]

Ok. Let C be a category and c an object in C.

Do you have a concept of an "automorphism" of c in this setting?

[quartz edge]

ahhhh right. we are dealing with abstract objects

not sets

oh

i see what you mean

[quartz edge]

there's the identity morphism

but that's all that's guaranteed to be there

[plain raven]

Would you let me call an "automorphism" any isomorphism from c to c?

does that sound reasonable to you?

[quartz edge]

yeah

[plain raven]

Like an automorphism of a vector space V is just an isomorphism V \cong V.

[quartz edge]

indeed

[plain raven]

So.. the automorphisms of c just consist of all isomorphisms c \cong c.

You can prove that this is closed under composition and each one has an inverse.

And the identity morphism id_c is definitely part of this family.

[quartz edge]

your explanation Hom(*, *) = G seems quite good. we don't consider the endomorphisms on G here in the groupoid do we?

[plain raven]

So we have a group of automorphisms of c.

[quartz edge]

like homomorphisms between G and some other group H ought to be encoded as functors?

in the groupoids corresponding

[plain raven]

Yes, from this perspective, if a one-object groupoid is a group, then a group homomorphism is a functor between categories.

[quartz edge]

got it

thanks

[plain raven]

One cool puzzle.

Ok. So we've agreed that if C is a one-object groupoid, then we can define G = Hom(*,*).

Then group homomorphisms G -> G correspond to functors from C -> C.

[gritty widget]

Why is there a talk on category theory here. How's this relevant to topology

[plain raven]

Let phi, psi : G -> G be regarded as endofunctors on a one-object category. Under what circumstances are these two functors naturally isomorphic? (In group-theoretic language)

Which homomorphisms G -> G, regarded as functors C -> C, are naturally isomorphic to the identity?

[gritty widget]

There doesn't even seem to be any topology motivation behind your question @quartz edge

[quartz edge]

ah yeah sorry. i was super tired last night and the question came from an algtop text

[gritty widget]

Oh, okay.

[peak crystal]

How to show that unit circle in $\mathbb{R}^2$ is not homeomorphic to [0, 1)?

[gentle osprey]

Giyu

[dim meadow]

Hint: think of removing points

(Alternatively you can do a compactness argument)

[peak crystal]

Yes if i remove any point from $S^{\perp}$ then it is connect but in interval if i remove only o then it is connected

[gentle osprey]

Giyu

[dim meadow]

Yes

[peak crystal]

So by this how do i claim that it is not homeomorphic

[dim meadow]

So suppose there was a homeomorphism

Remove a non 0 point from the interval and remove the image of that point from the circle

The restriction of the homeomorphism is still a homeomorphism

But being connected is a homeomorphism invariant

[peak crystal]

Alright got it i try from circle to interval then i remove point from circle and take image of that removing point is zero so it is not working i need to do in reverse

[dim meadow]

Yeah, or just take a point on the circle who's image is not 0

There are many of them

[peak crystal]

Yes got it, thank you.

[unreal stratus]

Another way is that e.g. the sequence 1/2,2/3,3/4,... has no convergent subsequence in [0,1) but its image in S^1 under any map will have one, in particular if it's a homeo we can pass to that subsequence and go back to [0,1) to get a convergent subsequence of the original sequence

Basically using (sequential) compactness

[peak crystal]

But i didn’t read compactness till now so

[unreal stratus]

Fair enough

[peak crystal]

How to show that any space is seperable or not i.e. they have countable dense set

[dim meadow]

What do you mean by that?

That's a bit of a broad question

[peak crystal]

I mean like l_1 is seperable how to construct dense set?

$l_1$

[dim meadow]

So you want to make a countable set with which you can approximate any other sequence with bounded absolute value sum

Usually for this you use rational approximations for reals

And you use finite sequences as an approximation of infinite sequences

Stuff like that

[peak crystal]

This is boring haha

[unreal stratus]

it can be kinda cute constructing such an approximation sometimes though imo

like it tests understanding too a bit

[peak crystal]

For construction of approximations i need measure?

[dim meadow]

You need to look at the metric

[unreal stratus]

this has nothing to do with measure rly (unless you meant L^1)

I knew somebody would say smth like that kek

[peak crystal]

Okay then i need read this deeply

[unreal stratus]

Just to clarify, you do mean the set of absolutely convergent sequences right

[peak crystal]

No there is space of sequence small ell

[unreal stratus]

ok sure

[peak crystal]

Yes

Another question: is there any Cauchy sequence which is not convergent anywhere, i mean $(a_n) \in X$ but not convergent in X so is it necessary that $a_n$ convergent in some other space?

