point-set-topology

[nimble portal]

Then consider (n+1)-balls of radius r and r', respectively, around x & y in R^(n+1), such that B_r(x) is disjoint from B_r'(y)

Then pi(B_r(x)) and pi(B_r'(y)) are disjoint, and since their preimages were open, that means they're neighborhoods of x & y?

Is that correct? If it is then I believe that would show Hausdorff right?

[coarse night]

no

how would you control the size of the balls

[languid patrol]

You can embed any manifold in R^n for some n. In particular RP^k embeds very easily into R^{(k+2)C2}

If it helps you visualize (probably not) it's the image of the sphere in $R^{k+1}$ under the map $(p_0, \dots, p_k) \to (p_0^2, p_0p_1, \dots, p_n^2) $

[gentle osprey]

Topos_Theory_E-Girl

[unreal stratus]

It doesn't immediately follow that the images are disjoint (indeed otherwise this argument would show every quotient of a Hausdorff space is Hausdorff)

But you can make that the case

[coarse night]

Shrink deez nuts

[peak isle]

is di suppose to be a face inclusion

and any know the difference of face inclusion and simplices i just know that face inclusions are combinations of smaller simplices

[languid patrol]

d_i is the ith face of an n simplex, which is an n-1 simplex

The idea is that the faces of an n-simplex are n-element subsets of {0, 1, \dots, n}, each one can be identified with the n-element subset which comes from forgetting the element i.

[peak isle]

oh ok thanks 1 question i just want to clarify cz it hasnt come up on defintions but i think i know what it is

the ker is just kernel right

[hidden crag]

Yes

[nimble portal]

Why not?

Or rather

OHHHH

I see

Okay sorry

Hm

My thinking was since the preimage of each on their own is disjoint

Then the preimage of their intersection is equal to that

[nimble portal]

This isn’t analysis why do I care about controlling the size

All I care is that they’re disjoint

Fine say r < r’ s.t B_r and B_r’ are disjoint

[nocturne basalt]

In Hatcher's proof of the ring structure of H^*(RP^n; Z_2) (denoted with P^n) he says that naturality of cup product makes this diagram commute but I am not sure why since it doesnt seem like it is induced from any map

like i guess the vertical maps come from the LES of pairs but they come from different long exact sequences right?

[languid patrol]

The cup product is pullback along the diagonal.

• Kunneth isomorphism.

So all morphisms in that diagram on cohomology come from a commutative square of maps of pairs.

[languid patrol]

The vertical maps are just (X, \emptyset) \to (X, Y) and the horizonal maps are just \Diag on the top and \Diag_X \times \Diag_Y on the bottom.

[nocturne basalt]

how is the bottom \Diag_X \times \Diag_Y?

[languid patrol]

(X, Y) \to (X \times X, Y \times Y)

The map of pairs I was writing as a product.

Perhaps you would like to think of it as (\Diag_X, \Diag_Y)

[languid patrol]

But it's just a notational difference.

[nocturne basalt]

ohh wait i get it nwo

thanks

[nimble portal]

I was never good at this stuff ugh, I can’t figure out how to get it

[nimble portal]

I still don’t understand why this is a problem, can someone explain? Isn’t the size of the balls controlled by saying “make them small enough so they’re disjoint”?

Or I guess that sets an upper bound on the radii, you can just say the radii can’t be zero to get a lower bound then it’s all gravy right?

[languid patrol]

Okay I think you are maybe missing geometrically what is going on with projective space

[nimble portal]

Indeed, I think I said that before

[languid patrol]

So in projective space we are taking points in R^{n+1} - {0} and identifying them if they lie on the same line

i.e. nonzero vectors x, y are equivalent if x = \lambda y for some nonzero \lambda.

[nimble portal]

Yep

[languid patrol]

So we can quotient in stages, first we say x ~ y iff there exists a positive \lambda such that \lambda x = y

Can you see what you get from taking that identification?

[nimble portal]

You get half the line

[languid patrol]

Hint: try to find a nice subspace of R^{n+1} such that every nonzero vector has a unique positive multiple that lies on that subspace

[nimble portal]

S^n? 🤨

[languid patrol]

Yes

[nimble portal]

Then you do the same with the negative lambda yes

[languid patrol]

So before we quotient to RP^n we can quotient first to S^n, so RP^n is just S^n with opposite points identified.

So think of the sphere but you identify the north and south pole, Ecuador with singapore, etc.

[nimble portal]

But why do we need to do that? What’s wrong with just considering it “the set of all lines that cross the origin”

[nimble portal]

Yep antipodal points are equivalent

[languid patrol]

I think this makes it easier to see what the charts are.

[nimble portal]

Okay

[languid patrol]

Because now a chart is just a neighborhood of a point on the sphere which is too small to contain any antipodal points.

So for instance any acute circular cap of the sphere containing a given point maps homeomorphically onto a coordinate patch in RP^n

[nimble portal]

Don’t you get the same thing if you just call it “the set of lines crossing through the origin”

Just slap a sphere on that bad boy and look at their intersection

[languid patrol]

I mean there are many ways to describe any space.

[nimble portal]

I understand this is for the purpose of making the charts but I haven’t even been able to show the other two conditions first, why am I worried about the charts right now?

[languid patrol]

I just think this way makes what is going on much more intuitive than taking a quotient of an n+1 dimensional ball in n+1 dimensional space minus the origin and arguing that that is an n-ball

[languid patrol]

What other two conditions?

[nimble portal]

Hausdorff & second countable

[unreal stratus]

Something much stronger than second countable is true, to be fair, which is good to show

[languid patrol]

Well okay I claim that both of those are very obvious from this description of the open sets as open sets of the sphere.

[unreal stratus]

You gave very nearly correct proofs that it's Hausdorff feather

[nimble portal]

Just the image disjoint thing is wrong right?

[languid patrol]

I've given you the tools to, given two points, explicitly construct two disjoint neighborhoods around those points using similar arguments to what you were saying before. Similarly for second countability.

[nimble portal]

Hm

[unreal stratus]

Well given distinct [x] and [y] in RP^n, you need to pick neighbourhoods of x and y whose images are disjoint

But that is okay if you think, say, of identifying antipodes on S^2 (then generalise)

[nimble portal]

If you have x & y

Consider one of their antipodes, say -x

Nope never mind

[languid patrol]

I would opine again that maybe with your background instead of reading a very difficult and abstract texts on manifolds to start learning differential geometry/topology it might be better to read something more down to earth like: http://www2.ing.unipi.it/griff/files/dC.pdf which will motivate the general definitions but be more accessible to just calculus and linear algebra.

[nimble portal]

[languid patrol]

It depends on where your interests in this subject really lie.

[nimble portal]

This is fun though

[languid patrol]

So open balls around [x] in RP^2 correspond to a pair of open spherical caps of the same size around a point x and it's opposite -x.

[nimble portal]

Mhm

And you have the same for y & -y

[languid patrol]

So can you see why if [x] \neq [y] you can pick the four caps to be disjoint from each other?

[nimble portal]

Yes

[languid patrol]

Okay then you're done

[nimble portal]

Wat

[unreal stratus]

Lol

[languid patrol]

My sister in christ maybe try to state clearly and in a totally dry, formal way what you are trying to prove.

[nimble portal]

being dry

[languid patrol]

Like unpack the definitions to make "RP^2 is hausdorff" a more accessible geometric statement.

Without turning it into something not even wrong.

[nimble portal]

Can I maybe

Draw a pic of what my intuition on it is and how I was trying to make a proof out of that

[languid patrol]

Okay post that, but first I think it would be a good exercise to state clearly in a simple and perhaps unintuitive way what you are trying to prove.

[nimble portal]

I’ll try again once I’m at uni

[nimble portal]

agh okay this is p whiny from me (to no one’s surprise) but I just cannot prove things if they’re not intuitive to me (and 90% of the time I can’t prove them even if they are ). I understand the point of a proof is just to show logical correctness, not necessarily reveal any insight or intuition, but if my proofs don’t do that then they just don’t feel satisfying to me you know?

