point-set-topology

[unreal stratus]

Well sorry yes I forgot like I mean the thing they drew with the obvious quotienting

Uh I mean I guess you wanna check like stuff is uniquely determined by vertices

That is probably it lol

But yeah weird

Well like so there is an implied like labelling of vertices etc

If you don't subdivide enough you could have distinct triangles w the same vertices or smth like that

As an extreme example, we can't take there to be just one 2 simplex for either shape since then you have a 2 simplex w one vertex lol etc

I think that is basically enough yes well you want to show that triangles/edges which share the same vertices are actually the same etc

So like for the torus it is implied the bottom middle edge is identified with the top one

But that is okay cause in our structure we are actually gluing those so they are the same vertices and same edge lol

But yeah this is weird

Of which shape

Uhhh

Okay so it seems fine to me at first glance lol

Wait no so

You have two triangles which share all vertices

But they are distinct

Ye

These r a meme tho cause if they ask you for a triangulation you can usually just cut up a ton around any places where stuff is identified and be gucci

Like the problem here is the identification on the boundary of the square

So we can just separate the opposing sides by subdividing further

[pastel linden]

reposting here in case someone knows: #diff-geo-diff-top message

[echo oyster]

How do I show that polynomials are continuous using product topology?

[gritty widget]

?

[unreal stratus]

Any anime recommendations?

[coarse night]

on the scale of 0-10 how tedious is it to calculate homology of $\mbb{R}P^n \times \mbb{R}P^m$?

[coarse night]

[gritty widget]

yeah, I have some

any particular genre?

[unreal stratus]

Hm tbh I quite like cute like slice of life type stuff but i really don't mind lol not watched much

Simplicial approximation theorem yes

Which maps into X are relevant when we're considering the fundamental group of X?

[gritty widget]

try Mikakunin de Shinkoukei

[unreal stratus]

oh hadn't heard of it

thanks

[gritty widget]

topology and anime is what I do

[unreal stratus]

aha

[lunar yoke]

Depends on the theorems you are allowed to use. If you already know the homology of RP^n, and can use künneth, its like a 3/10 i‘d say

[coarse night]

Using Küneth btw

To me it’s 10/10

[unreal stratus]

this is making me realise how lazy i have been with actually computing stuff in alg top

[lunar yoke]

Have you ever done spectral sequence computations

[coarse night]

No

[unreal stratus]

Eh Künneth is just a spectral sequence anyway

[coarse night]

Not touching spectral seq then

[unreal stratus]

(...which i need to look at at some point oop)

[coarse night]

[pearl holly]

Keep it that way, you don’t want to go insane

(Im joking ofc)

[unreal stratus]

Go to the dark side

not just that

well what is the fundamenta l group

well

homotopy classes

so you aren't just interested in maps [0,1] -> X

yup

so the point is that any map and any homotopy can be assumed to go into the 2 skeleton

Or I guess to be slightly more rigorous you can show the inclusion of the 2-skeleton into the whole thing induces a bijection on pi1

(surjectivity is easy by homotoping loops, but injectivity needs you to consider maps from the square)

oh like

i mean so this is just simplicial approximation theorem

so are you asking for intuition for that theorem

[0,1] is naturally a 1 dimensional simplicial complex

so you can take any map [0,1] -> X and homotope it to a simplicial map [0,1] -> X (for some new simplicial complex structure on [0,1])

And then since it's simplicial, the image will lie in X^1

simplicial maps map the k skeleton to k skeleton

So in particular the 1-skeleton of [0,1], that is, [0,1], is mapped to the 1-skeleton of X

okay maybe you use a different definition to normal lol

which definition are you using

Well okay so it maps vertices to vertices but importantly the images of the vertices of a simplex span a simplex

So in particular, given any k-simplex in the domain, the images of its vertices span a simplex, which is necessarily of dimension <= k just because there are only <= k distinct vertices