[gentle osprey]

Giyu

[dim meadow]

This is the definition of the completion of a metric space

[peak crystal]

a_n is Cauchy

[dim meadow]

Essentially

[peak crystal]

Here X is not complete

[dim meadow]

The completion of X is the smallest space which contains X in which all of the Cauchy sequences in X are convergent in the completion

[peak crystal]

Oh got it any space X have completion right so any Cauchy sequence converges in some other space

[dim meadow]

Yeah

[peak crystal]

Thanks

[dim meadow]

You can even make any individual Cauchy sequence converge by throwing in the point it should converge to

[peak crystal]

Didn’t understand

throwing in the point is not clear

[dim meadow]

So if you have a particular Cauchy sequence you want to converge you can add a point to your space and extend your metric to that point

[peak crystal]

Yes

How to relate sequentially closed set and closed set

I mean S is sequentially closed set then for any x in S there is sequence which converge to x but i think it is also true for closed set?

[dim meadow]

Yes so there is an implication in one direction (closed implies sequentially closed)

[peak crystal]

So in general other one is not true right how to distinguish them

[dim meadow]

And in a lot of nice spaces we have the converse

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

[peak crystal]

If X is metrizable and S is subset of X then is it iff?

[dim meadow]

Okay that article is probably not that helpful actually

[dim meadow]

Yes

Here's a nice class of spaces with that property

https://en.m.wikipedia.org/wiki/First-countable_space

[gritty widget]

Or just use https://en.wikipedia.org/wiki/Kuratowski_embedding

To obtain completion of X

[peak crystal]

Thank you so much

Is there relation between first countable and second countable?

[dim meadow]

Yes first countable is much weaker

[peak crystal]

weaker mean more restrictions right

[gritty widget]

No, less

[peak crystal]

So second countable imply first countable?

[dim meadow]

Yes

[peak crystal]

I have one doubt Lower limit topology is second countable?

I don’t think so

[dim meadow]

Lower limit topology on what space?

[peak crystal]

R

Real number

[dim meadow]

Yeah it is not

Maybe you should try proving that

[peak crystal]

Yes i proved this one and $\mathbb{R}_l$ lower limit topology is first countable by setting $\left [x, x+1/n \right)$ so this one is first countable isn’t?

[gritty widget]

[x, x+1/n)

[gentle osprey]

Giyu

[peak crystal]

Ohh sorry I’m little bit confused there is space which is not second countable but it is first countable,

[gritty widget]

Those are common even among metric spaces

Any metric space is first countable, but there is a lot of metric spaces which are not separable

[peak crystal]

$\ell_3$ is first countable but not separable

[gentle osprey]

Giyu

[gritty widget]

Usually the space of continuous bounded functions over a non-compact metric space is not separable iirc

l^infty is not separable

[peak crystal]

So if we imposed one more restrictions that totally bounded imply separable

[gritty widget]

Well, such space would contain a copy of l^infty I think

[peak crystal]

Oh yes sorry

[gritty widget]

Totally bounded implies separable, yes

[peak crystal]

So then we can claim that l^infty is not totally bounded

[gritty widget]

Yes

[peak crystal]

How to prove that l^infty is not separable

We need to find a point and then claim that there is no sequence which convergence to that point

[gritty widget]

Some kind of diagonal argument probably

[peak crystal]

Okay thank you.

[gritty widget]

There is uncountably many pairwise disjoint open sets in l^infty @peak crystal

Consider x in {0, 1}^infty and B(x, 1/2)

[peak crystal]

Not understand

[gritty widget]

Any sequence of zeros and ones and a ball in l^infty centered at it and with radius 1/2

[peak crystal]

I mean I’m not getting your point, by this what we conclude?

[gritty widget]

If l^infty were separable then there would be dense countable set

So any non-empty open set has element of this set

If we find uncountable family of pairwise disjoint open sets, then we need to have uncountably many points in our dense set

[peak crystal]

Ohh got it then its not remain countable

[bronze lake]

Any knot theorists here familiar with 2-knots? Different sources are giving me different presentations for the two-twist spun trefoil, and idk which is right

[cursive vigil]

Lol

I’m not familiar with knot theory

[marsh forge]

#chill

[cursive vigil]

Rate my pun from -i to ij

[marsh forge]

Have you confirmed these groups are actually different

[arctic relic]

Stupid question incoming

[bronze lake]

To the best of my ability, yeah. I'm not the best with working with group presentations though

[marsh forge]

Hm, not sure then.

