point-set-topology

[gritty widget]

the proof is quite simple

[coral pawn]

How do you prove it?

[gritty widget]

there's a proof on the wiki page

[coral pawn]

I feel like there should be a way to do this using Taylor's theorem

[coral pawn]

@gritty widget How does Hadamard's lemma give us what we need?

Ah wait

I see it

I think

[gritty widget]

the statement kinda leaves out that you can choose g_i's which equal the partials of f at the point

that's what you need for your problem

[rancid umbra]

true or false : finite intersection of compact sets is compact

[orchid forge]

if X is hausdorff, then compact sets are closed, finite intersections of closed sets are closed, and a closed subspace of a compact space is compact

so if you want a counterexample you need to look in non-hausdorff spaces at least

[rancid umbra]

yea, was just asking. it seemed like it should be true at first, but trying to think of counter examples is kinda fun

plus i really dont want to type up my analysis hw rn

[orchid forge]

Non hausdorff spaces don't exist imo

[rancid umbra]

lol

[tough imp]

Spec A

Don’t say that you’ll scare him!

[orchid forge]

[honest narwhal]

Spec(A) is a set with a topology

That's different from a space

[tough imp]

Go fuck yourself pretender king

The monkeys will usurp the throne and claim the crown they deserve

[honest narwhal]

It only becomes a space when you start doing etale stuff

[orchid forge]

Shëf

[winter prawn]

why the topology we define in complex plane have to be a disk but in real line we just use interval...?

is it because of the i?

[empty grove]

the i?

The euclidean topology is the one induced by the distance function on R^n, because the topology is supposed to capture closeness etc

and the standard distance function on R^2 makes the open balls circular

[uncut surge]

Exactly, the real line is fundamentally a one-dimensional object, whereas the complex plane is two-dimensional -- essentially "because of the i"

@winter prawn

[winter prawn]

thanks both!

[night pivot]

Is a second countable, bounded metric space compact?

I think the Hilbert space is second countable, then set the metric to min(1, sqrt(sum((xi-yi)^2))) and then it's bounded but it's not compact?

[wise sigil]

You can do that with non compact, second countable space

replacing the metric with min(1,...) i mean

[night pivot]

it must be metrizable first to create a counterexample tho

but yeah I get what the idea is

[wise sigil]

oh right ofc

you can simply take (0,1) also

[flint cove]

Sorry if this is a stupid and boring question, but it comes to my mind frequently.
Let $M$ be a compact connected orientable MF. I know that the volume form represents a nontrivial cohomology class because its integral does not vanish.
But how do we know that that's everything, i.e. that $H^{\mathrm{dim}(M)}(M, \mathbb R)$ does not have dimension greater than one?

[gentle osprey]

lux

[flint cove]

(I probably should start doing a comprehensive re-learning session of cohomology to fill gaps like this)

[night pivot]

same too lol don't worry

[flint cove]

If in my world MFs are without boundary, does that circumvent the problem? :D

But fair point

Does poincare duality work regardless of the coefficient group?

ie Z, Z/p instead of R

[empty grove]

Compact connected orientable motherfucker

[flint cove]

Alrighto, tyvm

[blazing geyser]

hey i was just wondering if real analysis 1 and abstract algebra 1 would be enough of a prerequisite for taking a course algebraic topology

i didnt take point-set topology but i did have a bit of exposure to it in analysis (e.g. open/closed sets, compactness, int/ext, etc.)

analysis and algebra are the prereqs for algebraic topology at my school but i was just wondering if i would be fine having done just those

[orchid forge]

theoretically it's possible but i would recommend taking a look at the/a book first

if you can make sense of the first couple chapters of the book on your own then you should be fine, otherwise i'd maybe talk to the professor

[blazing geyser]

alright yea ill definitely ask about it thx

is the general sequence point-set --> algebraic topology or are they kind of unrelated

[orchid forge]

that is the usual sequence but you don't use very much point set topology anyway

you just need to be familiar with the basic concepts (topological space, continuous map, open / closed sets, compactness, etc) and maybe a couple theorems

[blazing geyser]

oh ok cool yea i do remember covering a lot of that in analysis but ill definitely review it

[pearl holly]

If we just forget about all the notational bullshit, why can an element in H^pn be written like that?

I understand that it has something to do with the isomorphism H(L x X) = H(L) x H(X) and so the element can be written as a linear combination of elements in H(L) and H(X) tensored with each other but why is the index like it is?

so like why can't I write the element like $\sum_i w_i \otimes \theta_i$ where $\theta$ is the generator of $H^{n-i}(X)$ and $w_i$ is some element of $H^j(L)$?

[gentle osprey]

Tokidoki ✓

[pearl holly]

and $\alpha \in H^n(X)$ and by some stuff Hatcher did before I get an element $\lambda(\alpha) \in H^{pn}(X)$

[gentle osprey]

Tokidoki ✓

[pearl holly]

also the indexes don't sum up to np but that should be the case, right?

[pearl holly]

never mind, I got caught up in the notation lmao

[cursive flume]

what's the difference between stiefel and grassmann manifolds?

confused

[gritty widget]

stiefel manifolds consist of orthonormal bases of subspaces and grassmann manifolds consist of subspaces themselves

@cursive flume

[cursive flume]

ahh

makes sense

[tight agate]

The grassmannian is a quotient of a stiefel manifold

[cursive flume]

how?

by which group action?

[tight agate]

Whatever the structure group is

[cursive flume]

what do you mean by structure group here?

[tight agate]

For example, if you’re looking at the stiefel manifold of orthogonal frames, the. The structure group is O(n)

[cursive flume]

right

[tight agate]

Because the O(n) action takes orthogonal frames to orthogonal frames

[cursive flume]

then stiefel isomorphic to O(N) quotient O(n-k)

but how is this related to grassmannian at all?

[tight agate]

By quotienting by this action, you’re identifying all choices of basis for a vector space

The grassmannian parameterizes subspace s of a certain dimension

[cursive flume]

with this I agree

[tight agate]

brb

[cursive flume]

but idk how this is related to the discussion with Stiefel

[tight agate]

@cursive flume the grassmannian G(n, k) is the space of all k-planes in R^n

The stiefel manifold V(n,k) is the space of all orthogonal k-frames in R^n

each k-frame defines a k-dim subspace of R^n, i.e. a point of G(n,k)

so we have a map V(n,k) ----> G(n,k)

The fiber of this map over each point in G(n,k) is the set of all k-frames which span the same k-dimensional subspace of R^n

any two elements in the fiber over a point are related by a unique orthogonal transformation

therefore the fiber over a point is isomorphic (as spaces) to O(k)

so the quotient of V(n,k) by O(k) is G(n,k)

intuitively, we're just identifying all choices of orthogonal bases for each k-plane in R^n

this can be rephrased in the following way

we have a principal O(k)-bundle O(k) ---> V(n,k) ---> G(n,k)

did that make sense @cursive flume?

you can do the same thing with all classical groups by considering frames of the right type

[digital wraith]

How should I think of singular homology

Like I know the definition, I just don't have any real intuition for it

[orchid forge]

Vaguely to me it's just the natural way to take simplicial homology, which is intuitive and geometric, and embed it into any topological space

[pearl holly]

I like to think that simplicial homology is for computations and singular is for finding nice theorems but in the end they are basically the same thing

But don’t listen to me I’m noob

[orchid forge]

It kind of skips the process of actually triangulating your space, which you know is arbitrary in the first place

By considering all the maps of simplices into the space at once

[flint cove]

the main advantage of singular homology is that you don't have to make this arbitrary choice liike in simplicial or cellular homology afaik.

or like the choice of good open cover you usually make in cech cohomology

[digital wraith]

Right I just don't have much intuition for it

Since you're considering every singular simplex at once, even degenerate ones, yet when you take the quotient you magically get a nice group.

[pearl holly]

oh crap I didn't see that you responded lmao

but yeah I see

wait did you respond to me?