[languid patrol]

Yes I think you have to make statements intuitive to you and in the end all mathematicians argue based on intuition. But if you are not also able to state things formally/simply you can easily fall into traps where you can't even apply your intuition because you don't actually know what you're trying to prove.

It's a bit like saying "I like music so I want to play the piano, but I don't like all of this stuff about wrist posture it's too boring."

You need to learn at least a little bit of rigor/technique to allow you to use your intuition or you'll just go in circles.

Like in the end the proof I gave you of this statement is very intuitive/nonrigorous, but because I know how to state and understand the formal statement of what you are trying to prove in my head I am very satisfied with that proof, and it would satisfy any mathematician.

Whereas if you don't write things out clearly you will constantly be making little mistakes or not understanding what exactly needs to be shown, even if your intuition/your heart is in the right place.

But it doesn't mean that mathematics has to be dry/unfun. Just that you maybe would benefit greatly by trying to do a couple of things in a slightly more rigorous way than you have been.

[nimble portal]

Something like that lol, if we make the balls around x & y small enough then all the lines going through the points in those balls would be disjoint from each other

[languid patrol]

Yes I think that is satisfying. As long as you feel like if pressed you could reduce this to something very boring about the triangle inequality or what have you.

[nimble portal]

Oh

Uh

Okay so let’s state the problem again clearly

Given two distinct elements [x], [y] in P^n, I want to show that I can always construct disjoint neighborhoods around them. That is, I can always find two disjoint sets U, V around [x], [y] such that U and V are open in the quotient topology; that is, the preimages of U and V are open

[languid patrol]

Yes so what does it mean that U, V are open and disjoint in the quotient topology?

[unreal stratus]

[x], [y] aren't representatives

[languid patrol]

In terms of subsets of R^{n+1}

[nimble portal]

Oh sorry x and y are

[nimble portal]

It means that we can find disjoint open neighborhoods of x & y in R^(n+1) that map to disjoint neighborhoods in P^n

And the way I wanted to go about this way by saying take x & y and construct n-balls around them that don’t intersect, which I could formalize as saying make a ball B_r(x) around x and B_r’(y) around y, with r < r’, right?

such that neither is within one another

[languid patrol]

I mean technically you are right, but what does it mean to map to disjoint open neighborhoods in P^n?

There is a simpler way to express this.

[nimble portal]

That the neighborhoods in R^(n+1) are disjoint, no?

[languid patrol]

No for instance a ball of radius a half around x and a ball of radius a half around 10000x are both disjoint in R^{n+1} but not disjoint in P^n

[unreal stratus]

Otherwise all quotients of Hausdorff spaces would be Hausdorff

[nimble portal]

But we’re doing it on separate x & y

We can’t use x and 10000x because [x] and [y] are distinct

[languid patrol]

You should be able to see that even then there are problems.

[unreal stratus]

What if you use the balls tteg gave but move y to like 10000x plus something smol

For example

[nimble portal]

Well you just need to make your radius for the ball around y smaller than the distance between 10000x and y

[unreal stratus]

Yes

[nimble portal]

Then 10000x isn’t in this ball

[unreal stratus]

That is the key point

[nimble portal]

Isn’t that what I said…?

[unreal stratus]

No

[nimble portal]

Why not?

That’s what I’ve been trying to do this entire time and what I thought I was saying lol

Damn it

[unreal stratus]

Because we've shown the neighbourhoods in R^n+1 being disjoint isn't enough

[unreal stratus]

Which is what you said was enough here

Like you have to make the balls small enuff in a suitable sense

[nimble portal]

Oh I see the difference now

Can I draw it just to make sure I see it right

[languid patrol]

Sure

[nimble portal]

I’ve constructed disjoint neighborhoods in R^(n+1) but clearly in projective space they’re not disjoint

[languid patrol]

Yes this is exactly the example that Potato was mentioning.

[nimble portal]

Yep

I see where I went wrong

[unreal stratus]

Kinetics and Dynamics

[nimble portal]

Kinematics

[unreal stratus]

Yeah idk why I said kinetics

Lmao

[nimble portal]

So I need to construct the neighborhoods so that the balls around lambda x and lambda y are always disjoint

Not just x and y

[languid patrol]

Well

No

Stronger you need that for every pair of points in those neighborhoods no nonzero scalar multiples of those points are equal.

[nimble portal]

Ah

[languid patrol]

So really maybe it's better to only look at neighborhoods in the shape of a pair of opposite cones minus the origin, i.e. the neighborhoods closed under scaling by nonzero scalars.

And this is why visualizing these things on the sphere is equivalent but much easier

[nimble portal]

Wait

Because if we’re on the sphere

Then x & y are a fixed radial distance away from the origin

So I can just create disjoint neighborhoods and not have to worry about the scaling issue

[languid patrol]

Yes then the only scalar you have to worry about is +/- 1

[nimble portal]

I see

The reason I’m scared of the sphere argument is just because it didn’t come to me naturally

[languid patrol]

So B_r(x) and B_r(y) are disjoint in the image iff B_r(x) \cap B_r(y) is empty as is -B_r(x) \cap B_r(y)

[nimble portal]

I see why it’s better now

[languid patrol]

It's just a nicer way to write these things because you remove some redundant degrees of freedom

[nimble portal]

It’s just scary to me to use a technique/perspective I couldn’t come up with on my own, the idea of quotienting with positive scalars first then negative

[languid patrol]

I'm sure you are capable of coming up with that on your own.

[nimble portal]

Maybe that’s because this is the first time I’ve seen a quotient space tbf

[nimble portal]

Yeppers

[languid patrol]

It's generally a good idea to factor quotients through several steps if you don't find them intuitive.

Or at least to try to find some way of breaking them up into more workable pieces.

[nimble portal]

Just trying to come up with those neighborhoods now

If x & y are in the same hemisphere you could just make the radius half their distance

But the trouble is like when x & y are opposite, then the balls around -x & y would intersect

Could you just say that lol? Suppose WLOG that x & y are in the same hemisphere, then choose r < (1/2)d(x, y)

[languid patrol]

Yes just take the minimum of d(x, y) and d(-x, y)

[nimble portal]

RGONSOURNGOUSRNGOJRSN

Why am I so stupid man fuck 😭 yep

[languid patrol]

Okay but now you should try to think why RP^n is second countable!

[nimble portal]

Yeah

Let’s see

Second countable means we have a countable basis for the quotient topology

Which means there’s a countable family of sets in the quotient topology such that every open set can be obtained as a union of sets in this family, right?

[languid patrol]

Yes that is one characterization.

[unreal stratus]

Every open set can be uniquely written as a linear comb- oh wait

[nimble portal]

An open set in projective space is a set whose preimage in R^n is open

So I need to find a countable set so that the preimage of any union in the set is open?

[languid patrol]

Yes so maybe can you see why it suffices to show that R^n - {0} is second countable by using the fact that the quotient map is open (sends open sets to open sets)?

[nimble portal]

Yes

[languid patrol]

Okay so why is R^n - {0} second countable?

I will point out the following helpful equivalence: on a metric space M a set of open sets U_i is a basis for the topology if for every x \in M and every r > 0 there exists a U_i such that x \in U_i \subset B_r(x)

[nimble portal]

Well R^n is definitely second countable and R^n\{0} is a subset of R^n so couldn’t you just use the same basis

[languid patrol]

Yes you can use the same basis and intersect with R^n - {0}, but why is R^n definitely second countable?

[nimble portal]

Well take our basis to be all balls with rational radii around rational points

This set is definitely countable and we can cover R^(n+1) with these balls

[languid patrol]

Just showing that you can cover R^{n+1} is not enough of course.

[nimble portal]

And since the quotient maps open to open

Then just take our basis for proj space to be pi(basis for real space) no?