[arctic relic]

It’s obvious that H_n(empty set) = 0 for singular homology for all n but i was asked to verify this. So I defined the chain complex in the usual way but the map phi: n simplex to empty set doesn’t appear to exist.

[marsh forge]

You should be able to compute this yourself

Or if you draw it for me I could figure it out

I don’t know what this knot is tho

[plain raven]

yeah so the chain complex is the free abelian group on the empty set in dim geq 1

[arctic relic]

So I don’t define the chain complex starting from n

[plain raven]

what?

[arctic relic]

There are no n simplixes that fit into empty set right?

[marsh forge]

Well

The chain complex is just 0

[arctic relic]

So the chain complex goes from 0 to -1

Ok thanks

[plain raven]

Yeah. the set of singular n simplices in the empty space is the empty set

and the free abelian group on the empty set is zero

[bronze lake]

It's a 2-knot embedded in 4-dimensional space, so it's not nearly as easy to just find the relations as with a traditional knot

[marsh forge]

ah

[arctic relic]

In order to show that H_0(X) is isomorphic to augmented H_0(X) + Integers, Hatcher claims that epsilon composed with boundary map induces a map from H_0(X) to Integers st kernel of this map is augmented H_0(X). Does this follow from functoriality? I already proved this using a dimension argument but I would like to know Hatcher’s way.

Since epsilon vanishes on Im(boundary map)

Never mind

[upper basalt]

Is S={1,x,x^2,x^3.......} is compact in C[a,b]? any hint how can i prove this?

[wintry heart]

what does the notation pi_1(X) mean for a topological space X? i thought fundamental groups need a base point

[shut moat]

are you familiar with stone weierstrass?

[shut moat]

pi_1(X, x_0) only depends on the path component of the base point (up to isomorphism)

so when you restrict yourself to path connected spaces you can drop the basepoint

[wintry heart]

pi_1(X, x_0) only depends on the path component of the base point (up to isomorphism)
sorry but i don't understand this at all

[upper basalt]

Yea,but can i use thay here?

[shut moat]

if you can connect two base points by a continuous path, the fundamental groups at the two basepoints are the same. If the path between $x_0 $and $x_1$ is $\alpha$, then if $[a]\in \pi_1(X, x_0)$, $[\alpha] \ast [a] \ast [\alpha^{-1}]\in \pi_1(X, x_1)$

[gentle osprey]

Buncho Spheres

[shut moat]

you can show that this is a group isomorphism

[shut moat]

if you're not allowed to make use of that, you can still use it to get some intuition for what you need to do

[wintry heart]

oh okay thanks

[shut moat]

namely that {1, x, ...} is not closed in C[a,b]

[gritty widget]

It's not compact in general, you can take a = b = 1 though

[upper basalt]

That will be an intervel then

[gritty widget]

How does Stone-Weierstrass work here?

[hollow harbor]

ummm

[upper basalt]

Are you using closed + bounded = compectness?

[hollow harbor]

this is just a sequence right

the problem is that it's not equicontinuous

[gritty widget]

Yes, it's just a sequence

[hollow harbor]

(if a neq b)

well

maybe it is if -1 < a < b < 1

[shut moat]

jfc lmao for some reason i interpreted that as span{1, x, x^2,...}

[hollow harbor]

then it would be compact, no?

or precompact 😵‍💫 idk if that's closed

[gritty widget]

It's not compact then because 0 is not in this set

[upper basalt]

Then i will be , because it will generate set of all polynomials

[hollow harbor]

ah yeah

great point

whereas if b >= 1 or a <= -1, you get failure of equicontinuity

[shut moat]

nah it's not closed or precompact, that's what i was trying to say with the stone weierstrass thing

[upper basalt]

Why we focusing on equicontinity?

[shut moat]

it's dense in C[a,b]

[gritty widget]

It's not

[hollow harbor]

arzela ascoli says that for subsets of C[a, b], precompactness (compactness of the closure) is equivalent to pointwise boundedness and equicontinuity

[upper basalt]

How? Can you please elaborate ?