[orchid forge]

Nah just talkin

[pearl holly]

oh lmfao

[flint cove]

For me this is like with the fundamental group: you don't really care about the degenerate ways you can squash an S¹ into your space, if there's a „qualitatively similar“ way that's way nicer

I wouldn't be surprised if with a smooth structure you would be able to restrict yourself to piecewise smooth simplices, but the quotient would be the same

[cursive flume]

yes,that makes sense

thanks a lot! :untilted:

[orchid forge]

Maybe it would help to make sure you understand the proof that homotopic maps induce isomorphics in homology. That reduces the apparent amount of information in the complex

[digital wraith]

Ah I've been putting off reading that one

Do you have a source you reccomend?

[orchid forge]

So like we know continuous maps f : X --> Y act on the singular chain groups in a pretty intuitive way, sending a chain (delta^n --> X) to (delta^n --> X --> Y). If F : delta^n x I --> X is a homotopy with F(x,0)=f and F(x,1)=g, then this delta^n x I itself is almost a simplex since delta^n x I can naturally be subdivided into a bunch of n+1 simplices (e.g., when n=1 you get a square, which is 2 triangles). So a homotopy f ~ g induces an n+1 chain whose boundary is the image of F on the boundary of delta^n x I, which is (delta^n x 0) \cup (delta^n x 1) \cup (boundary of delta^n x I)

That is, the boundary of F is just im f - im g + a boundary

Uh I have no idea about sources tbh, I was supposed to read hatcher but didn't like this part much

[digital wraith]

How did you learn it then?

[orchid forge]

Osmosis primarily

[digital wraith]

The best way

[orchid forge]

So I guess the intuition is that homotopies are themselves (1 degree higher) singular simplices (chains) that respect the boundary map like this

So the singular chains can be safely quotiented out by homotopies

If you have homotopy equivalent spaces X --> Y, this gives you that the induced maps on singular chain are inverses (hence isomorphisms)

So you can even deform your space itself up to homotopy (equivalence) without any consequence to the homology

Now the complex doesn't seem so bad, because we can homotope our space to resolve many kinds of confusing pathologies, and choose representatives of simplices which are nondegenerate except where constrained by the space (by obstructions to homotopy)

This also justifies our intuition that singular homology reduces to simplicial homology with respect to a triangulation (which is ultimately arbitrary), where we think about a triangulation (if it exists) as a bunch of choices of representatives of singular simplices

So a plausible way you could have "invented singular homology yourself!" would be to start with simplicial, observe that continuous functions X --> Y send an embedded simplex in X to a (not so embedded) simplex in Y, make the observation about homotopies being n+1 simplices with nice boundaries, and then say "oh, I don't need a triangulation at all; homeomorphisms of delta^n to a subset of X can be replaced by any homotopic map." Then you'd check formally that this new definition works even for non-triangulable spaces, and conclude this is better

[pearl holly]

this is so nice

I will now proceed to invent singular homology myself!

[orchid forge]

Not if I do it first

[pearl holly]

[flint cove]

Semi-related, is there a wiki / website / book that kinda has a list of spaces together with their homology / homotopy groups, precalculated?

[pearl holly]

There’s topospaces.subwiki.org

but I don't know if its good

https://topospaces.subwiki.org/wiki/

boom

[light sky]

there's also https://topology.pi-base.org/spaces

[pearl holly]

oof

[light sky]

but I don't know if it lists homology groups and whatnot -- it's open source though, so you can file a feature request 🙂

[flint cove]

I know about π-base, which was kind of a motivator

[pearl holly]

here you just search in the top right corner for the space you want. So if you search for the sphere the (co)homology groups will be listed there

[flint cove]

Hm, but already the „projective space“ article has nothing like that

[pearl holly]

oops

[onyx crow]

The biggest table I saw is in the Wikipedia page for homotopy groups of spheres

[flint cove]

Mrh I know I should just have written „(co-)homology groups“

Because to be really honest and speak from the bottom of my heart, I have yet to find anything that I care less about than homotopy groups of spheres, or of anything, for that matter

(inb4 I radically change interest in a few months and do homotopy theory)

[pearl holly]

I'm literally getting stuck on every fricking sentence

and there's so much notation everywhere that I can't ask here because it would take me 5 years to explain the notation

[flint cove]

you talking homotopy theory? or am I missing context

[pearl holly]

no I'm just talking randomly lmao

it's called procrastinating

[flint cove]

LOL what a chad article name
https://topospaces.subwiki.org/wiki/Reduced_homology_of_wedge_sum_relative_to_basepoints_with_neighborhoods_that_deformation_retract_to_them_is_direct_sum_of_reduced_homologies

[pearl holly]

daim the article name is so long

[flint cove]

lmaoooo

why thank you very much for this elaboration

[fair idol]

Say I had a euclidean ball. How can I define a vector field that maps any point p to any other point q on the ball?

[gritty widget]

the vector field takes p to q?

or do you want something like its flow

[haughty anvil]

Hey so I got #1 here, it seems to just be Van Aubel's theory. But I'm struggling on b and c.

I have a guess for c but b I'm not really sure. My guess is that nothing would really change right? Because they're being flipped on the 2 perpendicular lines establishe din 1a.

[candid hatch]

(TIL "quadrangle" is a word...!)

[haughty anvil]

Same.

This is a very old book too. 1994.

<@&286206848099549185>

[haughty anvil]

Okay on a new question now. I did the first one, struggling on the second.

[obtuse meteor]

The homotopy groups of spheres are cool okay ;-;

[bleak path]

Hey! Can I steal the channel a bit to ask for thoughts on a T/F question? My gut feel says that this is false, but I don't quite know where to begin attempting to justify it, if anyone could give a pointer or two on what direction to start that would be great

X and Y here are generic non-empty topological spaces

[tough imp]

It’s totally true

The product of continuous is continuous and by definition of the product topology the image of open is open

[bleak path]

What about in the case of closed maps then?

closed continuous* maps

[tough imp]

I’m not sure

I don’t know how closed sets work off the basis for opens off the top of my head

[bleak path]

Alright, much appreciated anyway!

Just been tricked recently in topology and started to doubt a lot of things

About how if we have the infinite product of (countable) discrete topological spaces, the open set is not necessarily discrete

[tough imp]

Hahahaha this is pretty epic ngl

It’s used to define a topology in ring theory for a completion of something

Or was it the indiscrete topology on the things you’re taking the product of I forget

[gritty widget]

geometry

[gritty widget]

question for you lie group lovers. I want to show that most pairs of rotations in SO3 generate a free group. but I'm only doing this to satisfy my curiosity, so I get to decide what I mean by "most." I don't want to use haar measure, but I would like to show that for fixed A in SO3, and some word w(a,b) in the fee group on a and b, the set of B's satisfying w(A,B)=I is either all of SO3 or it is nowhere dense. hope this makes sense. any ideas?

then the pairs that generate the free group would be a thick subset of (SO3)²

[fair idol]

Remind me of the definition of a free group plz

[rancid umbra]

if X and Y are compact and hausdorff and f : X x Y --> R is continuous then is h : X --> R, h(x) = min_{y in Y}(f(x,y)) also continuous?

[gritty widget]

what is min_y(f(x,y))

[rancid umbra]

the minimum value of f(x,y) over all y in Y

[vast estuary]

Homological algebra goes here or in #groups-rings-fields?

I'm trying to understand where the induced map is coming from, what it is, and therefore I shall compute its ker and coker

[wise sigil]

yes i think, youd have to use the tube lemma

[rancid umbra]

damn forgot about that lemma. lemme pull out munkres real quick

[rancid umbra]

hmm. how are you applying it?

[wise sigil]

we want h(U_x)) subset of (h(x)-epsilon, h(x)+epsilon)

[rancid umbra]

wait... how are you adding things in X...

[wise sigil]

Ah shit

[rancid umbra]

lel

[wise sigil]

nah still works

every point (x,y) in x x Y is contained in an open set U_(x,y) such that f(U_(x,y)) subset of (f(x,y)-epsilon, f(x,y)+epsilon)

[rancid umbra]

okay, i buy that

[wise sigil]

tube lemma goes there

you produce a tube around x x Y

and values of f in this tube vary little

[rancid umbra]

i see

[coral pivot]

Hom alg

[rancid umbra]

this was actually a question asked on mse and is apparently supposed to be proved with like, basic definitions of continuity

[rancid umbra]

wait i have a question. is U(x,y) a subset of X here?

otherwise, how are you looking at the image of that set under h?

because it seems like its a subset of X x Y rn. did you just mean to look at the projection of U(x,y) on X?

