point-set-topology

[languid patrol]

Like if p is distance d from the boundary

[odd flame]

it's the d that's important

sorry too easy (i am a 7 year old trapped in a college student's body)

[languid patrol]

Then just declare p \to q and points P of distance x from p move as P \to P + (q - p)*(1 - x/d) if q is further from the boundary than p.

[silver umbra]

i think the intended method here wouldnt involve passing to a metric on the space

[languid patrol]

I mean why not? You have to write down a function somehow.

[silver umbra]

namely, we havent proved that all manifolds are metrizable yet

[languid patrol]

You don't need a global metric

We're doing this locally on a chart

The fact that the euclidean n-ball is metrizable is much easier to prove.

[silver umbra]

sure so we're working in the Euclidean n-ball

and the function u give is a homeomorphism fixing the boundary and sends p to q?

first of all how do we compute d

the distance of a point to the boundary

[languid patrol]

You don't need to compute it.

[odd flame]

i'll be back w more questions when pdk is done but thanks again

[silver umbra]

also, how is this function well defined if x is on the boundary

then d = 0

[languid patrol]

x is not a point it is a number

[silver umbra]

wait what

[languid patrol]

P is a point in that formula and d(P, p) = x

[silver umbra]

oh

[languid patrol]

Anyway the idea is:

Take a small disc around p which doesn't contain the boundary and such that the radius of the disc is less than the distance from q to the boundary

(you don't care how big the disc is, the only important thing is that it exists)

Call this disc D.

D has radius r

Actually here's a better way to do it

[silver umbra]

mhm

[languid patrol]

So take a homeomorphism from the interior of the disc to R^n

then p, q map to two vectors in R^n

So if this homeomorphism is \psi, we do P \to \psi^{-1}((\psi(q) - \psi(p)) + \psi(P))

[odd flame]

ohhhhh i kinda see why my question before was true

those regions in the equator are just being divided into four regions

[languid patrol]

the result is that we push p to q, then the only thing left to do is to use a linear decay function to make sure the homeomorphism fixes the boundary.

[odd flame]

did that post

oh there it is

[silver umbra]

uhh i'm not sure that i'm entirely following

[languid patrol]

Which part is confusing? We constructed the homeomorphism on the interior, the only thing left is to engineer it so that it fixes the boundary.

[silver umbra]

so psi is some artbirary homeomorphism from D to R^2

[languid patrol]

If your manifold is 2-dimensional, yes.

[silver umbra]

sure, let's reduce the case to surfaces

[languid patrol]

Yes I think you're getting this, so that heuristically should tell you that the H_1's are the same

[languid patrol]

Okay so yes we pick some arbitrary homeomorphism

Then we use the vector space structure on the target to just translate from one point to another

[odd flame]

yeah i see the general idea now i need to squeeze MV in there too

[languid patrol]

So this gives a homeomorphism on the open disc D which takes point p_1 to point p_2

[odd flame]

sorry i'll let you finish that other q tho

[silver umbra]

gotcha, i see what's going on here

the vector space structure on R^2

makes it possible to do these translations easily

and represent them by just adding vectors

[languid patrol]

Yeah, really it's just a dumb hack so you don't have to write a bunch of coordinates.

[silver umbra]

sure

[languid patrol]

So anyway once we've done this and we get our homeomorphism f on some open disk U, we fix an \epsilon > 0, for points of distance 1 - \epsilon or less from the center of U, g(x) = f(x), however for points x of distance d >= (1 - \epsilon) we define g(x) = f(x)*(1-d)/\epsilon+ x(d - (1 - \epsilon))/\epsilon

This looks like an annoying mess but all this is doing is dampening the effect of the homeomorphism as you get closer to the edge of the disc, that way the boundary is preserved

We're picking a ring of width \epsilon around the outside of the disc and decreasing the effect of the homeomorphism as you go further out along that disc

[silver umbra]

hmm ok

i'm trying to think of how to prove that this is a homeomorphism

[languid patrol]

Well let's just think about the same problem but in R^n since we're in that context

We want to construct a map from R^n \to R^n which is a homeomorphism, takes p to q, and does nothing for points very far away from the origin.

[silver umbra]

sure

[languid patrol]

Ah wait here's an even better way, which is really obviously a homeomorphism with no work

Take the midpoint of p and q and rotate around it, but rotate points less and less as the distance from the midpoint changes from |p - q|/2 to N, where N is sufficiently large.

This actually switches p, q which is kind of neat.

So wolog (p - q)/2 is the origin

Then let R(\theta) be a family of rotations by angles \theta around the origin such that R(\pi/2) takes p to q and vice versa. Then we do f(P) = R( (2(N - |P|)/|p - q|) * \pi/2) when N>= |P| >= |p - q|/2, f(P) = R(\pi/2) if |P| < |p - q|/2, and f(P) = P if |P| >=N

This one is clearly bijective since you can just rotate back. And it's very easy to check that it's continuous since it's clearly continuous on each region where it's piecewise defined and it matches on the boundary.

[silver umbra]

ok so R(theta) is counterclockwise rotation by angle theta

what is N here?

[languid patrol]

Sure in 2-dimensions yes

[silver umbra]

yeah lets just reduce all of this

to two dimensions

for simplicity

what is N here?

also why is it ok to take the midpoint of p and q to be the origin

[languid patrol]

N is just some very large number

Arbitrarily large

Just a big distance from the origin, and outside the sphere of radius N we want the homeomorphism to be trivial

[silver umbra]

um

[languid patrol]

Because it's just as easy to describe this rotation around the origin as it would be to describe it around any other point.

If you like just replace "the origin" in what I said with "the midpoint of p and q" if it makes you uncomfortable.

[silver umbra]

wait so is this map defined on the closed disk? or R^2

[languid patrol]

I'm defining it on R^2

But the point is that because it is trivial outside the sphere of radius N, that means that when we translate back to the disc, it will be trivial outside of some open subset of the disk which doesn't touch the boundary.

[silver umbra]

but how are we translating back to the closed disk

[languid patrol]

We're defining a function on the interior of a disk, which is trivial near the boundary

so that function can be glued with the identity on the boundary to make a function on the closed disk

[silver umbra]

but how is triviality near the boundary defined for a function on R2

[languid patrol]

(trivial meaning is the identity)

[silver umbra]

like doesnt it depend on what homeomorphism we choose from the open disk to R2

[languid patrol]

Not really, being trivial "outside a compact set" is well defined for any homeomorphism

[silver umbra]

i think this description is going over my head a little bit; would it be hard to prove that the previous map u suggested is a homeomorphism

[languid patrol]

Like it makes sense on both sides, on the R^2 side that means outside some disk of radius N, inside our open disk thats some wobbly looking compact subset.

I dunno none of it is hard, but you might find it easier if you construct your own function inspired by this method.

[silver umbra]

i mean its not hard viewing it from a purely visual high level perspective but getting down to it and trying to like

actually find the inverse

[languid patrol]

The method being: do whatever you want inside some open disk, then make it trivial outside of some compact subset by continuously reducing the effect of the homeomorphism.

[silver umbra]

espcially when ur dealing with distances and stuff

kinda hard

[languid patrol]

The nice thing about the rotation one is that the inverse is explicit.

Since it's just rotating back basically.

I wouldn't calculate out the distances explicitly, that's a waste of time. Just write dist(blah, blah)

The distance function is obviously continuous so any expression involving it and other continuous functions will also be continous.

No need to write any square roots

[silver umbra]

sure but its not continuity thats the issue

its trying to find the inverse

[hallow moss]

my homework asks that I classify the alphabet into homeomorphism classes surely I do not have to find an explicit homeomorphism right?

[silver umbra]

like the previous suggestion u gave, how would one even try to go about determining the inverse of that map explicitly

[languid patrol]

Lmao

[languid patrol]

Which one? The translation one or the rotation one?

[silver umbra]

the translation one

the inverse in the interior is obviously easy

but when u start like massaging it so that it becomes trivial toward the boundary

what even is the inverse

[languid patrol]

Yeah I think you have to be more careful about the way you massage it than what I described, the nice thing about the rotation one is that the massaging can be expressed cleanly in terms of distance from the midpoint.

Like I suspect what I described takes even more mess to turn into a real solution, whereas this rotation one just works.