[shut moat]

(i'm talking about the span in response to ryc, not the correct interpretation as a sequence)

[upper basalt]

Ohhh i forgot Arzela Ascoli

[hollow harbor]

if -1 < a < b < 1, then x^n converges to 0 uniformly

yeah try to work this out using AA

[gritty widget]

I mean, you can check this by hand

If |a|>1 or |b|>1 then it can't even converge pointwise

So we are left with -1 <= a < b <= 1

And here you have that it converges to something that's 0 on an interval if anything

So no polynomial of the form x^n

(Then it would be 0 polynomial, which is not in the set)

So only a = b = +-1 work

[upper basalt]

Can you please tell how can we show this is not equicontinious?

[hollow harbor]

the best way to see this is at x = 1

at x = 1, the derivative of x^n is n

this lets you say that, for epsilon = 1/2, if you give me any delta, we have |x^n - 1| > 1/2 for large enough n even when |x - 1| < delta. in particular, pick x = 1 - delta/2 and pick n big enough so that (1 - delta/2)^n < 1/10 (this surely will happen, since (1 - delta/2)^n converges to 0 as n goes to infinity)

i didn't really use the fact that the derivative is n

but you can also use that and the mean value theorem somehow

[gritty widget]

I feel like using equicontinuity is overcomplicating this

We have a sequence, and it's enough to show there is a subsequence with no convergent subsequences

[hollow harbor]

Sure, that's another way to do it. And you can do it for, say, [-1, 1] by saying that you do have pointwise convergence, so subsequences cant converge uniformly (or they would be forced to the pointwise, discontinuous limit)

(i dont know if equicontinuity is complicating it necessarily though - when I think about compactness of sets of functions in uniform norm, I start thinking about derivatives growing before I start thinking about literal sequential compactness. But that's just a perspective thing).

[midnight echo]

is a net just a map that takes one directed set to another?

and a subnet another map that takes a directed set to the directed set of the net?

[gritty widget]

A net is a map on directed set into our space

And subnets are analogues of subsequences, but they are slightly different

[midnight echo]

so is a net a way to order the space?

[gritty widget]

No, it's just a more general sequence

[midnight echo]

oh right ok

[gritty widget]

For example, partial sums of a Riemann integral are a type of net

I think this example is illustrative

Nets are one of two main ways of generalizing convergence, the other being filters

Both are pretty good

[midnight echo]

yes I can somewhat understand that a net is a less restricted definition of sequences, just having a hard time visualising what's going on

[gritty widget]

Directed sets, as name suggests, impose some sense of direction, without it being a well-order

Do you understand the Riemann integral example?

[midnight echo]

so an element can have two or more predecessors?
x_1,x_2 =< x_3

[gritty widget]

Yeah

[midnight echo]

not really

[gritty widget]

Well you have one partial sum, and a second one

Now you can say the first one is < than the second one if you're using all the points from first one, and then some

And for any two partial sums there is their common refinement

Now if the integral exists, the partial sums will converge to it

Precisely in the sense of convergence of nets

[midnight echo]

ahhh ok, yeah so basically the sums get finer and finer until you have the exact value of the integral

[gritty widget]

Yeah

And this is basically how nets work

[midnight echo]

cool jazz, thank you so much

[arctic relic]

Does the n+1 skeleton of a CW complex completely determine the nth homotopy group in an extension of the manner that the 2 skeleton completely determines the fundamental group?

[lunar yoke]

I think so. By cellular approximation, every map from S^n already lands in the n-skeleton and a homotopy starting in S^n x I lands in the n+1 skeleton

[fading vale]

No it needs to be n-1 connected

e.g you need pi_i(X) = 0 for i < n

Oh wait i am illiterate, n+1 skeleton yes, n and n+1 cells no

[arctic relic]

Would the homotopy have to be defined from S^n x I^n for the nth homotopy group?

[lunar yoke]

no

homotopies are always of the form X x I -> Y

sometimes you have some extra conditions that some things are preserved

but still the domain is of that form

you can think of the interval as a time parameter

[gritty widget]

There is a concept of $K$-homotopy for compact space $K$, which is a function $H:X\times K\to Y$

[gentle osprey]

Blitz

[gritty widget]

I'm not saying what chernberries said is right of course

[marsh forge]

is there a point to this? i've never seen it

Does K come equipped with like some structure

i.e. the inclusions of 0,1 into I

[gritty widget]

well, the only application I know of, is this simple theorem

[gentle osprey]

Blitz

[gritty widget]

I read it from a book and not sure if there are other applications, seems like a definition which is too basic

it exists in literature though

[winged badger]

Is there an open set in R^n which is connected but not path connected? Just out of curiousity

[gritty widget]

No

This is because of local path-connectedness

If U is open connected, its path-components are open and path-connected

Since U is connected, there is just one path-component

Is the proof

Interesting that Poincare duality is your favourite theorem, but you ask such basic questions

People learn every day ig

[fading vale]

They are mostly active in #algebraic-geometry so I imagine they learned it out of an AG book

[winged badger]

I read Hatcher before getting into AG :)

[fading vale]

That would also explain it

[unreal stratus]

Quick q - the cone of S^n is homeo to D^n+1 just because of essentially spherical coordinates right?