[wise sigil]

sorry U(x,y) is a subset of X x Y

Once you have the tube (in the form of V x Y for some open set V in X), just project it to X

[rancid umbra]

gotcha

ill try and piece this together tomorrow. thanks

i thought it was false for a while, so i spent some time working with some weird but cool compact hausdorff spaces to try and get a c.e.

[wise sigil]

I edited my msg

i meant f(U(x,y))

[rancid umbra]

i wonder where we're using the hausdorff assumption

[vast estuary]

for a moment i thought you're talking about me

[rancid umbra]

lol

[wise sigil]

yeah doesnt look like we use it

[rancid umbra]

dammit now i wanna go back and look for counter examples for some reason haha

i just dont like that this is true ig

[west delta]

T^T help me pls how??

[empty grove]

#geometry-and-trigonometry

[vast estuary]

Hey

[gentle osprey]

Hausdorff

[empty grove]

Whenever you have A subgroup B and C subgroup D,
A map f: B → D that restricts to A → C,

It induces a map B/A → D/C

assuming everything abelian

This is because you take the composite B → D → D/C where the first map is f and the second is the quotient

and this maps A to 0

so factors through the quotient by A

[gentle osprey]

Hausdorff

[empty grove]

B is larger in my example

wait

yeah it is in yours too

good

[vast estuary]

hmm why

[empty grove]

A maps into C which then maps to 0

under the composite

[vast estuary]

ahh true

[empty grove]

also learn to use universal properties already lol

[vast estuary]

i don't understand this line

haven't seen "factors through quotient" kinda language before

[empty grove]

That's the universal property of quotients

So whenever you have a map X to Y

such that A subset X maps to 0

[vast estuary]

i haven't learnt these well :/
they didn't do them in our algebra course sadly
where can i read them

[empty grove]

there is a unique map from X/A → Y such that the composite X → X/A → Y is the original map

[empty grove]

I learned from a dedicated cat theory book but I don't think that's the best way to learn if you just want universal properties specifically lol

someone might have good recommendations

[vast estuary]

ah okay! which book did you pick tho

[empty grove]

but one thing that is always recommended is looking at the tensor product

[empty grove]

categories for the working mathematician, mac lane. Riehl is the standard recommendation though

[vast estuary]

hmm so in our case Y = D/C and X = B?

[empty grove]

yes

[vast estuary]

oh ok. riehl was scary for me

[empty grove]

lol mac lane might be scarier

but you need not learn cat theory just to learn about univ props

you will, if you want to prove theorems about uni props in general, since then you need a precise definition of what a uni prop is, but you can get a feel for it without categories and still make your life a lot easier

people usually suggest learning about tensor products to learn how to work with uni props

[vast estuary]

ohh okay!

[empty grove]

because it is an example where you really can't work with the explicit construction

[empty grove]

but this is just first isomorphism theorem lol if you wanna think like that

The first isomorphism theorem is the universal property of the quotient

[vast estuary]

not really?

A is a subset of the kernel

A may not be the whole kernel

[empty grove]

Ye it's slightly different

the phi tilde map exists whenever A is in ker phi

it is an isomorphism iff equality

[empty grove]

some books state first isomorphism theorem with this part included

some exclude it

[vast estuary]

ohhh okay cool stuff

ah man i'm afraid if i don't do anything about it soon, i might just be weak at algebra for life

[empty grove]

dw once you have universal properties you'll be op

algebra is just universal property spam

[vast estuary]

btw instead of abelian groups here we can think of modules right

more generally

(Z-modules are abelian groups)

[empty grove]

yes

Any abelian category in fact

[vast estuary]

ooo okay!

[empty grove]

that's not any more general than what you said but sounds cooler

[quasi forum]

And that's all that matters!

[unreal stratus]

metallica moment

[quasi forum]

Okay, so I am learning about covering spaces and I have a question.
The definition says E is a covering space of B if every b in B has a nbhd U evenly covered by the continuous surjection p:E->B.

Out of curiosity? Why does it not say all neighborhoods of b are evenly covered? Why is this distinction in the definition relevant here?

[empty grove]

If every neighbourhood of b is evenly covered, then B is an evenly covered neighbourhood, so E is a disjoint union of copies of B

Not much to study when the situation is that simple

[quasi forum]

Ah, I see. Yeah, not very interesting

Okay, final question. If we lift the loop f from (B,x) onto E, then the end points of that path should be in p^-1(x) right? But the endpoints need not be the same point.

[empty grove]

Yes

[quasi forum]

Okay cool. That makes sense. Ty

Algebraic topology is a completely different beast from point set. So me asking these simple questions is just me trying to grasp what in the blazes is going on.

Okay, so here is a question. I know the reals evenly covers S^1.
But wouldn't something like [0,1] or [1,3] (really any integer intervals) also evenly cover S^1?

I know the fibers of a point is isomorphic to Z (which is also isomorphic to pi_1(B,b)).
So what makes R so special here?

[empty grove]

[0,1] doesn't, because the image of 0 doesn't have an evenly covered neighbourhood

[quasi forum]

Ah, so would it just be half open intervals?

[empty grove]

yes

There are other covers, like the S^1 → S^1 given by the circle winding around itself n times

Or a disjoint union of any other covers

[quasi forum]

Okay. But could we still determine what pi_1(B,b) is with any other cover?

[empty grove]

Connected covers are more important though, because if you have a disconnect cover then its a disjoint union of covers and you can study each on its own

And R is the best connected cover because it is simply connected

And simply connected covers make it really easy to figure out the fundamental group

[quasi forum]

But isn't something like (0,1] also simply connected?

[empty grove]

but it is not a cover

(0,a) union (b,1] is not isomorphic to a disjoint union of open intervals

A space can have at most 1 simply connected covering

and for nice enough spaces there is exactly 1

Such a covering is called a universal cover

and it covers all other connected covers of the space

[quasi forum]

Sorry, but why does this matter? I am not following

[empty grove]

basic open neighbourhood

[quasi forum]

Oh, I think I understand

[empty grove]

if there is an evenly covered neighbourhood, there is an evenly covered basic neighbourhood because open subsets of evenly covered neighbourhoods are also evenly covered

by the same decomposition essentially

[quasi forum]

That certainly makes sense

[pearl holly]

Can I ask a dumb question real quick or should I wait?

[quasi forum]

I can take a break. Go ahead

[empty grove]

only smart questions are allowed to interrupt

[pearl holly]

okay bye then 👋

[quasi forum]

Oh, in that case. Sorry, you can't ask

[empty grove]

lol ask

[pearl holly]

lmao okay wait

[quasi forum]

Wow, how exciting

[empty grove]

In fact those are all the connected covers that's even more exciting

[quasi forum]

Wait, that's it??

How do we know this?
Let me rephrase: how do you know this?

[empty grove]

yeah, circle winding around itself and R. Connected covers of a nice enough space are classified by the Galois correspondence between connected covers and the subgroups of the fundamental group

Any covering p: E → B induces an injective homomorphism on the fundamental groups

So fund grp of E is a subgroup of that of B

And if a universal cover exists, then there is exactly one connected cover corresponding to each subgroup, you can check that the above cover all cases

[pearl holly]

okay so I have $(S^1 \times K(\mathbb{Z_p}, n)^{\wedge p} )/\mathbb{Z}_p$ where $\mathbb{Z}_p$ acts on $S^1$ by rotating each point by $2 \pi /p$. The action on $K(\mathbb{Z_p}, n)^{\wedge p}$ by $\mathbb{Z}p$ is generated by $T([x_1, \ldots, x_p]) = [x_p, x_1, \ldots, x{p-1}]$. So now, why is $(S^1 \times K(\mathbb{Z_p}, n)^{\wedge p} )/\mathbb{Z}_p = (I \times K(\mathbb{Z_p}, n)^{\wedge p})/(0, x) \sim (1, T(x))$?