[rapid lagoon]

Excuse me for barging in, but I have a question regarding this problem, specifically 4b. Specifically, I am really not sure how I would construct a "smallest unique topology that contains all of T_alpha"

[silver umbra]

also what exactly do u mean by N sufficiently large

like N sufficiently large for what

[languid patrol]

Just N larger than say 2|p - q| suffices, it doesn't really matter beyond that

You just don't want to make the cutoff so small that it gets in the way of what you want to happen with p and q.

[silver umbra]

im also very confused

[languid patrol]

It follows from part (a)

[silver umbra]

R is rotations such that rotating by pi/2 sends p to q?

im not understanding what that even means

[languid patrol]

Yep, so it's rotating around the midpoint of the line between them

If you rotate by pi/2 you switch them, if you rotate by 0 degrees, you leave them alond

[silver umbra]

would maybe reducing to the case that, say q = 0

and the disk is the unit disk

make any of this easier

im just having a very hard time visualizing rotations about lines between points and stuff

[rapid lagoon]

Does it? Because $\bigcup T_\alpha$ is not neccesarily a topology, although it is the smallest set that contains all of $T_\alpha$

[gentle osprey]

dulg

[languid patrol]

I think you can just imagine that p = (1/2, 0) and q = (-1/2, 0) if you want, that's probably maximally easy.

[silver umbra]

but i cant just take fixed p and q

then the whole problem is moot

[languid patrol]

Yes

Which is why I didn't make any particular assumptions about p, q.

I thought you were asking for intuition, but to write it up you have to describe an arbitrary pair of points. Taking p to be the origin or something doesn't make anything simpler imo.

[languid patrol]

The smallest topology containing \bigcup T_\alpha is the intersection of all of the topologies containing it. There is at least one because of the discrete topology.

[rapid lagoon]

Ah!

Thats really neat! Thanks!

[languid patrol]

No worries.

[hallow moss]

bruh I just drew pictures for some of those alphabet things

if i get points off for no explicit homeomorphism im gonna die

[languid patrol]

I think that's fine

I dunno it depends on your TA, but it would be ridiculous to expect you to write down coordinates for all the letters

Just like some would say it would be ridiculous to stress too much about writing an explicit homeomorphism on the disk in enough detail 😉

[willow viper]

hello how can I start proving the first direction?? I know that f closed means that for all U ⊆ X closed there exists an f(U) ⊆ Y closed?

[abstract saffron]

If my memory is of any use, f is closed if it maps closed sets to closed sets.

[willow viper]

Is my definition different?

[abstract saffron]

It follows from two things, $f(\overline{A})$ is closed, and $f(A) \subseteq f(\overline{A})$.

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

Then $f(\overline{A})$ also contains the closure of $f(A)$. QED

[gentle osprey]

Megumi_Tadokoro

[abstract saffron]

It doesn't seem so, but I just find it weird

[willow viper]

maybe.. I tried to write it out to make things easier maybe

[rapid lagoon]

sidenote, but topology is genuinely kinda nice

beats real analysis

XD

real analysis, more like applied liminf and limsups amirite

real analysis, more like applied topology XD

[willow viper]

I don't get this

Oh Im dumb

because cl(f(cl(A))) = f(cl(A))

[willow viper]

Is the reverse direction then:
cl(f(A)) ⊆ f(cl(A)). So cl(f(A)) ⊆ cl(f(cl(A))) => A ⊆ cl(A), so A is closed.
Since A ⊆ cl(A) <=> f(A) ⊆ f(cl(A)). So f(A) is closed??

[odd flame]

going back to this question

for that last part about the homeomorphism is it enough to say that (1,0,0) has 1 line going to it in X and 2 in Y so it cannot be homeomorphic

or maybe better that removing a nbhd of (1,0,0) will make X and Y deformation retract to different spaces

X to like a disk with a cross up the middle

but disk deformation retracts to a point so just the cross?

and im p sure Y deformation retracts to a circle when such a nbhd is removed

walter button :(

,av walter

[gentle osprey]

walter#1555's Avatar

Click here to view the image.

[odd flame]

this needs to be an emote

different question - how is this obvious

[fading vale]

f induces a homeomorphism on the pairs of spaces

alternatively take the maps induced on the long exact sequences of the pair and apply the five lemma

[odd flame]

my idea was to define some g: Y -> X to be the inverse and somehow use that

is that what you meant?

[odd flame]

could you elaborate on this one pls i think we're doing five lemma next lecture

still curious

[coarse night]

you know what five lemma is?

[odd flame]

i looked ahead in hatcher

[coarse night]

it's actually one lemma but called 5 lemma

misnomer

[odd flame]

classic

[coarse night]

anyway given you have a homeo, X → Y you also have homeo X-x to Y-y which gives the maps
Hn(X-x) → Hn(X) → Hn(X, X-x) → Hn'(X-x)→Hn'(X)
|| || |? || ||
Hn(Y-y) → Hn(Y) → Hn(Y, Y-y) → Hn'(Y-y)→Hn'(Y)

? now becomes an iso by S5L

[odd flame]

so it's just

automagical

[odd flame]

y should be f(x) no?

f being the homeomorphism between X-> Y

[coarse night]

yes

that was implied

[odd flame]

silly question but

n' is n-1?

[coarse night]

also yes

[shadow charm]

Really don’t need 5-lemma here though 🤔

[coarse night]

wasn't enough space

at least that's what I thought

[shadow charm]

Homotopy equivalence of paire gives isomorphic relative homology, and in this case you have a homeomorphism of pairs (X\x maps homeomorphically through f to Y\f(x))

Okay I should’ve read through your full answer didn’t realize you were justifying why it induced an iso on relative hom

[odd flame]

is the excision part of this just as immediate?

[coarse night]

yes

if you have seen excision

[odd flame]

reading back rn

what's the point of considering pairs of spaces like this

actually is excision "part of the point"

that's probably a really vague question

[odd flame]

ok it was literally just restating excision lol

now i have this fun lil guy tho

[odd flame]

bump to this if you're around topos

wait wroong thing

i'll just repost in case anyone is around

intuitively i can see why C and Y have the same sequence but ive been unable to do it properly

[languid patrol]

Are you struggling to justify why the maps are the same? The groups that you get are obviously the same because you get homotopy equivalent spaces showing up.

[odd flame]

this is what i came up with after messing around w it w a friend

[languid patrol]

Something you wrote there is impossible

[odd flame]

i did write this at like 3am so i could be wrong

[languid patrol]

This long exact sequence where you compute H_0(U) you have written H_0(U - {x})

is Z/2

but I think that is untrue since youve written that you have an exact sequence 0 \to X \to Z \to 0 so X must be Z

[unreal stratus]

Inch resting

[odd flame]

hmmm

[languid patrol]

In general H_0 can never have torsion.

Or no what am I saying.

It is.

\pi_0 is not.

[odd flame]

hows that related

[languid patrol]

It's not I was responding to something that was deleted

From potato

[unreal stratus]

Yeah my half awake brain said smth very stupid

[odd flame]

im gonna switch to another MV question for the practice then

i already did a and b

[languid patrol]

Anyway your cover is by A, which retracts onto the sphere, and B which is just two lines crossing in both cases.

[odd flame]

wait quoi

[languid patrol]

I'm just saying in the previous problem

[odd flame]

oh is that about the one with the two spheres?

[languid patrol]

Yes isn't that what you were asking about?

[odd flame]

oh i just realized i posted the picture

[odd flame]

this was for a different question... my bad

[languid patrol]

I see.

[languid patrol]

Anyway yeah for part (c) what are you struggling with?

[odd flame]

is the right approach to break down E into the sphere parts and line parts and then use what we know about H_n(S^2)?

[languid patrol]

You should follow the approach outlined in the problem

so just calculate the H_2 for the interval [-n, n] with spheres attached at the integer points, then take the limit as n \to infinity

[odd flame]

slightly unrelated but what is d) hinting at? isnt pi_1(S1 v S2) nontrivial, making that hint not helpful?

iirc that has fundamental group Z

[languid patrol]

No but E is simply connected

because E is the universal cover of S1 V S2

[odd flame]

ohhhhh

and universal cover has trivial fundamental group

so the first nontrivial htpy group is pi_2

so this means that pi2(E) \cong H2(E)...?

and we get that H2 from MV sequence

[nocturne basalt]

well universal covers are simply connected so yeah

[languid patrol]

yep

[odd flame]

and how do you get the MV sequence

this goes back to my first question

when i asked if you started from H0 and built up

[languid patrol]

It goes H_2 (A \cap B) \to H_2(A) \oplus H_2(B) \to H_2(X) \to H_1(A \cap B) \to ...