Sure yeah

[unreal stratus]

that should imply this

ye

show kernel is 0

well so we can view it as being the same space [0,1] but with a different simplicial structure

but yes, you have a path f: I -> X, then you can take some homotopic g; I -> X whose image is in X^2

Then restricting that, we get a map h: I -> X^2, such that the map on π1s sends [h] to [g] = [f]

ye

[unreal stratus]

more generally just like subdivision preserves dimension and so u can use this

np owo

worth pointing out maybe that the same thing occurs with CW complexes

since there's the analogous cellular approximation theorem

[unreal stratus]

Oh rip

Uhhhh

Okay so like basically

A cw complex is a space $X$ which has subspaces $0 \subset X^0 \subset X^1 \subset \dots \subset X$ where $X^0$ is just any discrete set and formed by taking some (discrete) set $X^0$ and given $X^{n+1}$ is obtained from $X^n$ by attaching disks $D^{n+1}$ to $X^n$ via maps $S^{n} \to X^n$

[gentle osprey]

potato

[unreal stratus]

(you then need to make sure X has the right topology etc but here let's focus on the finite dimensional case, where this process terminates after finitely many steps and we only attach finitely many things at each step)

But yes the key idea is basically it's a space formed by attaching disks (well, 'cells') and it's built up dimension by dimension

So like we take some points, attach some 'lines' between points (or loops from points to themselves), then add in some disks whose boundaries are those loops/lines, and so on

[unreal stratus]

From that kinda handwavy description it includes simplicial complex as a special case, cause ur taking points, adding edges, then triangls etc

The thing is that this is a lot looser than simplicial complexes lol

like u can glue stuff with fewer restrictions

and ye cw complexes are very general which is nice

like i fink any compact manifold has the homotopy type of a cw complex and for every space Y there's a cw complex X and a map f:X -> Y inducing isomorphisms on all homotopy groups (CW approximation) and stuff like that

[the thing i am working on rn is 99% CW complexes lol]

[jagged sage]

Im trying to prove that for x_1,...,x_n distinct points in a Hausdorff space, that there exsits disjoint open U_1, ..., U_n where x_i in U_i. I tried proof by induction but dont know how to conclude that the n+1 set is disjoint

[gritty widget]

since the space is hausdorff, for each i=1,...,n you have a V_i open around x_(n+1) that is disjoint with U_i

can you use this to finish the proof?

[unreal stratus]

also like you can think about it diagrammatically if ur not sure

[jagged sage]

Yeah, that part i understand that but not sure how to use it to prove the U_n+1 disjoint

[gritty widget]

you don't have a U_(n+1)

it's your job to construct it

[jagged sage]

how can i use the v_i disjoint to construct u_n+1

[unreal stratus]

View it as two glued triangles i guess

or actually as just one triangle lol up to homeomorphism

But yes it is similar

[gritty widget]

heres a diagram to help you visualize the situation

V_1 is disjoint from U_1 and V_2 is disjoin from U_2

you want a open U_(2+1) around x_(2+1) that is disjoint from both U_1 and U_2. Can you think of what to do in this case?

[jagged sage]

oh take intersection right

[gritty widget]

Yes!

[jagged sage]

got it thanks picture was super helpful

[gritty widget]

Every element is a face of itself, here proper means it is not a face of anything bigger

[calm current]

Hi, I have an assignment to write a couple of pages on metric spaces in regards to topology (we have not yet had metrics be mentioned in the lecture thus far) and so I am experimenting a bit.
It's pretty clear that every topology induced by a metric is hausdorff, but I can't really figure out whether for every hausdorff topology there is a metric which induces it

Does anyone have an idea?