[quasi forum]

That was not a smart question

[pearl holly]

yo what is the latex command for ~

[empty grove]

sim

[gentle osprey]

Tokidoki ✓

[pearl holly]

oh I wrote sym

when I write ^p I mean the smash product p times

and Z_p acts "diagonally" like Z_p(x, y) = (Z_p(x), Z_p(y))

so the relation I get is $(x, y) \sim (e^{2 \pi m/p}x, T^m(y))$ for $m = 1, \ldots, p$. Now I can replace $T^m(y)$ by just $T(y)$ because that's equivalent. But I don't see why I should have $(0, x) \sim (1, T(x))$

[gentle osprey]

Tokidoki ✓

[pearl holly]

I almost want to say that S^1/Z_p is just S^1 because then I get the relation but that's false right?

[empty grove]

This is true

[pearl holly]

bruh

okay then I understand

[empty grove]

The map S^1 → S^1 given by x maps to x^p maps points of the same equivalence class to the same point, so it factors through the quotient to give a continuous bijection f: quotient to S^1

[pearl holly]

yeah so you first have an arc and then you identify the endpoints right?

[empty grove]

quotient is compact (quotient of a compact space) and S^1 is hausdorff so this is a homeomorphism

[empty grove]

yes

[pearl holly]

yo daim

[empty grove]

intuitively

😎

[pearl holly]

yo that's really nice

so like continouys bijections in compact hausdorff spaces are homeomorphisms?

is that the thing here?

[empty grove]

Yep

from compact to hausdorff

[pearl holly]

oh okay

[empty grove]

so yes in particular when you have compact hausdorff spaces

This is why CompHaus is nice category

I mean one of the reasons lol

[pearl holly]

yeah I think I remember this from munkres now when you mentioned it

[empty grove]

and compactly generated spaces capture a lot of the properties of compact hausdorff spaces

and a lot of AT happens in compactly gen spaces

which includes all CW complexes

May has a whole chapter on this

before any homie alg

[pearl holly]

yeah okay I see. I always thought that S^1/Z_2 is RP^2 because you just identify opposite points but that's also S^1?

oh daim

but that sounds like a lot of point set

[empty grove]

ye that is S^1

[pearl holly]

yeah okay I see

[empty grove]

RP^1 not 2

[pearl holly]

oh lmao yeah that's what I meant

[pearl holly]

but yeah this is really nice

thank you so much!

I just relized that slim'

[empty grove]

Ye I just joined lol

[pearl holly]

s talk already beghan

oh shit okay imma join now

[empty grove]

only just began

[pearl holly]

📺

[tawdry widget]

Is there someone familiar with knot theory? I am struggling to understand the alexander polynomial. It has something to do with the cyclic covering of S^3 “cut along a seifert surface “.here I don’t understand why S^3 cut along F into 2 pieces. Like R^3 with a disk removed remains 1 piece right?

From gtm 175 Lickorish’s An introduction to knot theory page 53. Also I don’t understand why any Seifert surface must have the following form:

[kind cedar]

Why is the interval (A,B) not open in the topology induced from f?

[empty grove]

induced topology being U is open iff its preimage is open?

[kind cedar]

I guess so

[empty grove]

The inverse image of (A,B) should be an open interval along with a singleton

There are 2 points that map to the origin

or wait

what exactly is f?

[empty grove]

oh

lol didn't notice the arrows

[kind cedar]

f is a one-to-one immersion

[empty grove]

ye so inverse image of (A,B) is an interval at the left end, an interval on the right end, and pi/2

do you see this?

[kind cedar]

Yes

[empty grove]

So pi/2 doesn't have a neighbourhood contained in the preimage

[kind cedar]

I'm having a hard time understanding what is the induced topology

[empty grove]

should be this

[kind cedar]

So, the preimage of (A,B) is not open in R, thus (A,B) is not open either?

[empty grove]

U is defined to be open in codomain iff f inverse (U) is open in domain

ye

[kind cedar]

Ah, Okay, thank you

Another related question then

If V = f(U), the book says the inverse f^-1(V) is not continuous

Once again, I see there is a problem in the point p, but I cannot express it in proper words. How is f-1(V) not continuous?

[empty grove]

what does (U) mean?

[kind cedar]

an open set containing p

[gritty widget]

f^-1 rip open loop and rip bad

[empty grove]

I mean the brackets

[kind cedar]

I meant f(U)

my bad

[empty grove]

ah

so you mean f-1(V) not open right?

[kind cedar]

Book uses the words "not continuous"

It's starting to make sense now

[empty grove]

but f-1(V) is a set 😵‍💫

How can a set be continuous

[kind cedar]

Sorry, f^-1 is not continuous

not f^-1(V)

[empty grove]

oh

right yeah the inverse image of U under f-1 is just the image of U under f

which is V

and V is not open in codomain

because there is no neighbourhood of p in it

[kind cedar]

That part is confusing

No open set V around f(p) can be contained within U

[empty grove]

within f(U)

[kind cedar]

Bad wording here

[empty grove]

ye that's the point

therefore f(U) not open

[kind cedar]

No neighborhood V of f(p) with f^-1(V) contained in U

[empty grove]

does o mean composition here?

oh lol

[kind cedar]

It seems my keyboard is eating the letter f

[empty grove]

yes

[kind cedar]

often

[empty grove]

but also no neighbourhood V of f(p) is contained in f(U)

and that's what's relevant here

[kind cedar]

Still confused. Sorry, I'm really slow rn

[empty grove]

see if this clarifies it

[kind cedar]

Not yet. I don't understand the relationship between "not containing neighborhoods" with being open in this specific case

What is an open set in the codomain?

In this case

[empty grove]

It has the subspace topology from R^2

basic open neighbourhoods are the balls

any ball around f(p) goes outside f(U)

[kind cedar]

the preimage of f(V) is not U, but U + something. But isn't this still an open set?

[kind cedar]

ah, okay, gimme a minute to think about it

[empty grove]

no need to look at preimage

[kind cedar]

Okay, f(U) is an open interval, containing f(p). V contains points outside f(U). I still can't relate this to V not being open. A set in codomain is open iff it's preimage is open. The preimage of V is U + something...

Not open?

[empty grove]

V is open by definition

f(U) isn't

You took V to be an open neighbourhood of f(p)

[kind cedar]

why?

[empty grove]

Does the codomain here have the induced topology?

it should have subspace top from R^2

not the induced one

otherwise I don't see why f inverse isn't continuous

[kind cedar]

I'm not sure, let me read it all again

subspace topology

So, in the subspace topology f^-1 is not continuous, but in the induced topology it is?

[empty grove]

yes because then f is a homeomorphism by construction

[kind cedar]

I think I got it

Need to digest it now

Thanks again

[quasi forum]

Okay, so I am having a really hard time understanding this proof. I understand the first bit, but I am a bit confused on the 2nd part of it.

Wait I get it. When it says take an element of the open cover, it is talking about one of the sets covering X. For some reason I thought it was talking about elements (which frankly does not make any sense)

[bleak helm]

Answering your own questions

[quasi forum]

The Lesbeque number is not at all a thing that is obvious to work with.

What are some good indicators for me to be looking at a Lebesque when writing a proof?

[bleak helm]

I had to google what it is, which is good because it gave you time to answer your own q

I guess the main idea is that if you consider small enough balls then they will always be contained in one of the cover sets, so look out for how you can use that

In this case you combined that with the fact that you can cover the space by finitely many balls of arbitrarily small radius

[quasi forum]

Here is another proof we did with it.
This is why I ask, because I was not expecting the Lesbeque number to appear here. But lo and behold, it has quite a significant role here.

[empty grove]

A typical example of its use is when you want to subdivide the unit square into squares, all with the same size, such that they are small enough that a certain condition holds on each

Take cover with small enough balls around each point and then take the uniform size to be a lebesgue number of the cover 🎉

Oh yeah what you're doing is also a good example

Replace square with interval

Well here all the things having the same size doesn't matter I think? I don't see why it does

[quasi forum]

The importance is that the preimage of U_b is in exactly one of the intervals

Which we can force with the Lesbeque number

[empty grove]

Do you mean that an interval is in exactly one U_b

Oh is this in the uniqueness check?