[gritty widget]

Yes

[unreal stratus]

Like if we embed it both R^n+1 we have the map S^(n) x [0,1] -> D^(n+1) just sending (x,λ) to λx and then ye that identifies all the points with λ = 0

compact hausdorff etc so it's a homeo

[gritty widget]

Yes

[unreal stratus]

Nice cheers

[gritty widget]

Np

[gritty widget]

More precisely, there's no argument with compactness, just quotient maps

[unreal stratus]

I just used that to show its inverse is continuous

But yeah the inverse can also just be explicitly written down anyway

[gritty widget]

Oh wait

You use compactness to show it's a quotient map

[unreal stratus]

Like continuous bijection from compact to hausdorff is a homeo i mean yeah

[gritty widget]

Here's where it's used

[unreal stratus]

Yup

[gritty widget]

https://math.stackexchange.com/questions/3064037/when-is-a-space-homeomorphic-to-a-quotient-space

I don't want to prove it, so here's a reference

[marsh forge]

In case you haven't noticed this visual: We can think of S^n x I as a copy of S^n, based at its rightmost point, living at each coordinate of [0,1]. In fact, we might as well consider them to be shrinking as we do so. Visually, this "fills in" the outermost unit sphere and looks exactly like the unit ball. The problem is that in the limit our shrinking wants a sphere of radius 0---but because we are coning, the last sphere gets to be squished to a point

[unreal stratus]

Oh i mean i've done this before lol dw

I was more just checking I had the right idea for a map/intuition

[unreal stratus]

Oh nice, that's a slightly different way of visualising to what I had, thanks

And yeah the interpretation of it being a 'cone' is seen more visually when we modify the map to, say S^1 x [0,1] -> R^3, (x,t) -> (tx,t) right? since then the image is literally just a standard cone and so CS^1 homeo to the cone

[marsh forge]

oh sure yeah

I mean the visual should always be like

you take X, stretch it to a "cylinder" of Xs, and then squish the final copy of X

i sent this like 5 times before my internet connected properly and discord decided to send them all lol

[unreal stratus]

Yeah

oop

but thank

[marsh forge]

also i meant to say something but this attitude has no place here, please refrain from saying dumb things like this. it's toxic and pointless.

[gritty widget]

I feel attacked here

[cedar pebble]

nobody is attacking you, just saying you don't have to act like that

no hard feelings just saying

[arctic relic]

“Induced homomorphism” just follows from functoriality of function right

It keeps popping up and I just assumed that was the definition

Can you give an example of the composition?

My understanding, for instance, is that the functor from based spaces that takes based spaces to groups and morphisms to morphisms st f(x0) = y0 is induced by the cont function from a based space to a based space

So like what you said w/ pi_1

What are you composing? The continuous function with the function that sends basepoint to base point?

[lunar yoke]

Given an element in pi_1(X,x) you can pick a representative a : (S^1,1) -> (X,x). If you now have a map f : (X,x) -> (Y,y) you can consider the composition f o a : (S^1,1) -> (Y,y), giving you a loop in Y. Taking homotopy classes then gives you an element in [f o a] in pi_1(Y,y).

so the induced morphism pi_1(f) : pi_1(X,x) -> pi_1(Y,y) just takes [a] to [f o a]

[arctic relic]

It’s weird to think that the composition is inducing functoriality vs the actual function

[lunar yoke]

drawing pictures helps

this is pretty common in algebraic topology though, you'll see similar things everywhere

[arctic relic]

Unfortunately I’m already knee deep in my algebraic topology course so I guess I need to revisit my intuition about induced things

[marsh forge]

Well, the composition includes the original function

We are just explaining how to get from a map X->Y to a function sending maps S^1->X to maps S^1->Y

which then gives us the algebraic map we need after taking homotopy classes

But really the original function X->Y is doing the heavy lifting here

[gritty widget]

Whatever, no use in arguing and I don't need to be here

[marsh forge]

lol what

you said something rude

and got called out

you aren't a victim here lmfao

[gritty widget]

This was already resolved btw (with me, max and Porphyrion in #discussion ) so if anything you are coming in for the drama, and I advise you to stop.