And the first map is injective when everything is 2-dimensional or less.

[odd flame]

oh it just always has that form?

oh i guess that's how it works for SVT too

and the groups themselves just vary per each space

[languid patrol]

Yes in general all long exact sequences on Homology or covariantly functorial algebraic topological objects should decrease degree, cohomological or contravariantly functorial objects will have long exact sequences in the opposite direction.

[odd flame]

so the nested subspace thing in the outline is just a hint about how to make it X = A U B

[languid patrol]

This comes from some general difference between the degrees in homological vs cohomological spectral sequences. Which is something that you won't understand now but perhaps later.

[odd flame]

we do cohomology soon yeah

[languid patrol]

So in our case A_n = [-n - 1/2, n+1/2] with spheres attached at the integer points

B_n is say (-\infty, -n - 1/4) \cup (n + 1/4, \infty) with spheres attached at integer points.

The intersection is just two line segments, which have no H_2

[odd flame]

we're taking this to be some interval in R right

[languid patrol]

This will show you that the maps from H_2(A_n) to H_2(X) are injective

[languid patrol]

Yes.

[odd flame]

including the respective spheres ofc

[languid patrol]

Because our space is R with spheres glued in at integer points.

So i'm using the coordinates on R

[odd flame]

[languid patrol]

Anyway so this is a somewhat trivial Meyer-Vietoris

But then you have to do another Meyer vietoris (which is similar) to compute H_2(A_n)

Alternatively you could just note that A_n is homotopy equivalent to a Bouquet of 2n + 1 spheres, and presumably you've already computed the homology of that in class.

If not you can also use Meyer Vietoris for that.

[odd flame]

we never really defined a bouquet

[languid patrol]

Just like 2n+1 spheres glued together at a single point.

It's usually important for defining homology to compute the homology of that space.

[odd flame]

and then using the nested subspace thing this build up to the whole space E?

[languid patrol]

Yes just to compute H_2 of the entire space.

[odd flame]

oh so it's still trivial then

[languid patrol]

What do you mean?

[odd flame]

have no H2

[languid patrol]

The space has lots of H_2

But yeah the intersections will have no H_2?

So because the map H_2(A_n) \to H_2(X) is injective this shows that H_2(A_n) \to H_2(A_{n+1}) is injective since A_n \to A_{n+1} \to X factors the map to X.

Then as your professor states if X = \cup A_n where A_n \subset A_{n+1} and the map H_i(A_n) \to H_i(A_{n+1}) is injective, then H_i(X) \cong \lim_n H_i(A_n)

[odd flame]

(i'll brb, relocating)

ok i think i see the MV stuff

how does it give us this

[languid patrol]

Because we just computed H_2(E) by hand

and showed that it was Z^{\oplus \infty}

Since \pi_1(E) is obviously zero (E is homotopy equivalent to S^2 V S^2 V ...)

By Hurewicz or whatever it's called this means that H_2(E) = \pi_2(E)

[odd flame]

[coarse night]

can you classify all maps upto homotopy from S¹×S¹ to S²?

[languid patrol]

Why is that relevant?

Or are you asking a question?

[novel acorn]

basepoint preserving or no?

[novel acorn]

because in this case we have the isomorphism $[S^1 \times S^1 , S^2] \cong [S^1 , (S^2)^{S^1}]$

[gentle osprey]

Irony Incarnate

[novel acorn]

which also holds in the basepoint preserving case

and in that case we have $[(S^1,x), (S^2,x)]^0 = \pi_1(S^2, x) \cong 0$

[gentle osprey]

Irony Incarnate

[novel acorn]

which then implies that $[(S^1 \times S^1,x), (S^2,x)]^0 \cong [S^1 , 0]^0$

[gentle osprey]

Irony Incarnate

[novel acorn]

which is just the trivial map

[coarse night]

But that’s not true

You can always collapse the 1-skeleton and get a nontrivial one

[novel acorn]

I'm not quite sure what ur going for here

[coarse night]

It’s not basepoint preserving but I believe you can make it basepoint preserving

Idea is using cellular approximation

[novel acorn]

I still don't get what exactly were talking Abt here anymore???

Like collapse the 1-skeleton of what
And why?

[coarse night]

so CW structure of S¹xS¹ is we have S¹vS¹ and a 2 cell attached by aba'b'. Now collapsing S¹vS¹ gives us a S² which is not nullhomotopiv

That is send S¹xS¹ →S¹xS¹/S¹vS¹=S²

[novel acorn]

Ahh wait

I know where I messed up yeah

[novel acorn]

Also so this isn't always the case

Mine is correct as well

It depends on how you look at the space

If you look at it as a product of pointed spaces

Then you get a smash product

But if you look at the torus as a space with a point as a base point

Then i think what i said holds?
Tho I'm not sure

Actually no it isn't lol

[languid patrol]

You are confused.

Homotopy happens on the source not the target.

All maps from S^1 \times S^1 \to S^2 are nullhomotopic by definition (based or not).

For instance the real numbers R have a quotient S^1, just as your space S^1 \times S^1 has a quotient S^1 \wedge S^1, this doesn't imply that there are nontrivial maps from R \to S^1 up to homotopy though.

[unreal stratus]

What do you mean by "by definition" here

[coarse night]

I just gave a nontrivial map

why is that nullhomotpoic?

can you explain

[languid patrol]

Why is it nontrivial?

[coarse night]

I'm asking why is it trivial

[languid patrol]

I had actually made a mistake but I think it's still nontrivial to show that it's nontrivial

[languid patrol]

I was mistakenly commuting products with Hom up to homotopy.

[unreal stratus]

Sure

[languid patrol]

I just don't see any reasoning you gave to show that it is non-trivial, there is a more canonical looking finite cover of the Sphere by the torus which is obviously nontrivial.

I guess probably the way to do this is to calculate the map on \pi_2 by expressing the torus S^1 \times S^1 as a quotient of a square by identifying opposite edges, and the sphere as a quotient of the square by identifying the entire boundary.

Then it's clear that the induced map on H_2 is nontrivial, so the map is not nullhomotopic.

[coarse night]

then why did you say the maps are nullhomotopic?

[languid patrol]

As I mentioned above, I was confused.

[coarse night]

okay

[languid patrol]

And thought you were saying something incorrect.

I still don't understand why it's related to the post that you posted this in response to.

[coarse night]

It's not, just gave as an exercise

I do that sometimes

[languid patrol]

lol

[coarse night]

in fact I believe the homotopy class of maps should be ℤ

given by the degree of the map S² → S²

[languid patrol]

Yeah, you can see this from the fact that the map on H_2 is just Z \to Z by the identity.

[coarse night]

But I don't have a proof

[languid patrol]

So, if I can redeem myself:

[coarse night]

or collapsing S¹vS¹

[languid patrol]

Yes that is what induces the map

[coarse night]

the whole idea was using cellular approx you get the 1-skelli of S¹xS¹ goes to a point in S² anyway as there's no 1-cell so you might as well look at maps from the quotient S¹xS¹/S¹vS¹ = S² anyway which gives the homotopy class to be Z

[languid patrol]

Yes I think that's a good argument, perhaps worded another way: up to homotopoy, S^1 V S^1 has no maps to S^2, so any map from S^1 \times S^1 to S^2 is homotopic to one factoring through this quotient, and there are Z of those as you say.

*homotopy.

But the main point is that the canonical quotient map IS nontrivial, because this argument only makes sense if you actually produce one non-trivial map.

For instance you could say the same thing about R mapping to S^1: certainly any map from Z \subset R to S^1 is homotopic to 0, but if the canonical quotient map isn't nontrivial the rest of the argument doesn't work.

^This was the mistake that I thought you had made.

[languid patrol]

^I think strictly thinking this is the problem with this argument as stated, you also need to compute what happens to H_2 or something to show that the quotient map isn't nullhomotopic.