[quasi forum]

No, existence.

[quasi forum]

Yeah, my bad

[empty grove]

I don't see why that's needed

Can't you just choose any U_b and work with that

[pearl holly]

Hopf male

[empty grove]

And also how does lebesgue number imply exactly one?

It just says that there is some

[quasi forum]

No no, ignore everything I said. I misread it

[empty grove]

And also you shouldn't write the lebesgue number of a covering, because any number smaller than it also works

It's not a unique number

[quasi forum]

True. Don't shoot the messenger. I just took notes :p

Hmm, so I am not sure why this diam{x_1,x_2} is relevant

There ya go. I forgot to show what I was actually proving.

Tbh, I'm a bit lost on everything after the Lesbeque number

[empty grove]

They're just saying that {x1, x2} is contained in an element of the cover

Because the set is smaller than a lebesgue number

[quasi forum]

Ah okay. So how did that jump to the conclusion that it was in an e/2 ball in Y

Ohhhh, pre images

[empty grove]

An element of the cover is the preimage of such a ball

[quasi forum]

I missed that word

So I imagine we want epsilon/2 balls because the diam of {f(x_1),f(x_2)} has to be smaller than 2(e/2)

[empty grove]

Yes

[quasi forum]

Alright cool. I am getting it

Kinda wish I had lesbeque numbers when I first learned about compactness (in my metric space only class)

It seems like it woulda been useful

[orchid forge]

My professor told me that the lebesgue number lemma is the single most important fact of point set topology

Then a year later they put its proof as a freebie on the topology qual

Very epic style

[quasi forum]

I've learned that Lesbeque is the guy who comes up with ideas you'd never think of but are incredibly useful.

[shy moss]

here what means that the group operation in $ pi_n $can be defined by $f+g(x)=\mu(f(x),g(x))$?

[gentle osprey]

Or x1

[shy moss]

i need to prove that $\pi_n(X)$ with the standar definition of $f+g$ is isomorphic to $\pi_n(X)$ with $f+g(x)=\mu(f(x),g(x))$ ?

[gentle osprey]

Or x1

[coral pivot]

Right so i dont think they want you to show these are isomorphic

its more like, this is an alternate way we could give this a group structure

Actually nvm these are isomorphic

[shy moss]

so what i have to prove is that $\pi_n$ with the operation $f+g(x)=\mu(f(x),g(x))$ is a group?

[gentle osprey]

Or x1

[coral pivot]

right sorry i take back what i said i was mistaken

both operations will turn out to be the same

the idea is that in general if you have [Y,X] with Y a coH space (like the spheres) and X a H space, then the operation induced by both are the same

and abelian

this comes from an eckmann hilton type arguement

so i think if you assume pi_n is abelian you can bypass and get an easier arguement

[shy moss]

thanks

[long coyote]

for a given space $X$, define $S_1(X)$ to be the free abelian group with basis all paths $\sigma:I\to X$ and let $S_0(X)$ be the free abelian group with basis $X$. If $x_1,x_0\in X$, show that $x_1-x_0\in im\partial_1$ iff $x_0,x_1$ lie in the same path component of $X$

[gentle osprey]

亜城木 夢叶

[long coyote]

the reverse direction is obvious, but i stuck on the forward direction

[coral pivot]

hmm so the hint id give here is

||consider parity of the sum of coefficients coming from boundaries||

[gentle osprey]

Nobody

[coral pivot]

I realize my hint might have been a bit vague, do you want me to clarify

[long coyote]

please

[coral pivot]

like, basically look at del_1 acting on some combination of paths. basically you cannot seperate the x_1 and x_0 into coming from two different combination of paths because del_1's image has things whose coefficient sums upto 0

[long coyote]

i see, thx

[quasi forum]

So I am learning about the 1 point compactification, and I am not so sure if I understand it visually.
Intuitively/visually, what is the 1 pt compactification of Z_+?

I know it's homeomorphic to that, but I don't understand why

I'm not sure I follow why that is

[coral pivot]

the way i usually think about these is like, ok the space is not compact because you can go off to infinity basically. So to eliminate that you add in a point at infinity so that if you draw a neighborhood around this point it captures all but finitely many points, and the remaining space is now compact. This is whats happening here, and then they are using the n-> 1/n map to make this look nicer ig

[quasi forum]

Why every open set containing 0 has infinitely many other points.

Ah okay, that part makes sense

Makes sense: that gives us a finite open cover

Okay, what's another recognizable space to do 1 pt compactification with?

[coral pivot]

real number line

[quasi forum]

Okay sure. The complex plane

[tight agate]

R^n

[quasi forum]

So say we add this point to the complex plane.
So how do we ensure that a neighborhood of that point contains all but finitely many points?

[tight agate]

?

wdym

a fact that might help: one-point compactifications are unique up to homeomorphism

[quasi forum]

I think my issue is I am not understanding the intuition very well

[tight agate]

can you think of a space of the form R \cup a point which is compact

[quasi forum]

Yea, I think so

[tight agate]

what is it

[quasi forum]

Oh, you want an actual result

[tight agate]

it's definitely a space that youve seen before

[quasi forum]

Honestly, I'm not seeing it

[tight agate]

alright try working through the construction

as a set, we have R \cup point

what do the open nbhds look like

[quasi forum]

Oh, this kinda reminds me of the cofinite topology on R

[tight agate]

what are the nbhds of the extra point

[quasi forum]

R\ finitely many points

[tight agate]

those are open, yes

[quasi forum]

With our extra point of course

[tight agate]

but that's not all of them

[quasi forum]

Well, you can have the space itself

[tight agate]

what about the space - [0,1]

[quasi forum]

Ah yes. The nbhds are R-compact set

[tight agate]

(complements of compact sets in R) \union point

yes

[quasi forum]

Yea, that's what I mean

I'll just say Y for the R u point space from now on

[tight agate]

alright, now say I have the sequence a_n = n n>=0

Y is compact, so some subsequence must have a limit

what is the limit

[quasi forum]

And the only way that'd happen is if it converges to our special point

[tight agate]

ok, now what about the sequence a_n = -n

[quasi forum]

Same situation

[tight agate]

so going in both "directions" takes you to the same point

[quasi forum]

So all properly divergent sequences now converge to our point

[tight agate]

can you imagine any space where this sort of thing happens

i.e. going off in both directions takes you to the same point

[quasi forum]

This makes me think of S^1. It's like wrapping R in a circle, and the thing that tapes it together is our extra point.

[tight agate]

yup!

the one-point compactification is S^1

now try proving it

we just explicitly wrote down the one-point compactification as some set with a topology on it

[quasi forum]

Well that seems like a step up :p

[tight agate]

can you write down a homeomorphism from S^1 to Y

(alternatively, you can just use uniqueness, but this might help when youre learning it for the first time)

[quasi forum]

x-> e^(i2 pi x), but that isnt injective (so not a homeomorphism)

[tight agate]

right

[tight agate]

try using this description

try doing it informally

writing down coordinates is painful

imagine a circle sitting on the real line

or a real line passing through the center of a circle

[quasi forum]

Okay, I am following

[tight agate]

imagine this

now consider all lines passing through the north pole

[quasi forum]

Ah, this idea again

[tight agate]

actually imagine this

[quasi forum]

Yea, so every line that passes through the north pole passes through exactly one point in the reals

[tight agate]

yup

sterographic projection

[quasi forum]

Oh, didnt know there was a name for it

[tight agate]

so send every point except for the north pole to the corresponding point on the real line

and send the north pole to the extra point in Y

convince yourself that this is continuous

it's clearly a bijection, domain is compact, codomain is hausdorff so it's a homeo

[quasi forum]

Oh yeah, that nice little theorem

[tight agate]

you could just write down the inverse

it's the same procedure

now find the one-point compactification of R^n, and try using the uniqueness argument this time

[quasi forum]

I know the answer is the same idea: S^n

[tight agate]

yup

[quasi forum]

This was extremely helpful. Thank you very much

[little hemlock]

How would I find a mobius transformation which maps the real line to the circle of radius 1 centered at i and maps the circle of radius 1 centered at -i to itself?
I asked this in #real-complex-analysis earlier but was advised to ask here instead

[quasi forum]

So here is a question. If two spaces are homeomorphic, do they have the same 1 pt compactification (up to a homeomorphism)?