[languid patrol]

Yes, I think it's good to verify from the definitions that the image of a basis under a surjective open quotient map is a basis.

[nimble portal]

Because then any open set in proj space can be written as a union of those guys which are all open and since the original basis was countable so is the new one

[nimble portal]

Will do

[languid patrol]

That is true but you should justify this rigorously just for the future.

[nimble portal]

By which you mean prove this right

(which should just be chaining together some stuff from definitions)

[languid patrol]

Yes it's a formal exercise but it's good to convince yourself that it's true.

Another thing that you can prove is that RP^n is actually compact (much stronger than second countable) from the fact that S^n is compact.

Which is what potato was alluding to

But I think it's good to practice proving that something is second countable at least once or twice rigorously.

[nimble portal]

I'm not too far from finishing the chapter

I think I'll get to that and then make some exercises for myself

Or should I just pause here before I get to smooth atlases and start exercises?

Ngl the problems at the end look really fun though

I'd rather do those 😛

[languid patrol]

Do what you want as long as you do a lot of examples and make sure you see abstract definitions "in the wild"

[tidal lynx]

Performing multiple quotients

[languid patrol]

For every point set topology thing you see you should try to think of one space that satisfies it and at least two that dont

And justify rigorously why

[nimble portal]

LOL

Oh no

It's analysis all over again

Counterexample after counterexample after counterexample just tearing apart any happiness I had

[nimble portal]

Is there such a thing as a manifold that evolves with time?

So as time passes the geometry of the manifold does too

The reason I ask is because my fluid mechanics professor does research with the velocity profiles of surface waves (or something like that)

[novel acorn]

✨ homotopy ✨

[nimble portal]

And I remember part of what he does is establish local coordinates on the surface of the wave

And then do everything wrt those (or something along those lines)

[novel acorn]

or you can also do surgery to change your manifold tho that is more complicated

[nimble portal]

So could waves be thought of as an ever-changing manifold in R^3?

[nimble portal]

Eh?

That's a...curve...oh wait

LOL

[unreal stratus]

Homotopy ain't just about curves

[nimble portal]

I wouldn't know

To me it's just rubber band go brrrrr

Pull one thing into another

Are analytic manifolds a lot nicer than just smooth manifolds?

Does this come from second-countability?

[novel acorn]

ermm no

[nimble portal]

[novel acorn]

it comes from the fact that it's locally euclidean iirc

[nimble portal]

I see

Mmm

[novel acorn]

Oh no

it does come from second coutability lol

I must have been thinking abt orientability

[hidden crag]

yes

[novel acorn]

sorry

[nimble portal]

Is it something like

[hidden crag]

but feather the proof for this is right below that proposition

[nimble portal]

I know

[hidden crag]

in the very doc you're reading

[nimble portal]

I want to try to think about it first before I read

[hidden crag]

and the first sentence relies on second countability

[nimble portal]

Noted

[hidden crag]

thinking about it -> reading ->asking questions

[nimble portal]

Alright 😛

[novel acorn]

I don't do the first step

[nimble portal]

Well

I got the first sentence of the proof 💀 but that was nothing like what I was expecting

I see why what I wanted to do wouldn't work though

[plush crescent]

60% 1st step,5% 2nd step and 35% 3rd step

[nimble portal]

What are the "components" of a manifold? The domains in a cover?

[fading vale]

Probably it's connected components

[coarse night]

Have you seen fundamental groups before?

[nimble portal]

Sort of

Not really

Just seen some examples with the torus & S^1

[nimble portal]

So all the path-connected points in the manifold?

[unreal stratus]

That isn't what a connected component is in general

[nimble portal]

for some reason I forgot "connected" was a definition in its own right lmao

So the subsets of the manifold that can't be represented as a disjoint union of two sets

[nimble portal]

But more generally what is a "component"

[sudden spire]

Yes this is the whole sub field of taking various flows

[nimble portal]

I think they just mean subsets that cover the space

[gritty widget]

"component" always means connected component

[pearl holly]

I prefer sleep -> sleep -> sleep

[nimble portal]

Isn't that circular 😭 what is the "component" in "connected component"?

I think d gives the definition

[gritty widget]

?

i see no circularity at all here...

d is not a definition

[nimble portal]

Component means connected component

Oh oop

So connected component means a subset that's connected?

[gritty widget]

it is more than that

[nimble portal]

What else is it?

[gritty widget]

open a topology textbook please

[nimble portal]

Well there we go

[plush crescent]

For someone that has only seen the basics of topology I can't think of something that's open and closed at the same time besides the empty set

[nimble portal]

I don't understand this part about special loops

[gritty widget]

the whole space, as well

[nimble portal]

Well open and closed depend on the topology

Take a set X with the discrete topology, if you take any subset of X then it's closed since its complement is in the topology, but it's also in the topology, so it's open

[plush crescent]

Yes I forgot about the 2nd trivial example ty

[nimble portal]

You can think of plenty of nontrivial examples with small sets too

[pseudo coral]

think of a 3 point set

a,b,c

[nimble portal]

Yep that was my idea too :)

[pseudo coral]

lol

[gritty widget]

this would be a good moment for you to give an example about connected components, feather

[plush crescent]

Lol

[nimble portal]

What

You mean like how the discrete topology isn't connected?

[pseudo coral]

i ALWAYS forget, discrte is every thing open and indisctret is only X and 0

right

[nimble portal]

Or the trivial topology

[pseudo coral]

discrete is all (sub)sets are open and thus closed right

and indiscrete is space where only X and \emptyset are open

[untold lily]

discrete topology is what is induced from the discrete metric, which is considerably more intuitive at least for me

[pseudo coral]

good way of viewing it

i have a question, how do i rigorously prove that a map S^m \to S^n is homotopic to constant loop when m<n ? Hatcher has the MOST hand wavy proof of all time for this fact i believe

not a big fan of Hatcher lol he leaves out tooo many details lol and does tons of proof by picture LOL

can I prove it by supposing merely that n=m+1?

i wanted a proof that doesnt rely on CW complexes if possible

[plush crescent]

Could you please share us the proof?

[pseudo coral]

yes here

[nimble portal]

So the idea is that we just take a loop inside every open coordinate ball covering our space (and their intersections)

Since there's countably many paths in any one ball then we have countably many special loops inside a ball

[hidden crag]

you'd need to share the proof below

[nimble portal]

And it doesn't matter which point in our ball we compute the fundamental group at because they're all isomorphic in the same ball

[hidden crag]

or what exactly are you confused about

[pseudo coral]

finding the homotopy from S^m \to S^n and the constant loop

[hidden crag]

if your map is homotopic to a cellular one then it is homotopic to a non surjective one in your case which means that it's homotopic to a constant map by factoring through R^n after removing the point that does not get hit

[hidden crag]

this makes it unnecessarily hard

the hard part here is proving cellular approximation

[pseudo coral]

theres a hint in intro to top by Gamelin & Greene that says any map f:S^m \to S^n can be approximated by a map g:S^m \to S^n such that the range of g is a proper subset of S^n

[pseudo coral]

is this ^^ what you meant

[hidden crag]

yes that follows by cellular approximation

S^n has higher dim. cells

so there are cells that can't be in the image of your cellular map from a space that only has lower dim. ones

Try proving that every non surjective map X -> S^n is nullhomotopic and then applying cellular approx. to show that every map S^m->S^n is homotopic to a non surjective one

[civic verge]

Let A be a compact set and $(x_n)_n\in\mathbb{N}$ be a sequence in A, then $(x_n)_n\in\mathbb{N}$ has a substring convergent to A.

Guys, could someone tell me what is the first step for this exercise?

[gritty widget]

checking your definition of "compact set"

[gentle osprey]

Awuita Fria

[gritty widget]

"convergent to A" is meaningless. do you mean "convergent in A"?