[noble osprey]

Hi can someone give me a hint as to how to find the 4 boundary points. Thank you

2a

[compact stump]

Suppose I have a closed hemisphere H ⊂ R³, then it's boundary ∂H is the hemispherical shell. But intuitively, the upper hemispherical shell also has a boundary, which is the circle. How do I denote the circle in terms of H, even though technically the boundary of the hemispherical shell is itself?

[coarse night]

the are sometimes distinguished as the topological boundary vs manifold boundary

your "circle" here is manifold boundary

[noble osprey]

Yea, i got the shape but i dont get where the boundary points come from, in the picture.

[nocturne basalt]

i think the dotted boundary is just to indicate that its open and doesn't contain the boundary

[noble osprey]

yea sorry, I mean that I dont get where the coord (3,0) came from. Would it not be (0,2) since the radius is 2? I thought to use the circle equation, is that something I can do?

[nocturne basalt]

its centered around x = (1, 0)

[noble osprey]

ah yes! i missed that point, thank you.

[next crystal]

what does it mean for a class of loops to generate a fundamental group?

[plain raven]

that every element in the fundamental group has a representative which can be expressed as the composition of elements in the class of loops, together with their inverses

[quick bough]

any hints?

i tried constructing such a J, but i came across a problem

so for every U_i, i can write it as a union of base elements, namely B_j with j \in N (since there is a countable basis), I choose the elements J to be the j such that U_{i_j} contains B_j. The problem now is that such a thing doesn't necessarily exist

[novel acorn]

Yes it does

A property of bases is that for any x in U where U is an open set, there exists a basis element B such that x in B subseteq U

So you can always choose such a U

[opaque cloud]

What more do you need for a topological space to be metrizable apart from hausdorff-ness?

or are every hausdorff spaces metrizable?

[languid patrol]

The one point compactification of R (or any uncountable set) with the discrete topology is not metrizable.

(Because every compact metric space is separable)

(i.e. has a countable dense subset)

[languid patrol]

https://en.wikipedia.org/wiki/Metrizable_space

[opaque cloud]

but it is hausdroff?

[languid patrol]

Yes, that's why I mentioned it

[languid patrol]

The one point compactification of a locally compact Hausdorff space is Hausdorff

[paper wedge]

hi

idk much topology

how do you attach a chain complex to a topological space X

[unreal stratus]

Uh you may wanna google singular chain complex

[paper wedge]

so u approximate it by simplicies?

and the map would be the boundary

cool

[unreal stratus]

Uhh approximate it by simplices is not really what you are doing

And "the map" lol

[paper wedge]

ig better is that ur forming a free group by just taking the vertices of ur simplex as the spanning set

[unreal stratus]

I would recommend just reading what this is lol

[paper wedge]

over X

[unreal stratus]

You aren't taking the vertices of a simplex

[cedar pebble]

well so if you have a triangulation of your space by simplices then you can build a chain complex from this triangulation and compute the simplicial homology from this

[unreal stratus]

Oky that too

Like there are a few different things here lol

[cedar pebble]

this is easier to think about than singular homology

[unreal stratus]

Yes sure

[cedar pebble]

where you are sort of looking at all possible ways of mapping in simplices at once which is a total mess

[unreal stratus]

(Isn't this simplicial rather than cellular though)

I end up forgetting simplicial even exists

[cedar pebble]

yeah oops

[unreal stratus]

When do you even care about simplicial homology tbh like I have never used it to compute anything lol

Other than intro examples in Hatcher

[cedar pebble]

I mean this should basically just be cellular homology

you're just being picky about the sorts of combinatorial descriptions you're allowing for your space

[unreal stratus]

I imagine it's like usable for computers and stuff though

[cedar pebble]

but like morally these are seeing the same thing in the same sort of way

[unreal stratus]

Yeah

[paper wedge]

cool boys and girls

triangulation is easier for me yea

makes more sense

ty al

[plain raven]

You could have come up with a good answer to this question by thinking for less time than it took you to write this sentence

One reason - computing the boundary map in cellular homology is harder than in the case of simplicial homology.

But another reason is that simplicial complexes appear naturally in numerous fields of mathematics (i.e., https://en.wikipedia.org/wiki/Clique_complex) whereas CW complexes only appear in homotopy theory

Even in homotopy theory it is very often useful to adopt a combinatorial perspective.
May wrote Simplicial objects in algebraic topology in 1960.
Goerss and Jardine wrote Simplicial homotopy theory in 1999.
Jacob Lurie wrote Higher Topos Theory in 2009.
Do you expect this to be the decade where we finally discard simplicial complexes?

I do not.

[unreal stratus]

Yeah Idk why I said that given I have used simplicial sets and stuff lol

[plain raven]

He probably meant connected.
But if the manifold is not connected, then each connected component will have a countable fundamental group.

[golden gust]

is there a name for the procedure (or final result) of gluing two copies of a manifold along its boundary?

[umbral panther]

Doubling / double

[golden gust]

thanks

[solemn oar]

If you are doing abstract homotopy theory you are far more likely to encounter simplicial sets and their associated simplicial abelian groups (which compute simplicial homology) than you are to encounter arbitrary CW-complexes.
The latter might be more common in manifold theory.

[noble osprey]

Hi, can someone explain to me in simple terms of open sets with different metrics. I cant seem to understand the sentence highlighted in terms of how the different metrics determine whether U is open or not? Thank you

[paper wedge]

open sets with different metrics

@noble osprey

an open set is a set where if you choose any point inside this set there is a ball around it that is contained inside the set

a ball of radius r centered at x is defined to be the set {y | d(x,y)<r}

if u change the metric the ball "sorta" changes

so u can have an set that is open wrt a metric abut is not wrt another metric

as the definition of the ball is different cuz the metric is different

do u get it

(this is all in the setting of metric spaces )

[paper wedge]

verify from the definitions the example given in the text

[noble osprey]

I get it thanks, so if the open ball had a radius r about (1,0) w.r.t the railway metric then the distance is 1 but from that point how do we go about determining if the open ball is open or not?

i mean if its less than r, but like how do we know?

[paper wedge]

how do we go about determining if the open ball is open or not? the open ball is always open under its respective metric

the point here is that under the railway metric this set IS the open ball

so its open by default ( u can try to prove it )

in general to determine if a set is open or not wrt a certain metric

it would be enough to show that all balls centered at some point of the set cannot be contained inside the set

for example consider [0,1] inside R with the metric defined to be d(x,y) = |x-y|

can u pick an element inside this interval and show no matter how small we choose our radius it still cannot be contianed in [0,1]?

[unreal stratus]

never heard of the railway metric lol

[paper wedge]

me neither

lmfao

i just trusted the text

[unreal stratus]

Oh wait no I did a question on it

But we didn't call it that lol

[paper wedge]

if you can do that then by definition u showed that [0,1] is not open wrt to the metric defined

[noble osprey]

ahh so its not open since it would be outside [0,1]

[paper wedge]

what do you mean by "it" here

i know this shit is confusing

but u must understand it as its very important later on

[noble osprey]

like if we choose r such that 0-r, 1+r is outside [0,1] ?

[paper wedge]

whats 1+r?

lets recap

choose an element in [0,1]

[noble osprey]

yea, this open stuff is confusing

[paper wedge]

such that an open ball around this element can never be fully contained in [0,1]

[paper wedge]

its ok

[paper wedge]

.

all u need to make sure u know fully are the definitions

once u have the definitions it just takes time untill u get it

but u will never get it if ur definitions are fuzzy

u will always be confused

[noble osprey]

so i know the def of open ball

and i know that an open set contains an open ball

[paper wedge]

okay lets be more accurate

an open ball has a center and an eleement

[noble osprey]

but i dont get the metric part

[paper wedge]

so you mean that an open set would alwyas contain a ball centered at any of its elements

[noble osprey]

yep

[paper wedge]

okay so

i want you to tell me

as accurately as yo ucan

the definition of an open set in some metric space X with a metric , call it d.

can u do that

[noble osprey]

a subset of X is open if there exists an open ball within that subset such that d(x,y) < r

[paper wedge]

hmmm

"within that subset"

its not quite

the word "open ball" is ambigious

what is its center

what is its radius

what are x and y?

a subset of X is open if for every element u can center a ball around it such that the ball is fully contained in that subset

so for example suppose u have some subset S

and this S is open

that means that for any random element

b for example

there must exist a ball centered around b such that the whole ball is contained

that is

there must exist some radius r; such that the set {a|d(b,a)<r} is a subset of S

better?

the set {a|d(b,a)<r} is called the ball centered at b with radius r.