[rugged rock]

Is [0,1] homeomorphic to [0,1)? I don't think so as $\mathbb{R} - [0,1] = (-\infty,0) \cup (1, \infty)$ which is open in $\mathbb{R}$ under the standard topology while $\mathbb{R} - [0,1) = (-\infty,0) \cup [1, \infty)$ is not.

[gentle osprey]

Évariste Galois

[rugged rock]

Although i am not sure if that is even an indication

[gritty widget]

It's not, you can prove it in few ways

Using compactness or connectedness

[rugged rock]

Is the way I showed it good?

[gritty widget]

A space can have homeomorphic sets which aren't both open/closed

[rugged rock]

I see

[gritty widget]

So your argument doesn't work

[rugged rock]

OK, so as you said ill show it using compactness or connectedness

Thanks 🙂

[gritty widget]

Np

[upper stag]

Any recommendations for introduction to TQFT? I have basic knowledge in algebra, algebraic topology, and category theory. When I look up the materials, I am always deterred by advanced topics like higher category theory

[jade stirrup]

[0,1) isnt compact

[jade stirrup]

there isnt anything to show

a compact space cant be homeomorphic to a noncompact one

[marsh forge]

The paper I always recommend starting with is the 2cob and frobenius algebras paper

[upper stag]

Thanks for your recommendation. Who is/are the authors?

[marsh forge]

https://math.uchicago.edu/~may/TQFT/Terry'sREUPaperOnTQFT's.pdf

That’s a good expository on the result

@upper stag

[upper stag]

Thanks. I will take a look

[gritty widget]

There is to show that they are not homeomorphic

I know it might sound obvious to you, but it might not be to someone who is starting with topology

[topaz plover]

I'm not sure if this goes here or in #real-complex-analysis, but I'm trying to figure out what the open balls of the uniform topology on $\mathbb{R}^{J}$ look like. In Munkres, the uniform metric is given as $\overline{\rho}(x,y)=\sup{\overline{d}(x_{\alpha},y_{\alpha})|\alpha\in J}$. My first guess is that the open ball $B_{\overline{\rho}}((0,0),1)$ would be $\prod_{\alpha\in J}(-1,1)$, but that doesn't feel right: If we let $\mathbf{a}=(\frac{1}{2},\frac{2}{3},\dots,\frac{n}{n+1})\in\mathbb{R}^{\omega}$, then $\overline{\rho}(\mathbf{a},\mathbf{0})=\sup{\overline{d}(\frac{n}{n+1},0)|n\in\mathbb{Z}^{+}}=1$ so $\mathbf{a}\notin B_{\overline{\rho}}((0,0),1)$, but $\mathbf{a}\in\prod_{\mathbb{Z}^{+}}(-1,1)$.

[gentle osprey]

Esix, Fractal Bloom
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty widget]

This is definitely topology

[topaz plover]

Yeah, it's just that it's specifically a question about a metric, and I've also done metric space stuff in real analysis

[gritty widget]

I'd go here for metric and metrizable spaces myself, and reserve #real-complex-analysis for integral stuff and so on

[topaz plover]

Fair enough

[gritty widget]

So you have $\sup |x_\alpha |<1$, right? Like you noticed, we need the sequence bounded uniformly by some $M<1$

[gentle osprey]

Blitz

[gritty widget]

So it should be $\bigcup_{0<M<1}\prod_{\alpha\in J} [-M, M]$

[gentle osprey]

Blitz

[topaz plover]

That looks believable. I want to make sure $(\frac{1}{2},\dots,\frac{n}{n+1},\dots)$ isn't in there, but I think that shouldn't be too difficult.

[gentle osprey]

Esix, Fractal Bloom

[topaz plover]

Yeah, you just pick $n>M$ for any given $M$

[gentle osprey]

Esix, Fractal Bloom

[gritty widget]

You can also write it as $\bigcup_{0<M<1}[-M, M]^J$ or $\bigcup_{0<M<1}(-M, M)^J$

[gentle osprey]

Blitz

[topaz plover]

Yeah, I was about to ask, actually, about open vs closed intervals

But because it's an infinite union I'm not sure how much it strictly matters

Infinite union of closed intervals can be open, etc.