[civic verge]

I was thinking well if A is bounded then be a sequence of A as the sequence is bounded in A then by Bolzano's theorem it has a convergent substring let's say x_n_j so the substring converges to a point "u" we can say that A is closed since it has a substring that belongs to A for all j that belongs to the naturals then u belongs to the set A so A is compact.

[nimble portal]

What is the point of the g_i business here?

[civic verge]

I don't know, if my idea is right :c

[civic verge]

Sorry, but I use translator not English
I hope you will excuse my lack of writing

[nimble portal]

Why is the first equivalence not enough? We showed that we can represent a loop in B as a finite product of paths of elements in B

Oh

I should've looked at the picture first

I was picturing it wrong

[tulip bluff]

Hi! Is there some easy way to see that R^k \times S^{n-k} is homotopy equivalent to S^{n-k}?

[gritty widget]

it seems pretty easy to check right from the definition of homotopy equivalence

[tulip bluff]

Is this due to R^k being contractible space? I did wonder whether there is a result on the product of contractible space and non-contractible space.

[gritty widget]

yes

[civic verge]

[gritty widget]

contractibility of R^k is useful. more generally, if X is any space and Y is any contractible space, then are X x Y and X homotopy equivalent?

that would be the immediate generalization of your question

[obtuse meteor]

I would use Sard's theorem to grab something useful

[gritty widget]

inb4 not smooth

(yes i know you can homotope to a smooth map)

[obtuse meteor]

¯_(ツ)_/¯

The fact you can homotope to a smooth map is an axiom

[tulip bluff]

Is this a true statement or are you conjecturing this?

[gritty widget]

why don't you find out by trying to prove it?

[obtuse meteor]

alternately you can use CW approximation

[tulip bluff]

I should. I'm being lazy and not wanting to prove something that has the possibility of not being true...

[obtuse meteor]

which is easier to prove sort of ~algebraically~

[cedar pebble]

use noggin

[gritty widget]

i would never say false things

never!

it's true, don't worry

[tulip bluff]

Off I go then haha

[cedar pebble]

[gritty widget]

ans: ||if the identity map on Y is homotopic to a constant y_0, then consider the maps X x Y -> X given by projection and X -> X x Y given by x -> (x, y_0)||

[pseudo coral]

ive heard of this it says something about the measure of critical point sunder a given map?

[pseudo coral]

im having a tough time wraping my head around this notion of a CW complex lol

does anyone have intro to top by gamelin and greene handy? If so, in chapter 4 section 4 are they using differential forms to prove the theorem????

[coarse night]

never heard of it

why not send the part where you're having problems with?

instead of asking ppl if they have a copy

[unreal stratus]

Ngl to be blunt like

If you are having a hard time with the notion of a CW complex why are you studying maps between spheres like this

[coarse night]

[obtuse meteor]

differential forms definitely not

[obtuse meteor]

if so they're doing this.

[gritty widget]

the set of critical values of a smooth map has measure zero

or of a continuous map which is sufficiently many times continuously differentiable (the amount depending on the dimensions of the domain and codomain manifolds), but smooth is good enough

[pseudo coral]

then whats the wedge represent

[pseudo coral]

what are the critical values of a map from S^m \to S^n

[pseudo coral]

wasnt sure if CW complexes are the only way about proving the results.

[pseudo coral]

youre right, sorry

[unreal stratus]

Simplicial approximation is another way

[pseudo coral]

like the notion of a simplex?

cause that I can see better

i think gamelin and greene uses this method

Hatcher uses CW complexes which man idk how to visualize or even grasp the concept its so intertwined

im also taking an intro to alg geo and that cours is easier than this alg top busineess

its no wonder pi_n(S^k) is uknownw for many values where n>k

im able to actually attack my alg geo problems but in my defense my alg geo prof is much better than my alg top prof

as you can all see, ive not learned much from alg top this sem..

[hidden crag]

i mean the intuition for CW complexes and simplicial complexes is pretty similar

You should read up a bit about CW complexes, the intuition behind it isn't hard to grasp once you see through the details

once you have that my proof idea will be more clear

[abstract saffron]

I still feel like CW complexes don't resemble too much of simplicial ones.

Iirc, we define CW complexes based on maps from balls to the space.

[bitter smelt]

Simplicial homology:

Cellular homology:

[abstract saffron]

You can literally just draw simplicial complexes, and see why they make sense. Not sure how you would draw a function. Technically it's just identifying boundaries, but heck no, it still took me quite some time to visually understand what was going on.

Like, if you draw a simplicial complex of a torus, ppl will immediately see why it's a torus.

[bitter smelt]

Can you? It you're going to do it on the fundamental polygon, you still need to glue things.

[abstract saffron]

Yeah, good luck explaining how you construct a cell complex. You have to have 2 S_1 to build the skeleton, then cover the surface with a disc by identifying the boundary.

Just, horrible

[bitter smelt]

How do you do the simplicial decomposition otherwise? Fix a triangulation? How is that any nicer?

[hidden crag]

Yeah you're being inconsistent here. You're going into detail why it would be tedious to explicitly give a CW structure but handwaving the process behind triangulation

[bitter smelt]

(Which is a non-trivial process)

[abstract saffron]

My point is, a simplicial complex, at least for beginners, is just a very edgy approximation of the surface. You can still see it resembles the original space, and at the same time a simplicial complex.

You can do that for CW, but it's less trivial to see how it is a CW complex

[hidden crag]

I disagree

[gritty widget]

are you asking how critical values are defined, or whether you can say something in general about the critical values of a map of spheres?

i don't really know of a general result. one immediate thing i can say is that, if m < n, then any such map will attain only critical values, and then you can use sard and stuff to get nonsurjectivity and nullhomotopy. this isn't really special though, since it works for any domain manifold of dimension m

there might be something relating to the hopf theorem but i dunno. skipped those lectures in my DT class

[cedar pebble]

god what a weird experience to have with AT compared to AG

[gritty widget]

i took AG and AT at the exact same time and AG took up like, way way way way more of my time

it was so much harder

[pseudo coral]

Whether you can say something about a given map of spheres from dimension smaller to dim bigger

[pseudo coral]

I feel it also has to do with the prof my alg geo prof explains everything FULLY while my alg top prof is hand wavy with almost all his proofs

[pseudo coral]

Tell me about it 🤦‍♂️

[gritty widget]

you can and i said a bit about it. it's a classic application of sard's theorem

[tidal lynx]

S^1 basis

[unreal stratus]

Lol

[novel acorn]

Cellular isn't that bad

It's formulated in such a way that you don't know tf ur supposed to do

[unreal stratus]

Cellular homology when the cell structure immediately gives you the cohomology

[novel acorn]

But then you realize there's a trick...

[unreal stratus]

Cellular homology when you have to compute the differetnials:

Lol

[novel acorn]

Yeah idk the first time i read what the differential was in hatcher i nearly bought a Jerry can of gasoline and burned it but then i realized the differentials are just related to the attaching maps of the n-cells so i could actually do computations 😎

[bitter smelt]

I tend to pull out every trick in the book to avoid computing differentials and finding degrees of maps

[civic verge]

guys could someone give me some ideas tips or tell me more or less a flowchart of how you could prove that the class of Q is R

pls

[gritty widget]

the "class"?

did you mean closure?