[noble osprey]

oh wait, so an an open set is an open ball?

[paper wedge]

no, but for every element u can form an open ball around it

so more ilke

a union of open balls

something like that

best thing to do now is to do examples

lets take our set to be R , the real numbers and the metric to be defined as d(x,y) = |x-y|

[noble osprey]

ahh ok i get the definition

[paper wedge]

the usual distance

this is a metric space.

so now

answer this question:

how does the open ball centered at 2 with radius 4 look like

describe it as a set

[paper wedge]

like we did here

[noble osprey]

a| d(2,a) < 4

[paper wedge]

yes

now

u missed something very small

or its just more pedantic really

but u should say {a is in R | d(2,a) <4}

cuz we are in R

at the end of the day right?

makes sense?

[noble osprey]

ahh yea, so if we are in R^2 then a belongs to R^2

makes sense

[paper wedge]

okay now

u can do go further now

this d(2,a) <4

can u unfold the definition?

u know what the metric is.

i defined it above

what does d(2,a) <4 mean

given the d i defined above

"expand" it

[noble osprey]

d(2,a) = |2-a| < 4

[paper wedge]

good

okay

so now lets look at extreme cases ig

from them we can understand a little bit better ig

u know what an open set is now correct?

[noble osprey]

yep

[paper wedge]

what isi t

what is an open set inside our metric space

define it for me

again

[noble osprey]

so an open set is when you take an element x , you have an open ball around that x which is the center of the open ball with a radius r such that d(x,y) < r

[paper wedge]

good

but

thing is that r could be anything

we cant choose it

the definition tells us that THERE EXISTS an open ball

idk how small or how large the ardius could be

but i know there must be atleast one

now lets look at this example

and i want u to prove or disprove that this set is open

for example

is the set {1,2} open in our metric space?

why or why not

[noble osprey]

Ok so this is what i dont get, if we are given a set we know that it has to contain an open ball in that set for it to be open. but with repect to the usual metric say, d(x,y) = | 1-2| < r i dont know how continue from here.

[paper wedge]

okay

so

we only have two elements right?

1 and 2

[noble osprey]

oh wait is it because d = -1 and its not in the set {1,2}

[paper wedge]

so can u form a ball around 1 such that the whole ball is contained in{1,2}

?

if u can do that for 1 and u can do that for 2 then that would mean this is open

if no matter what radius u choose , a ball centered around 1 would always contain an element other than {1,2} then this is not open

do u get it?

[noble osprey]

yea i think i get it, as long as its centred around 1 then if we pick a radius around r that still contains in {1,2} then it would be open? i guess

around 1 sorry

[paper wedge]

well try it

try to pick a ball

choose any radius u like

and see if its still contained inside this set

[noble osprey]

so if i choose u=2 then {1-2, 2+2} = {-1,4} is not contained in {1,2} ?

[paper wedge]

whats u

[noble osprey]

r sorry

radius

[paper wedge]

okaay so u want to look at the ball centered at 1 with raidus 2

so {y | |y-1| < 2} right?

this is your ball?

[noble osprey]

yes

[paper wedge]

now is this balll contained in {1,2}?

by contained i mean is every element of this ball an element of {1,2}?

if no then show me one

pick an element in this ball that is not an element of {1,2}

[noble osprey]

oh so {1,2} is an open set

yes i think

[paper wedge]

i want u to justify it for me

for {1,2} to be an open set there must be an open ball around 1 such that the whole open ball is in {1,2}

and as well as for the element 2

but lets pick in general any open ball around 1 with any radius

[noble osprey]

cos if we pick y=1,2

it would be in the ball

i guess

[paper wedge]

okay

so ur saying that

both 1,2 are in your open ball

rihgt?

[noble osprey]

yep

[paper wedge]

now are those the only ones in ur open ball?

are 1 and 2 the only solutions to the inequality |y-1| < 2 ?

what about y = 0

ur confusing the set definitoins

the condition is that

EVERY element of this ball IS AN ELEMENT of the set

so if ur set is just {1,2}

then ur ball's elements must CONTAIN either 1 or 2 or both

its a SUBSET of the whole set

but u know now that y=0 is inside your ball

so the ball around around 1 with radius 2 IS NOT a subset of {1,2}

because 0 is inside that ball but its not inside {1,2}

hence it is not a subset

do u get it?

[noble osprey]

0 would fail since it’s not contained in the ball

[paper wedge]

no

0 is not contained in the set

{1,2}

it is in the ball ( satisifies the condition ) but its not in the set

hence the ball is not contianed inside the set

contained*

do u get it?

[noble osprey]

Ahh so {1,2} is an open set with open ball around 1 or 2

But 0 is not contained in {1,2} so we can’t form a ball around 0

[paper wedge]

no

an open ball around 1 with radius 2 is not contained in {1,2}

do u see why

[noble osprey]

Because it’s not contained in the set {1,2}

[paper wedge]

yes

it has an element that is not in {1,2}

so that would only SUGGEST that it is not open

it would not show that is not open cuz that is just one ball

who knows if maybe we chose too big of a radius

do u get it?

to show that this is not open we would want to argue that for any choice of radius

an open ball around 1 would always contain an element that is not 1 or 2

remember that we are over the real numbers

that is alot of numbers

haha

[noble osprey]

Ahh so we always look at the end points 1 or 2 and form a ball around it

I get it now [tears]

I’m so sorry for being stupid and thank you for your patience haha

[paper wedge]

no need

u werent stupid

anyways

is the interval [1,2] open?

[noble osprey]

No because we can find an element that is not contained in the set {1,2}

[paper wedge]

not {1,2}

[1,2] is the closed interval

this is another problem now

[paper wedge]

and that doesnt make much sense

u must mean that u can find an element such that there is no ball around that element contained in the set

[noble osprey]

Oh ok so what exactly is [1,2] ? Is that a subset of R ( real line) ?

[paper wedge]

its the interval

All real numbers between 1 and 2 including 1 and 2

[paper wedge]

yes

the dotts are the ball

And the center is 1

U can see no matter how we shrink the radius

It will always be outside the interval

No matter the radius

So its not open as for all radii ucant center a ball around 1 such that the WHOLE ball is contained in the interval

Got it?

[noble osprey]

Yup get it, does that apply to open interval as well (1,2)

I mean in this case the centre can’t be 1 or 2

I’m I right in saying that?

[paper wedge]

Yup

So what does that mean

[noble osprey]

Then it would contain within (1,2) thus it’s open

[paper wedge]

Yes

u got it

now notice that this depends on the metric choice

now

As for the finisher

lets now look at the same set but different metric

let it be the same set (R)

But define d(x,y) = 1 if x=y

d(x,y) =0 if x !=y

i want u to try examples with this metric

How does the open ball around any point look like? What are the open sets?

Good luck

this should answer ur original question on how like different metrics induce diff open sets

different*

[noble osprey]

In this case would it be open

[next crystal]

by composition do you mean the product on loops?

[plain raven]

Yes.

Of course this is not actually function composition. But the term "composition" is common in group theory to refer to the multiplication of a group.

[tulip bridge]

Does someone have a visual example of cell decomposition, For example, if I was starting out with a disc from which two discs were removed (sort of like a torus but with 2 holes instead) how could I decompose that?

Then from that how could I think find the euler characteristic?

[languid patrol]

Okay sorry that this looks like it was drawn by a child, I did it with a trackpad

Here is a decomposition of your space with 8 vertices, 10 edges, and 4 faces.

[languid patrol]

The red lines are to indicate the 2 disks which were removed.

The orange arrows are orientations

The green arrows indicate the orientations of the edges.

[tulip bridge]

Thanks! And no worries about the drawing, it helps a ton regardless. Do you mind describing how you went about doing it like that?

[languid patrol]

I basically just did it randomly, you can decompose it in any way and you'll get the same answer.

[tulip bridge]

Oh ok, my initial attempt was just straight across the middle with 6 vertices instead of 8, but I was unsure if it was necessarily “correct”

[languid patrol]

I think that should work as well afaik.