[gritty widget]

It doesn't matter because M is from (0, 1), this is what makes the set open

[topaz plover]

Oh, okay

Thanks

[gritty widget]

Np

[surreal ocean]

munkres theorem 17.11 states that every simply ordered set in the order topology is Hausdorff. However consider the set {a,b} with the order relation defined by a < b. This satisfies comparability non-reflectivity and transitivity. If i’m not mistaken the order topology on this set would be { {empty}, {a,b} }. Clearly this is not Hausdorff. What am i missing?

[rancid umbra]

the order topology is generated by the open intervals and open rays here, so sets of the form

{x : a < x}
{x : x < b}
{x : a < x < b}

for all a,b in your set.

this says that the open sets are
{}, {b}, {a}, and {a}U{b} = {a,b}

[surreal ocean]

aahhh thanks

[pulsar lagoon]

Could someone help me get an intuition as to why these spaces are called "hedgehogs"?

[hollow harbor]

They're spiky

[plain panther]

sorry to ask such a basic question but:

I'm a bit confused by this because from my understanding [0, 1] is a compact set

actually hmm, i think I might've figured it out

my problem with it was for example if the open sets got smaller and smaller somewhere. but it clearly neats to have an open set containing the limit point, which if it got smaller and smaller would necessarily contain an infinite number of those sets, and reduce it to a finite number.

at least I think this is correct

[hollow harbor]

exactly!

trying to prove that the set {1/n : n in N} union {0} is compact will give you an idea of why this works.

[plain panther]

lol literally the example it gives later on

[hollow harbor]

for open intervals it's a tiny bit more involved to show it rigorously but your idea is 100% correct.

now maybe this makes it clearer why things like (0, 1) are not compact

[plain panther]

yea it does. thanks! :)
idk why but there's so many definitions in topology that have always confused me on if it's actually sound or not.

[hollow harbor]

(and in general, why compact sets should be closed - but you have to be careful, this can fail in non-hausdorff spaces)

[hollow harbor]

a lot of them are very confusing at first and only become more intuitive the more arguments you do with them and the more equivalent ways of stating them you see.

[plain panther]

any reccomendations on a text? the text from above is just Calculus on Manifolds, so def not actual toplogy, but I'd like to kinda like to read along and I don't think I have any texts for it yet

[hollow harbor]

The classic text people like to use is Munkres's topology book.

[plain panther]

thanks so much! :)

[river gull]

Does the topology $\tau$ as defined in this statement have a general name? It sounds interesting and probably merits a name because if $\tau_1$ and $\tau_2$ are induced by metrics, $\tau$ is induced by the metric given by the maximum of the two metrics.

[gentle osprey]

Mistral

[shadow charm]

https://math.stackexchange.com/questions/1999946/hatcher-question-1-2-10-show-that-the-loop-gamma-is-nullhomotopic in the answe to this they say that keeping track where gamma goes gives gamma = ab but when i try to visualize this i keep getting gamma = aba^-1b^-1: any nice way to visualize to figure out why im wrong ?

[coral pawn]

Question about this definition: is ii) really what is stated here? I ask this because most of the time when we define a delta complex structure on a space, ii) isn't exactly satisfied. However, a relaxed version of ii) where we ignore the ordering of simplexes is satisfied

Here's an example

This is the Klein Bottle

L restricted to the bottom edge of the triangle gives us the map -b instead of b

Where by -b, I mean the map which takes an element t of [0,1] to b(1-t)

Strictly speaking, this map is not one of the maps from Δ^1 to K

the only two maps from Δ^1 to K that we have in our structure are a and b

Can someone verify this?

[marsh forge]

Restriction of L to b is the exact same continuous map

er

let me think harder

yeah wait why do you think its -b?

@coral pawn

[coral pawn]

Well let's say we order L by [x_0, x_1, x_2] where x_0 is the bottom-left vertex, x_1 the top-right, and x_2 the bottom-right. Then restricting to the lower edge (which is [x_2, x_0]) gives us the map -b right?

@marsh forge

I guess if you orient L the other way, the restriction to the bottom edge would be b

However, then the restriction to the right edge would be -a

[high hill]

My (Masters) course in Algebraic Topology more or less consisted of basic Point Set Topology and Chapter 1 of Hatcher

https://i.imgur.com/Yb3GXmG.png

Is this typical content for a course, or is this quite short? (this thought has been bugging me for a while )

[shadow charm]

Dont know anything about a masters degree but it does seem very short

From what I can tell a masters alg top course at least goes through homology

[high hill]

sigh guess I just have to readup on extra stuff

[shadow charm]

If I’ve got an identification space can I always say that it’s homotopy equivalent to connecting the identified points to an intermediary contractible space?