[civic verge]

yeh

[nocturne basalt]

assume the closure is not R, the complement must be open and therefore must contain some epsilon ball around some point

[civic verge]

I am saying that if Q is contained in the closure of Q , we also say that Q is dense therefore if a belongs to an irrational then for all epsilon greater than zero there exists an m that belongs to the naturals such that n is greater than m this implies that the sequence is contained in the ball of center a and radius epsilon, suppose a is an adherent point of the set {xk/k belongs to the naturals} if, there exists a convergent subsequence x_k(_j) of x_k, where the limit of the subsequence converges to , we can conclude that a is an adherent point of Q and the subsequence converges to an I therefore the "closure" of Q is equal to R

[gritty widget]

i am sorry but this is completely incomprehensible

[civic verge]

could you help me

[nocturne basalt]

to expand on this, if the closure is not all of R, it's complement is nonempty and has some p

but since the closure is closed, the complement is open

so p must have some epsilon ball around it contained in the complement

but this contradicts Q being dense in R

[gritty widget]

"Q being dense in R" is literally the statement we're trying to prove

you may want to rephrase that last step

[nocturne basalt]

I guess if you have the property that for any real number you can get rationals arbitrarily close

but then you could also use some limit points argument

[tidal lynx]

completion =? closure

[urban zinc]

is this just because you can map |{n}| to n and then |{n,n+1}| to the line segment between n and n+1

[odd flame]

CP refers to complex projective plane right

working on this

[gritty widget]

CP^n means the n-dimensional complex projective space

[odd flame]

ok i'll think about this more then come back with real questions

actually my dominos is ready

,ti

[gentle osprey]

The current time for stμ₂dying is 12:17 AM (EDT) on Thu, 04/05/2023.

[odd flame]

breakfast

[urban zinc]

classification of surfaces is so coooool

[next crystal]

why does surjectivity of q imply that f is unique?

[gritty widget]

let f and f' be functions with f o q = f' o q. for each y in Y you can find an x in X with y = q(x). then f(y) = f(q(x)) = f'(q(x)) = f'(y)

[bitter smelt]

Yep

[civic verge]

name of book?

[next crystal]

my final exam

[odd flame]

[next crystal]

but topology by munkres

this one

[civic verge]

I'm studying for my exam of analysis in several variables, I hope I do well, it's tomorrow.

[next crystal]

shit good luck

[civic verge]

the book really scares me :c

[next crystal]

me too

i got a 48 on the final

[odd flame]

takes some getting used to but it's a good book

[odd flame]

mood

but

[next crystal]

median was 53 so hoping for a B fr

[next crystal]

what book do you guys use

[civic verge]

topology, the same as you from what I have noticed from my colleagues.

of analysis in several variables
the poderos de analisis real y analis de varias variables de elon lages lima, we also use sergio plaza's, I have all those books in physical form, shit.

[next crystal]

physical form the way to go i hate online textbooks

[next crystal]

they got legos on the right book

[civic verge]

I have several books but well, knowledge does not come for free.

[next crystal]

libgen

my intro analysis class is using rudin next semester ive heard mixed thoughts about the book

[civic verge]

my professor said there is no perfect book, plus it also depends on the professor who is giving analysis though, see topology and then analysis eh, that's weird.

[next crystal]

yeah i wasnt even supposed to be allowed to take topology this semester

hopefully it makes analysis easier though lmao

[nimble portal]

It will

[civic verge]

ne, if you have already seen topology, it will be even easier, and you will realize that it is similar.

[next crystal]

lets go

the topology of R chapter wont have shit on me

[civic verge]

$if A\subset X\
int(A)=X-\overline{X-A}$

[gentle osprey]

Awuita Fria

[civic verge]

it's true?

[nimble portal]

So the idea for this is that if we look at a point in our manifold that's in the intersection of two chart domains we can still look at that space wrt. the manifold by using the transition map

[gritty widget]

we can still look at that space wrt. the manifold
this doesn't make any sense to me

[nimble portal]

Sorry like

[gritty widget]

the idea is to look at how you transition from one coordinate chart to another

[nimble portal]

We can look at the intersection on the manifold through the homeomorphism to R^n given by the transition map

[nimble portal]

I think I get it

Any two charts includes a chart and itself?

[gritty widget]

what do you think? see if it's true

what's the transition map from a chart to itself

[nimble portal]

The identity no?

[gritty widget]

so is any chart smoothly compatible with itself?

[odd flame]

can i get a nudge on this pls

[nimble portal]

Well yeah the identity map in R^n is smooth

[gritty widget]

there you go

[nimble portal]

I see

That's a cool way to do it

Wait so does every equivalence relation just induce a quotient space

[gritty widget]

yes

[nimble portal]

That's interesting

[gritty widget]

quotient spaces are typically terribly behaved when it comes to topology

but some are nice

[nimble portal]

I see the term come up a lot in algebra discussions, are they nice there?

[odd flame]

quotient spaces?

[nimble portal]

The only example I have is projective space on R^n

[odd flame]

quotient spaces pop up everywhere

quotienting is a fancy way of saying gluing shit together

the sphere and torus are quotients of a unit square for example

[nimble portal]

Yeah I don't get that at all

Like I saw a demonstration on how we make a torus out of a square (and I can see how to do the same for a sphere)

[odd flame]

that's quotienting

[nimble portal]

But I don't understand how quotienting is gluing

Is it because of the equivalence relation?

[odd flame]

yeah

[nimble portal]

We have two separate things

BUt because of the equivalence they're the "same" so we can "combine them" ("glue them")

[odd flame]

quotienting is adding a relation to the space

[nimble portal]

I see

[odd flame]

which is gluing things together

[nimble portal]

Oh

Duh

That makes sense LMAO

[odd flame]

all good lol

[nimble portal]

Hm

[odd flame]

wiki good as always

tterra button if you're there lurkiing

[nimble portal]

I'm trying to generate an equivalence relation for the torus out of the unit square but I'm having trouble

Two points are equivalent if they're on "opposite sides" of the boundary

[odd flame]

with the unit square it's easier to do things visually

[nimble portal]

So (a, b) is equivalent to (c, d) if a = c and b = 0 or 1 & d is the opposite but that sounds so convoluted

[gritty widget]

i think that it would be a good exercise for feather to explicitly write out the equivalence relation

[odd flame]

torus

[nimble portal]

Yes

[gritty widget]

visually is good, but you need to be able to translate things to mathematical language

[nimble portal]

I just don't know how to write that

[odd flame]

true

[nimble portal]

Like everything on the bottom edge is of the form (a, 0)

Everything on the top edge is of the form (a, 1)

[odd flame]

i guess for me the picture makes the actual equivalence relation make sense

[odd flame]

bump teaterra

[nimble portal]

So (x1, y1) ~ (x2, y2) if x1 = x2 and (y1 = 0 & y2 = 1)

But that's so weird to read

Seems like there's gotta be a nicer way to say that xd

Can you define an equivalence relation in "parts"?

Wait

[odd flame]

integration by parts

[nimble portal]

Yeah I think so? That's what we did with projective space to see that it was homeomorphic to the sphere

[odd flame]

un dia vi una vaca sin cola vestida de uniforme

wdym parts

[nimble portal]

Like

For projective space you say x ~ y if x = lamba*y for nonzero real #s lambda

But if you consider that in two “parts”

so x = -lambda * y and x = lambda * y for positive real numbers lambda

Then you can see that you get antipodal points on a circle of radius lambda

So in the same way

Could you just say that (a, 0) ~ (a, 1) and (0, b) ~ (1, b)?

So you define your equivalence relation in two different “parts”

[nimble portal]

Though it feels like these are different parts from the kind we used here

Idk how to explain it but this feels like an intersection of relations whereas that felt like a union of relations?

So is that valid?

[odd flame]

what is lambda supposed to represent here

sorry feather im also down cataclysmically rn

[gritty widget]

α is a generator of H^2 and f*α is also in H^2

it should say f*(α^i) = λ^i α^i. then the lefschetz number is the sum of λ^i

many typos

[odd flame]

oop

is lambda still just any integer?

[gritty widget]

it's some integer

depending on f and α

[odd flame]

hm

cohomology hard

ok my question is a little different

but the same reasoning applies no?

[gritty widget]

its not just asking you to show that f has a fixed point, it's asking you to find one

the lefschetz fixed point theorem doesn't seem particularly useful

[odd flame]

https://tenor.com/view/hng-gif-4501621

,ti

[gentle osprey]

The current time for stμ₂dying is 02:42 AM (EDT) on Thu, 04/05/2023.