It depends on what you mean by cell decomposition. Sometimes there are extra rules if you want to compute simplicial homology etc.

[tulip bridge]

I mean I guess ultimately all the matters is I get the same characteristic, then they’re just homeomorphs of each other

Nah nothing that fancy

You had the right idea

My professor went at a snails pace so we didn’t get that far lol

[languid patrol]

Yeah, and in the end even that doesn't matter. For a "reasonable" space (i.e. anything that you can draw) you can just make the cells basically random and everything will work out.

In terms of computing homology or just the euler char.

[pseudo coral]

hey

for higher homotopy groups

all

[gentle osprey]

MyMathYourMath

[pseudo coral]

all

$\gamma:[0,1]^n \to X$ such that

[gentle osprey]

MyMathYourMath

MyMathYourMath

[languid patrol]

Yes, it is those things up to homotopy which fixes the boundary

[pseudo coral]

yes so all boundaries are x_0

[gentle osprey]

MyMathYourMath

[languid patrol]

The boundary of the image of \gamma is x_0

[pseudo coral]

yes

what i meant to say

[languid patrol]

but a homotopy is a map $H: [0, 1]^{n+1} \to X$ and it fixing the boundary means that $H(\partial [0, 1]^n \times [0, 1]) = x_0$.

[pseudo coral]

ok im doing a project on higher homotopy theory so i may be in this chat room a lot these next few weeks

[languid patrol]

ok

[gentle osprey]

Topos_Theory_E-Girl

[steel glen]

im not entirely comfortable with the construction of cohomology groups yet \
is there a general result we can get from $H_n(X) \approx H_n(A) \oplus H_n(X / A)$, given that $A$ is a retract of $X$ about the cohomology groups $H^n(X)$

[gentle osprey]

maximo

[steel glen]

im trying to show $H^n(X) \approx H^n(A) \oplus H^n(X / A)$

[gentle osprey]

maximo

[steel glen]

actually can i just use mayer-vietoris?

those are meant to be (X,A) not X/A sorry

[chrome ridge]

I am trying to show the bijection of isomorphism classes of covering spaces with isomorphism classes of $\pi_1(X, x_0)-sets$. I know that for any given a covering space we can get a $\pi-$set by considering the action of $\pi_1$ on the fiber $p^{-1}(x_0)$. Now I am trying to show bijection. In particular I am struggling with surjectivity. Any ideas?

[gentle osprey]

ru0xffian

[nocturne basalt]

i think quotienting by the action should give you a covering space

[thorn citrus]

I'm not sure there is a simple way to show the surjectivity.
Maybe the actual construction is needed.

For a given $\pi_0(X,x_0)$-set $A$, consider it as a discrete topology,
Let $Y$ be the set of all paths in $X$ starting from $x_0$, with a topology defined to satisfies
the map $\gamma \mapsto \gamma(1)$ from $Y$ to $X$ is locally homeomorphic.

Now my suggestion is:
take a quotient topology of $A\times X$ by the equivalence relation
$\gamma(1) ~ \gamma'(1)$, where $\gamma'$ is the path obtained by:

1. first lift any element of $\pi_0(X, x_0)$, starting from the $(id, x_0) \in A\times X$,
2. and then lift $\gamma$

I believe this construction might work, but please check it.

[gentle osprey]

etale

[fading vale]

hypothetically you could just take the universal coefficients theorem and use that it splits and that Ext and Hom factor through the direct sum but you can also just prove it directly using a basically identical proof to that of homology and its probably faster

[cedar pebble]

moth use ur szamuely knowledge

[fading vale]

EVIL

demon

[thorn citrus]

Agree

[chrome ridge]

could you elaborate more on that? Quotienting what exactly?

[cedar pebble]

it may help to talk about the universal cover

[fading vale]

im sure getting the equivalence between Cov(X) and pi_1(X, x)-sets via locally constant sheaves will really elucidate the subject 🥰

[thorn citrus]

Sorry for few mistakes in my writing.

Consider $(a, \gamma(1))$ and $(b, \gamma'(1))$ in $A \times Y$ whenever the following holds:

1. There exists $g \in \pi_1(X, x_0)$ s.t. $ga = b$.

2. Whenever you lift the a representative of $g$ to $A\times Y$ and then lift any element homotopic to $\gamma$ (with the same end point), the resulting path is homotopic to $\gamma'$.

[gentle osprey]

etale

[steel glen]

i did end up using the universal coefficients theorem with mayer vietoris

[thorn citrus]

I think everyone are saying the similar strategy, but the actual construction might bother.
It's hard to me to remember the exact one.

[fading vale]

yes this is the quotient by the universal cover in question

you could also go via gluing which might be slightly more explicit

[thorn citrus]

Right

[gentle osprey]

Moth

[fading vale]

This has not been topologized yet, to be clear!

[gentle osprey]

Moth

[fading vale]

Ok tbh im realizing I dont know quite how to make this construction work without the fundamental groupoid instead of the fundamental group Im sure i can but I will think about it for a bit instead of posting something wrong

[long grail]

An open set of a compact hausdorff space X can't be dense if it is missing isolated points, correct? For instance, in X=[1,2]U{3}, (1,2) cant be dense because the closure of (1,2) is [1,2], which doesn't include {3}. The reason I ask is because the boundary of a closed set is nowhere dense. But I read the complement of a closed nowhere dense set is open and dense. Where I'm having trouble is that the boundary of X is {1}U{2}U{3} the complement of which is (1,2) which is not dense from what I said above. Can someone tell me where I'm going wrong?

[thorn citrus]

I believe boundary of $X = cl(X)-int(X)=X-X=\phi$.
I guess you got the boundary of $X$ as a subspace of $\mathbb{R}$, which seems not relevant in this context.

[gentle osprey]

etale

[languid patrol]

So an easy argument shows that you can reduce to connected $\pi_1$-sets (i.e. those sets on which $\pi_1$ acts transitively) by taking disjoint unions of covers. Can you use the universal cover to construct all of those?

[gentle osprey]

Topos_Theory_E-Girl

[gentle osprey]

Moth

Moth

Moth

Moth

[long grail]

For a compact Hausdorff space, If C_1 > C_2 > ... is a nested sequence of closed sets, the intersection is non-empty, is that right?

[bitter smelt]

Yes. You don't need hausdorff. This is cantors intersection theorem

[noble osprey]

Hi for this prop 9.11 if X\V is open is that the same as X being open? ( but 1. says X is closed) Thanks

oh wait X\V a collection of open subsets yea?

[long prairie]

X\V is an open set, for some closed set V

if you put $V=\emptyset$, then you can see $X \backslash \emptyset=X$ is indeed a closed set.

[gentle osprey]

cflau_

[long prairie]

in fact, $X$ is both open and closed, and there's nothing wrong with that

[gentle osprey]

cflau_

[noble osprey]

Get it, thanks a lot

[abstract wigeon]

a grad student at my school told me that topology allows us to use open shapes to determine which points are near to each other without specifying distance, can someone help me think through why this is the case?

i would assume that two points are "near" if we can find an open set that contains both of them, but the whole space is open, and obviously the whole space contains any two points in it

so im not exactly sure the intuition behind this notion

[fading vale]

It's a heuristic notion of being near rather than a formal one

If I take the points 0, 1, and 2 in the real line then any open ball about 0 containing 2 contains 1 but not the reverse so in this vague sense 1 is "closer" to 0 than to 2

[abstract wigeon]

hmm okay

[fading vale]

But there are uncountably many open balls about 0 containing both so it's not really a strict definition

A useful intuition might be to think about Hausdorff spaces vs non Hausdorff spaces

In a non Hausdorff spaces you have points x, y such that any open ball about x contains y

So in this sense the points x and y are "so close together you cannot distinguish them topologically"

[abstract wigeon]

oh okay that makes a bit more sense

so in a general case, how can we determine whether two points are "near" or "far"?

or is this not the correct way to think about it?

[fading vale]

It doesn't really mean anything in an absolute sense beyond this separated vs non separated thing

[gaunt linden]

It's not really "nearness" you get to model -- it's more accurate that you get enough of the properties of distance to be able to define limits.