[lunar yoke]

My 1st sem masters course in algebraic topology had ch 1-3 and basic htpy group stuff as prerequisites. We started with hurewicz

[shadow charm]

Sounds like something you might be able to say about CW complexes

[lunar yoke]

There were 2 bachelor courses preceding this though, covering mostly chapter 1-3

[shadow charm]

From what I can tell our courses cover higher homotopy groups and cohomology in masters and do point set, fundamental group and Homology in second and third undergrad year

[shadow charm]

https://www.math.umb.edu/~oleg/hw_2_solutions.pdf i don’t quite get in 1.2.11 why they attach two 2-cells by the given maps instead of a single 2-cell by the map abcf(a^-1)f(b^-1)c^-1

[marsh forge]

I think the issue here is that the orientation is like a formal algebraic choice rather than something actually describing the order of the map but maybe that also doesn't work

I havent thought about this in a bit and it is indeed confusing at times

Like the identifications of the edges via the directional arrows and the map representing b in the chain complex need not coincide i think

so as a map [0,1] the "direction" of b might not agree with the identification pattern

Like we could reverse the orientation, get a different shape, but keep the delta complex structure the same, I think

Hm. But then maybe you are messing up the boundary maps or something

[viral yoke]

Let $\mathcal{C}, \mathcal{D}$ be two chain complexes and let $f: \mathcal{C} \to \mathcal{D}$ be an arbitrary chain map. Show $f: \mathcal{C} \to \mathcal{D}$ induces a sequence of group homomorphisms $f_* = {((f_q)*) , | , q \in \mathbb{Z}}$, where $(f_q): H_q(\mathcal{C}) \to H_q(\mathcal{D})$ with $[\sigma] \mapsto [f_q(\sigma)]$ for each $q \in \mathbb{Z}$. (We write $f_: H(\mathcal{C}) \to H(\mathcal{D})$.)

[gentle osprey]

eM

[viral yoke]

Am I showing the map $(f_q)*$ is well-defined and is a group homomorphism for each $q \in \mathbb{Z}$?

[gentle osprey]

eM

[marsh forge]

Sounds right to me

Although the latter is pretty easy if you assume the former

[viral yoke]

Yeah the well-defined part doesn't seem as straightforward as I thought 🤔

So I let $[\sigma], [\tau]$ be in $H_q(\mathcal{C})$ such that $[\sigma] = [\tau]$, and therefore $\sigma = \tau + \theta$ for some $\theta in \mathrm{Img} , \partial_{q+1} = B_{q+1}(\mathcal{C}) \subseteq C_q$. Recalling $f_q: C_q \to D_q$ is a group homomorphism, from our map as defined previously, the induced map gives $(f_q)_*([\sigma]) = [f_q(\sigma)] = [f_q(\tau + \theta)] = [f_q(\tau) + f_q(\theta)] = [f_q(\tau)] + [f_q(\theta)]$.

I'm figuring out how to "delete" $[f_q(\theta)]$, which can be done so by showing $f_q(theta) \in B_q(\mathcal{D}) = \mathrm{Img} , \partial_{q+1}$, where the term $\partial_{q+1}: D_q \to D_{q+1}$. I think commutative of $f_q$ might come into the picture but I'm trying to see how to put this together.

[gentle osprey]

eM

[cedar pebble]

the map you should get sends a chain a+im∂ of degree n in C to f(a)+im∂ of degree n in D. It's well-defined since if you take a+im∂=b+im∂ then (a-b) is in im∂, but since f takes boundaries to boundaries we get f(a-b) in im∂, so f(a)+im∂=f(b)+im∂

so f takes kernels to kernels and images to images, so you get an induced map from kernel mod image to kernel mod image

(I'm being sloppy with omitting indexing/lower star)

[viral yoke]

I literally just saw where theta was living lol how did I miss that

[cedar pebble]

all good! 🙂

[viral yoke]

I understand

Also, how did you get that partial character to show up?

[cedar pebble]

option + d

on mac

[viral yoke]

Oof

[cedar pebble]

I know there are some fancy math keyboards but

[viral yoke]

I have windows

Gonna look it up

[ivory dragon]

consider trying wincompose

[arctic relic]

If A is contained X, is the fundamental group of A contained in the fundamental group of X? I found a counterexample but I’m not too sure

[coarse night]

if you have already found a ce, why are you still unsure?