[odd flame]

i might just quit on this homework then

im so fucking tired and i have three other questions

[odd flame]

for once it's not entirely my fault

[gritty widget]

sleeping is usually good for you

[odd flame]

my personal life decided to go up in flames right before finals

i will keep trying

bc i wanna get into grad school

[odd flame]

nvm

cant fucking do this shit no more

[odd flame]

i have risen to try once more

[odd flame]

bump

if not lefschetz then what

[novel acorn]

just find the fixed point directly by calculation?

[languid patrol]

If you have an automorphism of projective space which comes from a linear map then the fixed points are just eigenlines, in this case they are very simple to find.

[unreal stratus]

First term I have heard the term eigenlines and it makes too much sense lol

[nimble portal]

LMAO Yeah I like that

[odd flame]

reminds me of 3b1b

[odd flame]

out of curiosity does this fail in even dimensions

seems elementary enough that it shouldn’t but the question is in odd dimension

oh wait

is it then just showing that rank is 2n-1

[odd flame]

https://tenor.com/view/yelling-meryl-streep-cheer-gif-8372999

topos theory egirl

[unreal stratus]

Lmao

[languid patrol]

Hello

Yes

what is your question exactly?

[odd flame]

boop

[languid patrol]

Is what just showing that the rank is 2n-1?

[odd flame]

findinf the fixed pt of the map before

but thats silly so nvm

doesnt actually find any points

[languid patrol]

The first map is just a linear map, so it has loads of fixed points.

[novel acorn]

I have a feeling like you're overcomplicating this problem by a lot

[languid patrol]

E.g. just find any nontrivial eigenvector.

[odd flame]

probably 🥲

[languid patrol]

The second map you can also just prove the lack of fixed points explicitly

By trying to solve for them and showing there are no solutions.

Note that it suffices to prove the existence of fixed points/nonexistence for P^1 by an easy induction.

This is basically reducible to just semilinear algebra, but you're trying to use very difficult topology to get around computing anything.

[odd flame]

didnt expect this much realness…

but youre righr

[languid patrol]

I mean we've all been there. But if you sit down and try to compute you'll realize it's even easier than making some argument about the lefschetz fixed point theorem.

[odd flame]

i mean im also like

not doing too hot so probably thinking less

butt

i can still do this

[languid patrol]

It sucks to have personal problems during exam season but I believe in you!

[odd flame]

yeah

gonna be a long fucking week

but thank you

might bother with another q later

ok wait

in P1 thats just the antipodal map

0 the fixed point?

[languid patrol]

For f or for g?

[odd flame]

f

[languid patrol]

For f it's the map f([X:Y]) = [-Y:X]

0 is not a point in P^1

[odd flame]

yeah i said that wrong

can we think in P3 instead

[languid patrol]

Can you see for what values of X, Y where X and Y are not both 0 we have (X, Y) = \lambda \cdot (-Y, X)?

[odd flame]

im gonna take a small break, dont wanna keep wasting your time

[languid patrol]

No worries!

[gritty widget]

but you're working in complex projective space!

[odd flame]

ok this time i do just wanna ask for a hint if possible please

[nocturne basalt]

its not clean but i think you can just compute it directly from the boundary map

[odd flame]

sorry i keep jumping around but different question

[languid patrol]

They want you to just compute it but you can explicitly homotope it to the wedge of S^2 with two circles.

Just to check your answer

(by contracting the face that they're glued along, then contracting v_2 -> v_4

[odd flame]

i dont immediately see that

hmmmm

ok wait sorry id gone back to a dif question

trying to show that X v Y has the fixed point property iff X and Y both have it

for backwards direction i was starting by assuming that if X and Y have it but we have a self-map of the wedge with no fixed points this would induce some kind of selfmap on X and/or Y with no fixed points too which is a contradiction

but that feels too weak

or maybe induced is the wrong word but just like

the restriction of the self map of the wedge sum to X or Y

[odd flame]

this is wrong

[odd flame]

are those cohomology groups

prof promised us no cohomology

[novel acorn]

they're homology

[odd flame]

oh so this is just the standard ker/im stuff

i will NOT overthink

[novel acorn]

yeah it's the only thing is that you're working with different coefficients so you just gotta account for that

[odd flame]

that notation is just saying what gets modded out right

like okay

[novel acorn]

yeah you mod out the coefficients

[odd flame]

trying one rn hol up

for Z/4

ker(d1) = {a1, b1}

im(d2) = 2a1

so H1 = Z/2 x Z...?

also im assuming Q probably means rationals right

[lean knot]

a topological group G acts on a topological space X, and the action is continuous. x is ana element of X, then, is the orbit of x a connected subspace?

x is an element of X

[gritty widget]

not necessarily

[lean knot]

Any simple counterexample? In what case it is?

[gritty widget]

a common example is G = {-1, 1} acting by multiplication on X = S^n. the orbit of any point x is {x, -x}, which is disconnected

a sufficient condition is for the group to be connected: the orbit of x in X is the image of the continuous map G -> X given by g -> gx

but it's not a necessary condition. take your favorite disconnected topological group and let it act on your favorite topological space trivially (every element acts as the identity). the orbits are points, so they're connected

maybe you could throw orbit-stabilizer at it and get something nice

[lean knot]

That helps a lot, thank you very much!

[nocturne basalt]

yes with Q coefficients

[abstract saffron]

Hmm... I'm still waiting for feathers' dose of question for today.

[nimble portal]

LMAO

I have a 2000 word essay due tonight and I’ve only written 150

My questions will come after 😌

[rough cedar]

ong me too

[thin scarab]

reading lee and he mentions that an open simplex (that is, without boundary) isnt open in R^n in general, is this just about the dimension mismatch? or am i missing something

like, would it not be true that every open k-simplex is open in R^k?

[languid patrol]

It’s just the dimension mismatch

An open k simplex is open

[thin scarab]

ok lol, the way he phrased it made it seem deeper than that

[nimble portal]

well\

since I'm almost done with chapter 1 (just smooth manifolds w boundary left, whihc looks simple enough)

I think it would be good to make myself some notes/prove stuff within the chapter as I re-review it before I start te exercises

What are some basic topology facts I should prove?

Like stuff you'll see pop up all the time, things about connectedness & compactness and whatveer else might show up

[gritty widget]

show that the continuous image of a connected/compact set is connected/compact

and tie these into things you know from calculus

[nimble portal]

Can I just use the EVT?

lol

[gritty widget]

this is the EVT

[nimble portal]

o I thought EVT was continuous on compact means attains max & min

(which in R means closed, bounded -> compact by heine-borel)

but I guess I'd want to prove it for any top space huh

[gritty widget]

do the exercise i suggested and then convince yourself that it is the generalization of the extreme value theorem you know

[nimble portal]

Sure

What else?

[gritty widget]

a continuous map from a compact space to a hausdorff space is a closed map

(i.e. takes closed sets to closed sets)

[nimble portal]

Oh

This one was the detail I missed in the coordinate ball covering proof

[gritty widget]

a space X is hausdorff if and only if its diagonal {(x, x) : x is in X} is closed in the product topology of X x X

[nimble portal]

Oooooh that's an interesting one

I like that

[quiet thorn]

i am actually just so hopelessly confused in homology

[thin scarab]

struggling to see why this would be true

[gritty widget]

skill issue

[thin scarab]

clearly

[coarse night]

connected locally path connected spaces are path connected

[gritty widget]

that one's important for manifold stuff

nice

[coarse night]

you can use to show connected manifolds are path connected.

There's this one, X second countable then any basis of X has a countable sub basis. Lee uses this to show manifolds have a basis with regular coordinate balls

[thin scarab]

i have intuition for pretty much every one of the basic properties of a manifold except for hausdorffness

[opaque cloud]

so how exactly does homology work? and what exactly are the elements and the operation of a homology group? all I know is it's a way to classify holes in a top space

[opaque cloud]

doesn't hausdorffness naturally follow from a manifold being locally homeo to R^n?

hausdorffness is a local property innit?