[fading vale]

At best one can get this relative notion of closeness as in this example (everyone open set about x containing z contains y)

Yes this is a better way of thinking about it

Honestly for most practical purposes whatever particular field you're working in will have it's own heuristic notions of what makes points close or near that you will practically use for intuition

So I think worrying about this nearness closeness idea is not super important in the point set context

[gaunt linden]

So in a topology you can say no matter how close you insist it has to be, there's something with such-and-such properties. But you cannot say "this is closer than that".

[abstract wigeon]

okay i think this makes more sense now

thank you both for your time 🙂

[chrome ridge]

Yeah I thought about this but also couldn't get the covering space from the universal cover; any ideas?

[languid patrol]

Did you at least show that you can reduce to constructing a cover for each G = \pi_1 set with transitive action?

[chrome ridge]

Isn't the idea if we show that for each orbit we get a path connected cover and then the cover for all of the set will be the space with all these path connected components ?

[languid patrol]

Yes exakt

*exactly

[chrome ridge]

Now wlog we can assume that our pi-set is transitive and we want to construct a cover (it's gonna be autmoatically path-connected)

[languid patrol]

So fix a base point for your G set

How can you write it using this?

[chrome ridge]

Not sure; but we want loops that lifts to loops in the covering space to have no effect on the base point. So do we consider the covering space whose fundamental group is the stabilizer of the point that we fix in the G-set?

[languid patrol]

Yes so this gives an equivalence between based coverings and based G-sets (which should’ve been part of your equivalence)

The other direction comes from looking at the monodromy action of pi_1 on the fiber, wrt the fixed base point in the fiber

[chrome ridge]

yeah but I don't see how the action that we get from the covering space that we found is isomorphic to the action that we're given on the pi-set

[languid patrol]

That’s almost by definition of X^univ/H (note that X^univ also comes with a based point in this based category)

The monodromy action on the fiber comes from the action of G on the fiber of X^univ above the base point of X

But if you didn’t prove this in class you can show it with path lifting

[chrome ridge]

I see. I like this approach better; we're utilizing the classifiaction that we have of path-connected covering spaces.

[languid patrol]

Yes. It's a nice way

Then the claim about isomorphism classes of unbased objects comes from forgetting the base points on both sides.

[solar grove]

İs every line in topology a set of infinite point with them and we technicaly can chance arrangement of point in set and set will technicaly be same.so same logic affect topology is it because why to diffrent lines with diffrent shape is counted as same

I think i wrote very complicated

But you got the point

Okey

İ will trying explain

So we have sets of 1,2,3

İf we mixed up like 3,2,1

it would be same set

So i am asking lines in topology is a sets of infinite coordinate of points

is that why it is diffrent lines in topology counted as equal

Not equals

Lets say only rule is a point not crosing with other point

[noble osprey]

Hi, for the preimage of e^3x it is h^-1({a}) <= 1 why is it not 2? Thanks

not <=2

[novel acorn]

what is a here?

[noble osprey]

so like the function X - Y ( In this case f : R - R) a is an element we choose in Y

[novel acorn]

OK but then how did you come up with the bound <=1?

[bronze wind]

Hey ! In topology we can define the notion of convergence in a very general setting (for a topological space (X,T)). But does all the type of convergence that we see during our studies always mean that there is an "underlying" topology within the space that we study ? For example we saw in class the convergence for distribution and our teacher said that indeed it corresponds to the notion of convergence that you would have if you put a (very exotic) topology on the space of distribution. In measure theory for example, is there a topology for almost everywhere convergence ?

I hope my question is clear

[novel acorn]

it actually isn't!

for other types of convergence (here specifically convergence in measure) you put topology of convergence in measure on your function space and that gives you said convergence
But I don't think there's a topology for almost everywhere convergence
This is because I think almost everywhere convergence has some issues if you try to put the topological form of convergence on it iirc

(But I think you can put a different structure on your space in that case but I'm not 100% sure what the details are)

https://mathoverflow.net/questions/255912/what-is-the-structure-associated-to-almost-everywhere-convergence

[shadow charm]

What are some homological tools to show two spaces aren’t homotopy equivalent when you know homology and fundamental groups agree? I don’t think something like local homology should be preserved under homotopy equivalences but can we do something similar that is?

[languid patrol]

Local homology isn't preserved by homotopy equivalences unfortunately.

[languid patrol]

Your last easy resort in this case is to use the ring structure on cohomology to distinguish them.

So show that the intersection pairing between cycles is different on one than the other.

[shadow charm]

Aight thought so

[shadow charm]

Hmm we haven’t started cohomology yet so I’ll have to think of smth else

[languid patrol]

Are you sure you're not being asked about homeomorphism? That would be easier and then local homology is preserved.

[shadow charm]

Yeah for sure, homeo would be easy in this case

[languid patrol]

Hmm shame that

[shadow charm]

Im looking at S^1xS^2 vs S^1 v S^2 v S^3

Intuitively speaking a homotopy equivalence would have to contract S^3

But idk how to phrase that

I was thinking maybe there was a way to say that a homotopy equivalence could be viewed as a homotopy in R^4 and use Jordan curve or smth to get a contradiction but that sounds whack

[bronze wind]

Thanks a lot !

[languid patrol]

Wait (I could be making a mistake) but it seems to me that here the universal covers are not homotopy equivalent: one is S^2 and the other is an infinite wedge of S^2's and S^3's?

[shadow charm]

Ah right im not used to thinking about covers

Isn’t the universal cover of S^1xS^2 RxS^2 though?

Not that it makes a difference

[languid patrol]

Yes, I was speaking up to homotopy equivalence.

[shadow charm]

Yeah okay

Alright that’s smart I’ll keep covers in mind from now on

We haven’t really looked at them yet so I’m not too comfortable with calling them up to my toolkit but by exam time we’ll have seen it so it’s aight

[languid patrol]

A fun example of spaces which have the same homotopy groups and homology but are not homotopy equivalent are a point and the "long line"

It is a stupid example though.

[shadow charm]

Ew long line

[languid patrol]

*Yay long line!

[shadow charm]

Good to know though

Thanks

[crisp helm]

To be honest, no idea where this belongs as it comes from a practical problem.

My friend is making a golf computer game with a twist: the user enters an RGB hex code, and that is where your ball gets putted to. You can see the flag you are trying to get to.

Since we are putting our ball on a 2D plane we want a projection from the cube of R,G,B values into the plane.
An important property that our projection needs to have is that if two colours are close in RGB space, they should be close in the 2D projection. It doesn't matter if far away colours are even further away than what they really "should be".
Is his game possible to make?
What sort of considerations need to be made for the projection?

Ideally he wants to have every colour represented in his game.

[abstract saffron]

Ever heard of space filling curves?

Hilbert's curve is your best friend

[solemn oar]

The next step would normally be to compute the cohomology rings and see if the ring structure agrees. If they do, you can try to distinguish them by looking at the action of the Steenrod algebra on the cohomology groups.

[obtuse meteor]

You can also look at the behavior of bundles and similar topological objects over these base spaces.

Ex. The topological K-theory

A general idea in this topic is: build an object homotopy-theoretically from your two spaces and then look at the pi1 / (co)homology of that instead

this is surprisingly powerful, because if the construction is complex enough, it might not be that the object's pi1/cohomology can be computed just from the pi1 / cohomology of the base space you started with

The simplest example being the loop space (but unfortunately this is terrible to compute, as it's higher homotopy groups)

[solemn oar]

You also have more complicated cohomology operations like the Massey product. For example IIRC the Massey product distinguishes the complements of Borromean and the Brunnian links, which the cup product is too coarse to detect.

[quick bough]

can anyone elaborate the definition of this, this seems so weird to me (at least the last part with the open cells)

[novel acorn]

take a bunch of points
Then attach lines between these points such that these lines are homeomorphic to open intervals
Then attach some open disks to this graph (you can imagine this as like gluing the ends of the disk along certain edges of the disk)
And continue this inductively

[quick bough]

yeah, i kinda got that part, but what is meant by the last point

(open cells)

[unreal stratus]

It's a definition

[novel acorn]

I think this has some formal consequences somewhere so it's better that the cells are open

something something this makes the pushout work

tho like idk what exactly is weird abt it

[plain raven]

is it confusing because they're not literally open

[abstract saffron]

That's Hatcher

I guess one of the many reasons is the glueing lemma

Similar to how you breakdown simplicial complexes into simplices. Here you break down the spaces into a bunch of balls. Why balls? Because they're simple and nice.