[little hemlock]

so, I can find a mobius transformation which maps the real line to the circle of radius one centered at i for example, but there are infinitely many of these mobius transformations, and I'm stuck on guessing which one fixes the circle of radius one centered at -i.

b = 0 is obvious, yea, but i don't think anything else is

[quasi forum]

Okay, I am having a hard time understanding this proof. Why does hausdorfness imply the closure of V is compact?

[orchid forge]

no idea it's a closed subset of a compact space so it's compact anyway right

the hausdorffness is used to say that C is closed

[little hemlock]

I don't think infinity can be fixed. The real line has to map to a circle in C

[orchid forge]

i think (by hausdorff) should be moved after the line $\bar V \subset \bar C = C$

[quasi forum]

What do you mean exactly?

[orchid forge]

C bar = C is because X is hausdorff

[quasi forum]

Doesnt c bar =C since C is closed?

[orchid forge]

C is closed because X is hausdorff

[quasi forum]

Oh duh... that's where hausdorfness matters

Also, I have never heard of sub nbhd before. Is it just a nbhd in a nbhd?

[digital peak]

I'm looking for a particular construction I guess... In the homotopy category, if we have a parallel pair of maps, we can form a coequalizer:
$$\begin{tikzcd}
P \arrow[r, "l", shift left] \arrow[r, "r"', shift right] & X \arrow[r, "f"] & {<coeq>(l, r)}
\end{tikzcd}$$
the initial object with morphism f such that $f \circ l = f \circ r$.

I'm looking for a sort of generalization: if we have a map $i : B \to J$ embedding a "boundary" into some index space $J$, we can talk about pair of maps going into $X^J$ such that $X^i$ coequalizes them:
$$\begin{tikzcd}
& X^B \arrow[r, "f^B"] & Y^B & B \arrow[d, "i"] \
P \arrow[r, "r"', shift right] \arrow[r, "l", shift left] & X^J \arrow[r, "f^J"'] \arrow[u, "X^i"'] & Y^J \arrow[u, "Y^i"'] & J \
& X \arrow[r, "f"'] & Y &
\end{tikzcd}$$
Then I'm looking for an initial object $Y$ with map $f$ such that $f^J$ coequalizes $l$ and $r$ and makes the square commute.

This object always exists (I think?): we can take an $I \times J \times P$ cylinder and $X$, and glue the $(0, j, p) = l(p)(j)$, $(1, j, p) = r(p)(j)$, $(i, b, p) = l(p)(b) \equiv r(p, b)$.

[gentle osprey]

mniip

[digital peak]

Does this have a name or maybe this is expressible through other more common constructions?

[digital peak]

motivating example I guess?

[gritty widget]

hi, do you guys have any advanced lie theory textbook recommendations?

emphasis on advanced

ping me

[rancid umbra]

what drawing app is this?

[digital peak]

it's just gimp

[gritty widget]

actually all lie theory textbook recommendations are welcome

[wicked bolt]

Hello there. Is it possible to inscribe a Sierpinski's triangle pattern into a circle and scale it accordingly to the radius of the circle?

[gritty widget]

depending on what exactly you mean, probably yes

sierpinski's outer boundary is a triangle

[wicked bolt]

So, is it theoretically possible to inscribe fractals into other fractals, if those are made up of regular shapes?

Sure.

I didn't really finish it.

[empty grove]

[gritty widget]

fractals and circles are not real

[wicked bolt]

Well, it doesn't have to be every non-void triangle;

Just that after n-many iterations, or sort of like, alternate between those.

Lets suppose I did inscribe a fractal into a fractal, but I don't want to define a lineral ratio at which those fractals scale. Like, I want to program two variables which I can control, that define how scaled each fractal is independantly (regarldess of one of them being inscribed). Is that possible?

And the final question is, how do I make iterations of that figure and how should I call it?

If I can do it in MS Paint, is it possible to write down a formula?

[viral atlas]

I don't think it would be easy to come up with a nice formula for everything you can draw iteratively

[wicked bolt]

Well, I just wondered whether I can inscribe that

Into the Sierpinski's triangle. Or at least, inscribe the inverse Sierpinski's triangle into the regular Sierpinski's triangle.

Would be somewhat useful for simulating quantum colours, or at least for ray coloring the neurally denoised game graphics.

[wicked bolt]

@wooden falconOkay, so suppose I'm inscribing a circle into every void triangle from stage n-2; is the volume approaching 0?

[stone cipher]

In Lee book Introduction to topological manifolds defines the regular coordinate ball in this way. I don't understand if r is arbitrary s.t. 0<r<r' or there must be r s.t. 0<r<r' satisfying...

ohh, nvm.. it is the second one.

[gentle osprey]

Martin Male

[stone cipher]

So $X$ is Baire space if, for every countable collection of open sets $(U_n)$ such that $\overline{U_n}=X$ for all the $n$, we have [\overline{\bigcup_{n}U_n}=X.]

[gentle osprey]

RaD0N

[gritty widget]

yes

[stone cipher]

but since $U_1\subset \cup_n U_n$ we have [X=\overline{U_1}\subset \overline{\bigcup_{n}U_n}\subset X,] hence all the topological spaces are Baire space?

[gentle osprey]

RaD0N

[empty grove]

It should say intersection not union

[gritty widget]

closure of infinite union is fucky

[stone cipher]

OHHHHHHHHHHHHHHHH

HOLY CRAP

[gritty widget]

also wow that's quite the typo

[gentle osprey]

Martin Male

[gentle osprey]

Martin Male

[digital peak]

wouldn't the coequalizer be just X with the two copies glued together? As for Hom I guess yea that's true

ok and?

[gritty widget]

hi, trying to prove this.

I started by just saying what the definition of compact is, which means when I choose a collection of open sets that cover K, and finitely many of those open sets still cover K.

The definition of open set involves open balls, but they're not in the direction I'd like, since open means there's an open ball that's a subset of each U_i (see 2nd pic), which is not the direction I need to prove the above statement. Since each ball is a subset than each U_i, I can't conclude that K is a subset of the union of those balls

[digital peak]

coequalizers

[empty grove]

Rather than trying to prove a general result here, you have to find a specific cover of K such that the given thing happens

The problem is asking you to prove that there always is a finite open cover of K of a certain form. K is compact. Then you should expect that you start by finding an open cover, such that any finite subcover of it is of the given form

You shouldn't start from a generic open cover and try to prove things about it, because a generic open cover need not have a subcover of that form

[gritty widget]

Okay, so K is a subset of R^d, which can be written as the infinite union of balls (each of the radius r in natural numbers), centered around a point such as 0... I'll think about how to make that finite. My thought is maybe find the smallest r in the infinite union such that K is a subset, which will make it finite.

[gentle osprey]

Nobody
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[digital peak]

are you saying X already satisfies the universal property for coeq(f, f)

I guess that's true

[frigid patrol]

I need help computing intersection products of explicit curves in a homology basis of a surface

My surface is the double pentagon which is a genus 2 surface and my homology basis is these particular curves denoted a_1, a_2, a_3, a_4 in this image

How can I compute the intersection product of $a_1 \frown a_2$ for example?

[gentle osprey]

nYaminoid

[frigid patrol]

I think $a_1 \frown a_2 =1$ because of the following picture:

[gentle osprey]

nYaminoid

[frigid patrol]

Is this correct?

Yep, the "matrix of intersection products" is

[bleak path]

Hey, looking for some intuition on Gauss and Mean curvatures of a surface

Firstly, I always see that we have 2 eigenvalues which we get from the first and second fundamental forms, and we use these eigenvalues to describe the 2 kinds of curvatures

First question - is a local surface always describable by 2 curvatures? Sometimes it can be 1, or sometimes none, but in general is it 2?