[thin scarab]

? no

[gritty widget]

line with two origins

[thin scarab]

what i mean is that i know how to use the definition but its not something theres a nice visual intuition for in my head

i guess shrinking open sets around two points or something until theyre separated idk

bit its never obvious to me how it relates to other concepts

[opaque cloud]

it doesn't? so if Y is a topological space that's hausdorff and if X is locally homeomorphic to Y then X is not hausdorff?

[gritty widget]

just gave you an example neamesis

[coarse night]

Non Hausdroff spaces are wwird, for example intersection of compact sets may not be compact for non HD spaces. So in a sense HD spaces are spaced where intuition work

[opaque cloud]

okay reading about it now

[thin scarab]

lee says that hausdorffness roughly means a space has "enough points"

[coarse night]

no

[thin scarab]

and second countable means it doesnt have "too many"

^ this one makes sense to me

i get what youre saying, weaker separation axioms can give strange results

my brother told me he had a prof give websites connected by hyperlinks as an example

but i still dont see hausdorff and it frustrates me a little

[opaque cloud]

in this stack exchange answer they're constructing the line with two origins

https://math.stackexchange.com/questions/709777/the-line-with-two-origins

people also denote the extended reals as R U {-inf, +inf} you're just adding in points just like how they're constructing the line with two origins, does that mean topologically the extended reals are like a line with 3 origins?

[thin scarab]

its not that youre adding in points randomly, its that youre saying theres more than one point that serves the same role

so subbing in either point you get open sets which are homeomorphic to R^n

but then the points themselves can't be separated

[opaque cloud]

But just taking the union doesn't give you any information of where you're adding the points or how it serves a particular role?

[thin scarab]

correct, thats why theres more to the definition

[opaque cloud]

ah okay

[thin scarab]

you take as basis:
(-a,0) U {one origin} U (0,a)

so for one origin you get something homeomorphic to the real line, youre basically just relabeling 0

[opaque cloud]

ooh I see

[coarse night]

say you want your sequences in the space to have unique limits. HD enforces this

this (and some countability axiom) allows you to give sequential arguments for continuity

At most you can motivate the definition and convince that it's the right kind of requirement to have on your space

[icy schooner]

For a fiber bundle do we need the projection to be smooth

[languid patrol]

If it's a smooth fiber bundle yes.

[icy schooner]

I feel like in Hatcher everything is smooth so

[languid patrol]

But there are, e.g., fiber bundles over topological manifolds and it doesn't make sense to require them to be smooth.

[icy schooner]

The question is if S^k -> S^m -> S^n is a fiber bundle then k = n-1 and m = 2n-1
if everything is smooth then n can’t be 0 because S^m is connected and S^0 is two points

[languid patrol]

That has nothing to do with smoothness, the continuous image of a connected set is always connected.

[icy schooner]

yes I am only using continuity

n \neq 0 gives k < m

Then the inclusion image of S^k is contractible in S^m

and the homotopy LES splits

[languid patrol]

It's not that easy.

E.g. consider the Hopf fibration

Just because the fibers of a bundle are nullhomotopic in the whole space individually it doesn't mean the fiber bundle is trivial.

[languid patrol]

Yes but how is that helping you? You need to make an argument for why that is impossible, e.g. why the bundles will be trivial.

[languid patrol]

.

[icy schooner]

the inclusion is nullhomotopic since its image is contractible?

Then I get $\pi_i(S^n) = \pi_i(S^m) \oplus \pi_{i-1}(S^k)$

[gentle osprey]

Iteribus

[icy schooner]

if I set i = n the first term on the RHS is 0

[languid patrol]

no

That was my point, that is not true necessarily

[icy schooner]

oof

[languid patrol]

So like I would keep in mind S^1 \to S^3 \to S^2

[languid patrol]

which is a fiber bundle which is not trivial between different spheres.

That's one of the examples of why m = 2n-1 k = n-1 is excluded.

[icy schooner]

ok let's look just at the LES around i = n which is where it might be interesting
$\pi_{n}(S^k) \rightarrow \pi_{n}(S^{n+k}) \rightarrow \pi_{n}(S^n) \rightarrow \pi_{n-1}(S^k) \rightarrow \pi_{n-1}(S^{n+k})$

[gentle osprey]

Iteribus

[icy schooner]

second term is 0
last term is 0

[languid patrol]

So k has to be equal to n-1 yes.

[icy schooner]

we can still get $\pi_{n-1}(S^k) = Z$

ok

[gentle osprey]

Iteribus

[icy schooner]

ty

[languid patrol]

Nvm lol

I was being dumb

[icy schooner]

I thought we get m = k + n from dimensions

[languid patrol]

Yeah lol

[unreal stratus]

owo

I am do fundamental group stuff again for exam and it makes we wish we did the fundamental groupoid too lol

feels much more natural in some cases

[hidden crag]

All spaces are path connected

[surreal hornet]

hey is S^2 x (S^2 - {p,q}) homotopy equivalent to S^2 - {p,q}?

[languid patrol]

no

[surreal hornet]

the second factor is homotopy equivalent to R^2 \ {p} which is S^1 so would this be homotopyequivalent to S^2 x S^1?

[languid patrol]

Yep, and then you have to show that that is not homotopy equivalent to S^1

[surreal hornet]

i'm doing this because i would have needed to get something simple out of S^2 x (S^2 - {p,q}) to calculate cohomology for it, but S^2 x S^1 is not simple enough for me

[languid patrol]

hmmm well do you know the Kunneth formula? Or do you know about higher homotopy groups?

[surreal hornet]

well a bit but i'm doing a mayer-vietoris construction here

can I get some direct sum out of S^2 x (S^2 - {p,q})?

[languid patrol]

That will also work! You can cover S^2 by the complement of the north and south pole, U, V. Then Meyer-Vietoris on U \times S^2 - {p, q} and V \times S^2 - {p, q} will work.

[surreal hornet]

well that is what i'm doing but the intersection of U and V comes out to S^2 x (S^2 - {p,q})

for context i'm calculating S^2 x S^2

[languid patrol]

Ah so you want to compute the cohomology of S^2 \times S^1

For that you can do another meyer vietoris on the left hand side

[languid patrol]

And you will reduce to (U \cap V) \times S^1 being a familiar space.

[surreal hornet]

i'm only doing this up to the case k = 2 so can I get away without m-v for S^2 x S^1?

[languid patrol]

It depends on what you use.

[surreal hornet]

okay so i've managed to conclude that H^2(S^2 x S^2) = R + R from the m-v construction. i would now want to figure out a base for this and i think that since H^2(S^2 x S^2) = H^2(S^2) + H^2(S^2) i could use the orientation forms on the spheres?

[coarse night]

why not Küneth?

[surreal hornet]

since i'm interested in the generators i think i need something out of the maps on m-v

[obtuse meteor]

Künneth also tells you something about generators when phrased correctly.

This is something I’ve realized as I’ve done more alg top. I don’t just wanna know there is an iso. I wanna know what the hell the iso is

[languid patrol]

The generators that you're getting from your argument are obtained in exactly the same way from Kunneth

This is esp. easy to see if you work over a field.

[surreal hornet]

how do you retrieve them from just knowing that H^2(S^2 x S^2) = R + R? i should be able to use the volume/orientation forms of the spheres to get the generators, but no idea how

[coarse night]

actually the cup product gives you the generator of H²

or wedge product in case of De Rahm

ok actually the pull back of the volume form does it

p1: S²xS² → S² and ω be volume form of S² or orientation form then you can p1* ω to get one generator

anyone can correct me if I'm wrong

[eager vigil]

Could anyone give a hint to this one?