But don't quote me. I've fought with this for days. Still haven't got it.

[crisp helm]

I'm not so sure, the hard part of this question is defining what colours go where since we want similar colours to be placed in similar positions spatially

[plain raven]

Is it?

[quick bough]

i’m a bit confused on what the quotient map should be in the first place

in this case (not the definition)

[nocturne basalt]

you quotient by the attaching map on the n-1 skeleton

[unreal stratus]

Basically the point is like if you attach a cell D^n to a space X to form Y, then there's a canonical map D^n -> Y

Since you attach it via the boundary, this map is a homeomorphism onto its image when you restrict to D^n \ boundary

[quick bough]

ohhh, okay, got it now

thank you

[civic verge]

guys a question is it true that the

$\partial X=X'-X$

[gentle osprey]

Awuita Fria

[civic verge]

I was thinking that no

[unreal stratus]

You're correct. For example note that if X is closed, then X' is contained in X, so X' - X is empty

[civic verge]

Thanks, I will do the demonstration

I'm just not sure about some things but I'm going to do it hehe.

[civic verge]

Let $A$ be contained in $R^{n}$, If $A$ is closed then we say that $a$ belongs to $R^{n}$, suppose that $a$ is an accumu
lation point of $A$ where for any given epsilon, there exists a y that belongs to $A$ such that y belongs to the Ball of center a and radius epsilon.

We know that A is closed, this implies that $A'$ is contained in closed $A$, so the boundary of A is equal to empty since $a$ is an accumulation point, therefore it is false

[gentle osprey]

Awuita Fria

[civic verge]

look

[grim knot]

Hey guys, can someone help me get an intuition of retracts, deformation retracts and strong deformation retracts? I somehow can't quite imagine how they could work

[languid patrol]

Have you never seen examples of them? An example which contains all 3 types of retract is the following, let $X = {(x, y) \in \mathbb{R}^2| 0 < x^2 + y^2 \leq 1 }$ then we can parameterize $X$ in polar coordinates by $f(r, \theta) = r(cos(\theta), sin(\theta))$ for $0 < r \leq 1, 0 \leq \theta < 2\pi$. Let $R$ be the map which sends $f(r, \theta) \to f(1, \theta)$, then $R$ is a retract of the inclusion of the circle into $X$. On the other hand $H(f(r, \theta), t) = f(r(1 - t) + t, \theta)$ is a (strong) deformation retract. Intuitively a deformation retract is a space where every point can be moved along a path to lie in a subspace in a way where the paths vary continuously, and a retract is a space where every point can be assigned a point in the subspace continuously, but not necessarily moved there along a path in such a way that the paths vary continuously. Perhaps the hardest thing to understand is a strong deformation retract, because most deformation retracts which are geometrically intuitive are strong deformation retracts.

[gentle osprey]

Topos_Theory_E-Girl

[lime sable]

hmm i'm wondering if there's some niceness condition on a pair for deformation retracts to automatically be upgraded to strong deformation retracts

nlab and hatcher are of no help, although nlab gives a fancy definition for a strong deformation retract

[languid patrol]

nlab is only good for learning about Hegel https://ncatlab.org/nlab/show/Science+of+Logic

[willow vapor]

how can I understand an identity map between 2 metric spaces?

like, I get that every element probably gets mapped to itself, but how do the metrics play a role?

like say from $(X,d_\infty)$ to $(X,d_2)$

[gentle osprey]

WasMachenWirDennDa

[lime sable]

lawvere was so weird

i guess hegel figured out what an adjunction was over a century before mathematicians did

[languid patrol]

I think that's pretty charitable, but what do I know.

[umbral panther]

Sure, if it’s a relative cell complex. X is a CW complex and A is a subcomplex. Or more generally if X is formed by adding cells to A without hypothesis on A. This is also studied under the name cofibration.

[lime sable]

i'm familiar with cofibrations, but what part of the cofibration is used here

[languid patrol]

The definition

[pastel linden]

Is there a global G-action on the associated bundle to a principal G-bundle?

[languid patrol]

The associated vector bundle you mean?

[pastel linden]

Yes, sorry

(Nontrivial action and associated bundle)

[languid patrol]

So given some representation (\rho, V) the associated bundle is V \times^G \mcal{G} so there is a canonical G action coming from the one on either side.

[pastel linden]

Could you elaborate? I was under the impression we have
((g_a ○ g_b^-1)(x, v)) ● h = (x, r(g_ab)v) ● h = (x, r(h^-1 g_ab)v)
(g_a ○ g_b^-1)((x, v) ● h) = (g_a ○ g_b^-1)(x, r(h^-1)v) = (x, r(g_ab h^-1)v)
which isn't global

[languid patrol]

I don't know what you mean by not global

Also it would be nice if you could TeX what you write so I can read it more easily

[pastel linden]

As in, if we define it as $(x,v) \cdot h = (x, \rho(h^{-1})v)$, this is not independent of the trivialization

[gentle osprey]

bacono

[lime sable]

as in the homotopy extension property, but for some stupid reason i don't know what i plug in

so that's what i'm asking

[languid patrol]

The retract

[pastel linden]

Also, the global G action on P becomes trivial when we pass it down to the associative bundle by construction

[languid patrol]

No

It is equalized with the left G-action on V.

The action which becomes trivial is the action $g*(v, s) = (gv, sg^{-1})$

[gentle osprey]

Topos_Theory_E-Girl

[pastel linden]

What do you mean by the "it is equalized by the left G action"?

I'm not familiar with that terminology, sorry

[languid patrol]

But yes this action is not a "global" one, in general there cannot always be such a "global" action even for $G = Gl_n$.

[gentle osprey]

Topos_Theory_E-Girl

[languid patrol]

I mean what I LaTeXed

[lime sable]

so it's actually "every retract is gives a strong deformation retract" not "every deformation retract is gives a strong deformation retract"

[pastel linden]

Thank you, so there is not a global action, makes sense

[languid patrol]

So if G is abelian this exists, but for general G even for G = Gl_n you can see that such an action can't exist in general

[pastel linden]

Thank you! This is what I've been thinking but i thought i was going insane

[languid patrol]

No worries! If you're working with a category of algebraic-looking objects (varieties, complex manifolds, etc.) I can even write down an example for you. But you probably already know one.

[lime sable]

so being a cofibration is sufficient, but what is equivalent? that's what i'm asking

[languid patrol]

Yep it is stronger!

[pastel linden]

Yeah, this came up in the context of nonabelian gauge theories in physics which involves connections on associated bundles to principal SU(3) things

Physicists like to pretend that global actions on sections exist (when in fact, they do not)

[languid patrol]

Yeah that is very strange. Perhaps they are too used to abelian gauge theories.

[umbral panther]

I was answering the second

Strong def retr is a combination of equivalence and nice. Equivalence is a substantial property, rather than a technical property. Asking for nice to imply equivalence is asking for a lot. If every retract of X is equivalent then every point is, so X is contractile. Probably “absolute retract” is a sufficient hypothesis

[chrome ridge]

Orbits of the action on deck transformations on the covering space are of the same sizes?

[languid patrol]

Not in general, what are your hypotheses?

For instance if X_2 is a double cover of X then X \coprod X_2 has two orbits for its deck transformations of different sizes.

[chrome ridge]

Very vague intuition coming from the fact that only the identity transformation fixes a point.

But I guess the set on which we are acting is not finite so it is not necessarily true that orbits have the same sizes

[solemn oar]

If your space is connected (plus the other usual assumptions) it's true.

[pseudo coral]

question, how do i prove the case riguourusly that $\pi_n(X)$ is abelian for $n \geq 2$ specifically the case $n=2$

[gentle osprey]

MyMathYourMath

[pseudo coral]

for $n=2$ can this be made rigorous easily?

[gentle osprey]

MyMathYourMath

[pseudo coral]

let $\alpha, \beta \in \pi_2(X)$ for some top space $X$ then $\alpha,\beta:(S^2,a) \to (X,x_0)$

[gentle osprey]

MyMathYourMath

[pseudo coral]

is it the extra dimention that allows for this "Hop" so i can commute them? i can see it but

[fading vale]

There are quite a few ways to do it