[worldly cradle]

Does anyone have insight how showing [0,1] is a retract of the Real Line?

[gritty widget]

define a retraction

[plain raven]

$\sin$

[gentle osprey]

diligentClerk

[minor hornet]

If I take two copies of a topological space and identify any point x in the first copy with the point f(x) in the second copy (for some group automorphism f:X->X), does the two copies of X join in a nice way? Ie. does the resulting quotient space have any nice properties?

My first instinct is that they will overlap exactly over each other (perhaps one is reflected/rotated relative to the other) and the resulting quotient space is just X, but idk if its true

[quartz edge]

am i wrong in believing that the set of all closed balls forms a basis for a topology in the plane (R^2)?

it's clear that there's a basis nbhd of every point in R^2

and every intersection of basis nbdhs contains another basis nbhd

(namely, the point itself)

this seems to satisfy the requirements for a basis

does the intersection have to strictly contain another basis nbhd

i.e. be a strict superset?

i suppose this isn't "closed balls" it's just closed circular regions

i guess this just generates the discrete topology

which is not a problem

[little hemlock]

do you mean a basis for the standard topology on R2, or do you mean a basis which generates some topology on R2?

im guessing the latter?

[quartz edge]

yeah

the latter

i don't care about recovering the usual metric on R^2

just trying to understand bases for topologies

[little hemlock]

The set of closed balls would contain all singletons, so yea u just get the discrete topology on R2

[gritty widget]

[quartz edge]

ok good

[little hemlock]

well, you would have to allow balls of radius 0

[quartz edge]

yes

[little hemlock]

otherwise no basis

[quartz edge]

right cuz

intersection of two closed balls that touch only at a point

but

i read that the discrete topology is metrizable but does not recover the euclidean metric on R^2

why is this

what metric does it produce

i haven't learned about metrization

[little hemlock]

the discrete topology can be generated by the discrete metric defined by d(x,x) = 0 and d(x,y) = 1 for all x != y. the topology on a metric space is the topology generated by a basis of open balls radius r > 0. so B(x; 1/2) = {x} is open for each x.

on the other hand, in the euclidean metric, every open ball containing x contains points other than x, so {x} is not open in R2 with the topology coming from the euclidean metric

[quartz edge]

metrics just have to be positive definite and satisfy triangle inequality right

[little hemlock]

they have to satisfy symmetry d(x,y) = d(y,x)

[quartz edge]

ah yeah

i knew there was another thing

kk

[gritty widget]

How do I show the surface $M:=S^2/\sim$ is not orientable, where $\sim$ is an equivalent relation on unit sphere $S^2$ in $\mathbb{R}^3$ defined by $(x,y,z)\sim (-x,-y,-z)$ for all $(x,y,z)\in S^2\subset \mathbb{R}^3$?

[gentle osprey]

weilam06

[cosmic beacon]

Let's say I have two spaces $X$ and $Y$ and I know all of their $n$-sheeted path-connected covering spaces, up to isomorphism with basepoints. Is there an ``easy'' way to enumerate the path-connected $n$-sheeted covering space of their wedge sum $X\vee Y$, similarly up to isomorphism with basepoints?

[gentle osprey]

Isaiah

[finite heath]

classic

👀

[gentle osprey]

Moldilocks ✓

[wise sigil]

yeah it depends on the chart

[empty grove]

Thanks

[gritty widget]

why are lines in projective planes required to have at least three points?

[gritty widget]

There are several different definitions of tensors right? do all those definitions describe the same mathematical object?

[winged badger]

to get rid of boring examples of projective planes, like a "plane
" where all points except one lie on a single line

[gritty widget]

I think there is just one general construction

[winged badger]

i can think of at least 2

[gritty widget]

idk I thought just construction for modules is the most general one

[cosmic socket]

I feel like there isn’t just based off S1 V S1

The free product makes things complicated

Essentially you are asking if there is an easy way to index n subgroups of a free product

[empty grove]

Which tensor/definitions of tensor are you referring to?

[shy moss]

if X is path connected what exactly is the homomorfism between $\pi_n(X,x_0)$ and $\pi_n(X,x_1)$?

[gentle osprey]

Or x1

[empty grove]

Take path from x0 to x1, conjugate any element of one group by this path (in the only way that makes sense) to get an element of the other group. Do this backwards to get an inverse map

[shy moss]

i think that for $f:(I^n, \partial I^n) \rightarrow (X,x_0)$ i can define $\varphi (f)(s_1,..,s_n)=f(3s_1-1,...,3s_n-1)$ when $1/3 \geq s_i \geq 2/3$ for all i

[gentle osprey]

Or x1

[weak narwhal]

probably using multilinear maps vs arrays

https://en.wikipedia.org/wiki/Tensor_(intrinsic_definition) oh theres also this (in which case yes, it corresponds with multilinear map defn.)

[sonic hill]

Given a finite set { V1, ..., Vn } of vector spaces over a common field F, one may form their tensor product V1 ⊗ ... ⊗ Vn, an element of which is termed a tensor.

Done, and done

[gritty widget]

#groups-rings-fields

[sonic hill]

Not so fast

[weak narwhal]

?

[sonic hill]

v

Differential geometry, physics and engineering must often deal with tensor fields on smooth manifolds. The term tensor is sometimes used as a shorthand for tensor field. A tensor field expresses the concept of a tensor that varies from point to point on the manifold.

So for example, the metric tensor in general relativity is actually a type (0,2) tensor field over spacetime

[gritty widget]

#prealg-and-algebra

[weak narwhal]

its not bad to ask it here imo

[gritty widget]

has any read chapter 1 of luries higher algebra

and if so

do you know of an easier reference

[weak narwhal]

since clifford algebra stuff gets moved into tensor territory

[weak narwhal]

https://math.stackexchange.com/questions/10282/an-introduction-to-tensors anyways this might help

[plain raven]

it's funny when people recommend higher algebra for no reason

I saw a stackoverflow post where someone asked about the Dold Kan correspondence and the answer was "Look in higher algebra"

😐

[tight agate]

yes

[tight agate]

no

[fading vale]

Is this true on a formal level or just like heuristically

(i assume this is a typo and they mean n > 1 btw)

(a K(pi, n) is an nth-eilenberg maclane space with group pi as you might expect)

[plain raven]

haha

that's interesting

i wonder what that would mean

[fading vale]

Idk

Maybe it means like

can be equipped with a riemannian metric

idek if thats true of CP^infty but its all i can think of

like even being a manifold is too weak to say this because loop spaces i think

Frétchet manifold structure moment

I guess thats only true of smooth loop spaces but same different i think

[sonic hill]

It's probably just saying that there is no simple description or construction of the higher spaces except by attaching arbitrarily high dimensional CW complexes inductively based on what the higher homotopy groups happen to be

Whereas $\bC P^\infty$ is just, well, the inductive limit of $\bC P^n$, or if you like, $\bC^\infty/\bC^\times$

[gentle osprey]

Icy001

[cosmic beacon]

What would be the induced subgroup of the given 2-sheeted covering space ~X of X = RP2 v T2?

Where ~X is the 2-sheeted covering space obtained by attaching two tori to the 2-sheeted cover of RP^2 (the sphere)

[cursive flume]

how could I prove ii) formally?

it seems very intuitive,but idk how to write it down formally,I don't have tools

for i) I proved that M1xM2 is a topological space,it is hausdorff,it is second countable and it is locally euclidean

for ii) idk how to write down that the atlas is smooth

[cosmic beacon]

It would be <b,c,aba,aca>, right?

[cosmic socket]

This looks right (mod commutator for b,c)

[cosmic beacon]

Thanks!

[gritty widget]

just show that the transition maps are smooth

(and that you actually cover M_1 x M_2, but that should be obvious)

[cosmic beacon]

Consider the torus with fundamental group Z^2 = <a,b | [a,b]>. It has 3 isomorphism classes of two-sheeted connected covering spaces, each associated to one of the groups <a^2, b>, <a, b^2>, and <a^2, ba^(-1)> (kernels of 3 possible non-trivial homomorphisms from Z^2 to Z/2Z).