[next crystal]

i had this problem for hw last semester, the way i did it was to just come up with a metric and show that it induces the dictionary order topology

it ended up working out nicely in that the metric was kind of "the most obvious one" you could choose

[unreal stratus]

Interesting

[next crystal]

if this is from munkres then idt theres any other way to do it? bc theres no theorems about metrizability introduced by this point

[languid patrol]

Yeah they just want you to construct a metric, it's easy enough.

[coral pawn]

What are some resources to learn homotopy theory post-Hatcher? I'm interested in learning motivic homotopy theory eventually, so ideally I want a book that develops tools found in both the classical (topologicak) and motivic case.

Is Simplicial homotopy theory by Goerss and Jardine good for this?

[solemn oar]

Yes, that's the standard resource.

[coral pawn]

That's what I've heard

Though Hartshorne is also a standard resource, but I don't think it is the best for learning AG

[solemn oar]

Supplement with May's Concise maybe.

[coral pawn]

Got it

Have you studied motivic homotopy theory?

[solemn oar]

I know some rudimentary definitions, but nothing substantial.

[coral pawn]

Do you know what sections of it I should pay more attention to for motivic homotopy purposes?

And what sections I should skip for now

I'm planning on working through it over the summer

[solemn oar]

I suspect getting a good understanding of G-J and then moving into motivic is a good idea.

Some key ideas, like grothendieck topologies, are not covered though iirc, so you'll need to fill that gap.

[coral pawn]

Oh I know a bit about that from stacks

Should I learn any more details or is a basic understanding sufficient?

What about elements of homotopy theory?

[solemn oar]

I think just getting comfortable with the material in G-J is sufficient.

[coral pawn]

Got it

Thanks

[solemn oar]

Good luck!

[coral pawn]

Thanks!

[mint rose]

Is there an example of a retraction $i:A \to B$, where both $A$ and $B$ are retracts of a $CW$-complex, such that $i$ is a deformation retract, but not a strong deformation retract?

[gentle osprey]

Digiteraat

[mint rose]

I'm trying to find a counterexample to the following exercise

[languid patrol]

The usual example is the space consisting of the straight lines connecting the origin to (1, 1/n) for all n union with the line connecting the origin to (1, 0)

Then (1, 0) is a deformation retract, but cannot be a strong one which is a nice exercise.

The topology is the subspace topology inherited from the plane, the idea is that the points (1, 1/n) get too close to (1, 0) to allow them to move without violating continuity

[mint rose]

But that's not a retract of a CW complex, is it?

because I need both A and B be bifibrant

[tidal cedar]

I think a lot of this is more inertia than actually being good

Then again I think AG has a kinda weird thing in that like

commutative algebra largely is motivated by AG and many commalg texts are written from that perspective

so it almost ends up being cyclical

almost because of course you can learn commalg without understanding the AG motivations (yet)

but I think there's still a lot of room for improvement in the pedagogy here and I wish there was a book that tried to do them both at the same time and maximize intuition

[tidal cedar]

Have you done SHGH?

[coral pivot]

https://people.math.rochester.edu/faculty/doug/otherpapers/Adams-SHGH-latex.pdf

have fun!

[tidal cedar]

ty

[pearl holly]

np

[novel acorn]

toki!!!
Have you started learning about infty cats?

I started reading HTT recently so idk if you wanna join me

[novel acorn]

Idk this is a personal thing
I enjoy it

of all the resources for AG i've found I think hartshorne + Gathmann's notes are perfect for learning

[pearl holly]

ayy hello! No I haven't started yet

[pearl holly]

oof sounds nice! I don't have a lot of time rn so I don't think I will be able to read it unfortunately

maybe in the future, but maybe by then you're already done lol

[novel acorn]

idk ifyou have plans over the summer but we could maybe read then

[pearl holly]

nah my summer looks fine, I should have time then and I would be down reading stuff about this

[novel acorn]

yay
Then I'll put off reading HTT for now and do some other stuff

[pearl holly]

let's dm when summer is approaching then

[coral pivot]

What math are you doing these days toki

It’s been a while

[pearl holly]

ye fr

but I've been doing the same thing over and over I guess lol

like stable stuff I guess

[coral pivot]

Oof I see, yeah that stuff is fun but also hard so makes sense

[pearl holly]

ye exactly. How about you tho john?

[coral pivot]

Oh still doing operator stuff, transition to more math phys type of stuff

Probably gonna do some actual physics once I get to grad school

[pearl holly]

oh damn I see

[coral pivot]

i forget, are you going to grad school too or are you applying next cycle

[pearl holly]

I'm going to apply next cycle

I think

I don't really know what a cycle is lmao

[coral pivot]

right so its members of the kerner of d

but i see, nice. Presumably going to target for homotopy theory stuff for grad school?

[pearl holly]

ye definitely

I barely know any other math at this point

[coral pivot]

oof i see. yeah I mean you know a lot of homotopy theory so the rest will seem low in comparison

[dusky stratus]

Hello. If I have an injective group morphism A -> B, does it imply that the morphism H_(A) -> H_(B) (between their group homology at constant coefficients) is also injective ? I don’t know examples of computation of group homology to find counterexamples. Thank you.

[languid patrol]

No.

Also the map is in the wrong direction.

It should go H^(B) to H^(A)

unless you mean for A, B to be coefficients for H^*(G; A)

[dusky stratus]

I wanted to put the star below but I don’t know how to do it on phone.

[umbral panther]

It’s not true for H_1, which is the abelianization. Take any nonabelian group and an element that doesn’t survive the abelianization

[languid patrol]

it's not even true for H_0

[dusky stratus]

Ok thank you.

[dusky stratus]

I think that it is true for constant coefficients since the action of group is trivial

But I’m maybe wrong, I’m new with group homology.

[languid patrol]

Ah just with constant coefficients yes

[odd flame]

to see that Rn and Rm have the same htpy type can i just say they both retract to the origiin

[languid patrol]

Yes

[odd flame]

[odd flame]

reviewing for a final so im going back to very basics and im just curious about how someone else might phrase thiis

what is the motivation for/what info do we get from homology

[winged viper]

Here, could it not be the case that K_xi is a nonorientable surface like a Klein bottle? For instance, if we take X to be a Klein bottle with the Delta-complex structure consisting of two 2-simplices A and B, then taking xi to be the chain A - B I think would result in K_xi being a Klein bottle

[hidden crag]

We've been over this before

[odd flame]

holes

yes

[hidden crag]

if you search for older messages in that convo you'll find answers from ryu as well

i think it was them at least

[odd flame]

i should rephrase then

[obtuse meteor]

You failed to take the signs into account when doing the identification

At least I think this is what’s happening

(I believe you get something like a punctured torus but maybe I am mistaken)

[winged viper]

Yes actually I just realized it’s more dumb than that, A - B isn’t a cycle

Thanks

[obtuse meteor]

Ah well I guess it isn’t a cycle because of the signs :)

[coral pawn]

I have not

Would you recommend doing that before G&J or after?

I feel like I should learn homotopy theory before stable stuff

[umbral panther]

Adams is easy but sloppy. He’s the motivation for learning homotopy theory, so you should read him first

[abstract saffron]

Has anyone tried May's More Concise Algebraic Topology?

I'm being invited to a reading group, but honestly the book itself scares me to death already

Yeah, probably not the best idea 😄 it's dense af

[umbral panther]

The first volume is very dense. More concise is not actually concise. I don’t think it’s dense, either

[languid patrol]

I think it's a good book

but it's a pretty mature perspective and ymmv whether or not it works for you

Also a bit light on examples

[coral pawn]

Motivation isn't a big problem for me

Will G&J be too difficult without working through Adams?

[solemn oar]

No, you don't need to read Adams to read G-J.

[rough cedar]

Ujs are an open cover of P^n. is the middle thing supposed to be u_{j - 1} / u_j, u_{j + 1} / u_j?

[languid patrol]

Yes

[rough cedar]

thanks

[rough cedar]

not sure if this is still the correct channel