My question is: what does the covering space associated to that last subgroup look like?

[gritty widget]

These are excellent thank you!

Awh thanks so much this has exactly what I looking for

[flint cove]

I didn't ask for this but it's amazing, thanks

[kind cedar]

What' the meaning of "As the zero set of finitely many continuous functions" ??

got it

[true garden]

Hello. Why is there no function $f\colon \mathbb{R} \to \mathbb{R}$ that is continuous precisely on a countable dense subset of $\mathbb{R}$?

[gentle osprey]

MathPhysics

[coarse night]

There is also another way of viewing it. The set of discontinuities of a function is necessarily a F_sigma set. If you want your function to cont only in a countable set (which is F_sigma) and disc at the complement (which is G_delta), is not possible because complement of countable dense subset of R cannot be F_sigma. I would suggest you to look into Abbott-Understanding Analysis for further reading

[shy moss]

when $A=x_0$, we have that $\pi_n(X,A,x_0)=\pi_n(X,x_0)$?

[gentle osprey]

Or x1

[hollow harbor]

Blatantly false, any step function is continuous everywhere but the steps. Uniform continuity on a dense set implies continuity on R.

[empty grove]

slim should hang out more in #groups-rings-fields

[coarse night]

did not get this, can you elaborate?

[empty grove]

step functions are continuous everywhere but the steps

[coarse night]

[empty grove]

everywhere but the steps is a dense subset

[coarse night]

No i mean there is a function that is cont in Q and disc in Q complement?

[empty grove]

no rice was contradicting slim who made a stronger statement

which was dropping countability

[coarse night]

oh i see

[cursive spade]

let ~ be the equivalence relation on $S^1$ that identify antipodal points

[gentle osprey]

bchaotic

[cursive spade]

I am having trouble writing down an explicit homeomorphism between $S^1$ and $S^1 /\sim$

[pearl holly]

yo moldi already answered this to me before hol up I will link the chat

[pearl holly]

the same argument applies right?

[empty grove]

also \sim for ~

[pearl holly]

don't do \sym

[gentle osprey]

bchaotic

[cursive spade]

@pearl holly can you explain it

[pearl holly]

So you have a map from S¹ to S¹ given by x to x² in the complex plane and then by the universal property of quotients it must factor through the quotient. That map will be bijective since the map from S¹ to S¹ is bijective and the projection is surjective and injective. Then you have a continous map from the quotient (compact) to the sphere (Hausdorff) so it is a homeomorphism but this was moldi's argument so maybe moldi can explain better

maybe that's wrong

[empty grove]

the map from S^1 to S^1 is bijective

[pearl holly]

wait so how do you know that the map froim the quotient is bijective?

ye bchaotic listen to moldi instead

[empty grove]

so you have s = tp, where s is the squaring map, p is the quotient map, t is the induced map

whenever s is surjective, t is surjective (general fact about such factorings)

[pearl holly]

oh lmao

[empty grove]

t is injective because we identified everything that maps to the same thing

[cursive spade]

I was thinking of first mapping to a helix covers the circle twice identify end points and then to the quotient space of S^1/~

[pearl holly]

ye I thought that this only holds when s is bijective lmao

[empty grove]

cat theory book chapter 1 exercises be like

[cursive spade]

why the squaring map though?

[pearl holly]

because then it will be constant on the quotient, right? So that allows you to use that property

don't listen to me tho, let moldi answer if moldi want's to

[empty grove]

ye constant on each equivalence class

squaring like in complex numbers

point at angle t from x axis goes to point with angle 2x

[cursive spade]

so the squaring so you rap around twice , so x and -x maps to same point ,

but how do we get the existence of t ? @empty grove

[empty grove]

universal property of set quotients

define [x] maps to sx

check well defined

prove continuity

but

you can prove universal property of quotients once and spam it whenever you see a quotient

10/10 would recommend

[empty grove]

if you don't wanna do uni props follow this route

[kind cedar]

What's the difference between an open neighborhood and a coordinate neighborhood? Can't they be the same

I'm often confused by demonstrations that take an open set and then inside the open set take a coordinate neighborhood. Like... why? Why not just taking the open coordinate neighborhood from the beginning?

Why must I take a V0 which is the intersection of W and V to make the chart? Why not just considering V entirely, or W entirely?

I've seen many other examples which confsued me

\iota is an embedding in V

why do we use the W?

[gritty widget]

open neighborhood = open set containing a usually specified set or point
coordinate neighborhood = open set on which a chart in your atlas is defined

[kind cedar]

V can be suficiently small to be identical to W, not?

Why not?

Humm, I'm having a hunch

V is small enough to allow adapted charts

those adapted charts may or may not be as big as V

maybe smaller, thus the Ws

But then why define V0 = W intersec V, instead of just using W? Since W must be inside V

[gentle osprey]

Jexus Smith

[kind cedar]

It's the same

I'm highly confused by some terms.

First, V was defined in S. Then all of a sudden it's a set in N?

Wouldn't it be \iota(V) ?

Neither did the book

Aaah, that helps a lot

that picture changed my life ahaha

I'm seeing everything different now

it's the one I'm taking the example from

hahahaha

Let me check

version 3.0 means???

It must the first edition then

😦

I feel decepted

[gritty widget]

you've been using the old version this whole time??

[kind cedar]

ahahah it seems so

[honest narwhal]

This is probably a draft version lol

[kind cedar]

Ah, that's why there were NO pictures at all and I was super confused

I was hating the book

Now that I have (totally legally) acquired the new edition, things look very different

[gritty widget]

[kind cedar]

ahahha

[pearl holly]

[pearl holly]

this might be a weird question but why would one ever want to work with the smash product instead of the Cartesian product?

like are there some "benefits" of using the smash product over the Cartesian product?

[empty grove]

I think it's the product in the category of pointed spaces Might be wrong

[pearl holly]

wait hol up I will be back soon

[empty grove]

Oh it's not

[pearl holly]

okay I'm back lol

[empty grove]

It's the left adjoint of the internal hom in nice subcategories of Top

[pearl holly]

😵‍💫

[coral pivot]

smash products is tensor product in pointed space i believe

[pearl holly]

oh crap

ye okay that might explain a lot actually

[empty grove]

I said the same thing but more precisely

[coral pivot]

you said product

smh

[empty grove]

Goddamn degenerates

[empty grove]

This lol

[coral pivot]

ah oof didnt see that

[pearl holly]

ye that's over my head lmao

[empty grove]

[pearl holly]

but yeah when Hatcher constructs the Steenrod operations he "randomly" switches everything up and starts using smahs products and instead of using the cross product he starts to use the tensor product from nowhere and it just seems weird

[coral pivot]

if ur that far into hatcher, you should start using a different book

[pearl holly]

and I went through the construction and it felt like nothing would break if I would just replace everything by the cross product and use the Cartesian product

[pearl holly]

I skipped a BUNCH tho

Max said this was cool so thjat's why

[empty grove]

prankd

This is why you trust no one. Let this be a life lesson

[pearl holly]

I mean it is cool and the construction is very nice but I just have to go through it a couple more times to truly understand it

[pearl holly]

because things like these appear all the time

He says like "the construction of a certain function will take place cell by cell, so we may get rid of all the unnecessary cells by replacing XxY with X \wedge Y"

something like that

[empty grove]

Hatcher 😩

[cursive flume]

so in this problem,I can show that there is a unique topological structure on t^n

do I have to prove that this topology is hausdorff/second countable?

[gritty widget]

yes

sincethat's part of being a manifold

[pearl holly]

maybe this is important for the cell structure

[gritty widget]

can u link the post

[cosmic beacon]

Figured it out (I’ve drawn the lifts of the generators of the torus)

[digital wraith]

Hatcher is fine

[fading vale]

Most of the time the main reason to care is tensor hom adjunction

[pearl holly]

I don't know what that is

but I think I see where the construction would break if we replaced the smash product wiht the Cartesian product now so it's fine