[gritty widget]

not every hausdorff topological space is metrizable

fun tool: https://topology.jdabbs.com/spaces?q=t_2 %2B ~Metrizable

[calm current]

Thanks

[pearl holly]

oh wow

[calm current]

This tool is actually really useful for my assignment

Thanks a lot

[gritty widget]

This is a hard problem actually

To give a characterization of metrizability

And there are Nagata-Smirnov or Bing's metrization theorems for that

For separable metric spaces we have a simpler Urysohn metrization theorem

[gritty widget]

They rely on paracompactness of metric spaces

There's also metrization theorems in terms of star refinements

Theorems by Morita, Stone and Archangelski

[gritty widget]

There's lots of properties satisfied by metrizable spaces that aren't satisfied by all hausdorff spaces, just pick one of them and find an hausdorff space not satisfying it

[hidden crag]

if only i found this when i was digging up counterexamples for my topology oral exam

[gilded oyster]

So I have a book in a semi rare speciality, it's called topological graph theory, studies embeddings of graphs onto topological surfaces

I have a decent enough background in graph theory, I've also gotten through point-set topology, i'm still finding it a difficult read past chapter 1 or 2, any ideas about prerequisites, here's a link to the book: https://www.amazon.com/Topological-Graph-Theory-Dover-Mathematics/dp/0486417417/ref=asc_df_0486417417/?tag=hyprod-20&linkCode=df0&hvadid=312130957577&hvpos=&hvnetw=g&hvrand=2282354625504542619&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9016510&hvtargid=pla-553995538720&psc=1&asin=0486417417&revisionId=&format=4&depth=1

[little hemlock]

dumb question, but let B be an open ball in R^n, and x in B. Why does the inclusion of pairs (R^n, R^n - B) --> (R^n, R^n - {x}) induce an isomorphism on H_n?

[unreal stratus]

Whilst there's probably an easier way to do this, it does follow from the corresponding morphism of LES i think

[little hemlock]

hm right, okay

[unreal stratus]

I think though that the second pair of spaces admits a deformation retraction into the first

Since you can send any of the points in B \ {x} into R^n \ B radially

[little hemlock]

okay, so ur talking about (very informally speaking) squishing B - {x} into the boundary of R^n - B?

so, i can see that would show that the inclusion R^n - B --> R^n - {x} induces an isomorphism on homology, but does that give you isomorphism on relative homology w/o going through the LES morphism?

[gritty widget]

I'd suspect a book like that to use tools from algebraic topology more than general topology

[coarse night]

Follows from five lemma?

[void gazelle]

Hi, guys, is there any hint for the proof of lemma

[gritty widget]

#real-complex-analysis

Anyway this follows from continuity

[void gazelle]

Oh, X and Y does not have to be Euclidean

[gritty widget]

what do you mean by euclidean?

it says "holomorphic" so complex surfaces are what they probably mean by "non euclidean"

[fossil spindle]

What are some of the things that make a fibration p : E -> B useful?

Does it let us study (some) homotopies from X to E in terms of simpler homotopies from X to B?

Or is it usually the other way around, fibrations are helpful because the base space B and homotopies into it are easier to understand when spread out in terms of the bigger total space E?

Looking at: https://en.wikipedia.org/wiki/Fibration

[glass bison]

the three topologies are the topologies of uniform, pointwise and compact convergence

I am unsure how to prove for uniform and compact convergences

[fossil spindle]

can you recall how to prove uniform or compact convergence for any simpler examples?

[glass bison]

There don't seem to be many examples but I'll check again

at least in my book

here's a definition for compact convergence

not def thm srry

uniform

I would assume x/n converges uniformly and in compact

actually not uniformly

since the case for n -> inf/-inf wouldn't satisfy that I think

since x can be arbitrarily large, but then so can n so I'm confuzzled lol

[fossil spindle]

@glass bison First, lets establish that f_n converges pointwise to the function f, where f(x) = 0. To see this, note that at an arbitrary point, f_n(x) = x / n, which tends to 0 as n -> infty.

Does it seem plausible that f_n -> f = 0?

[glass bison]

Yes

does this idea work for uniform?

[fossil spindle]

looks like a start

Now, lets suppose that f_n converges uniformly.

Then we should be able to globally bound |f_n(x) - f(x)| for all x and for all n suitably large.

But in our case,

|f_n(x) -f(x)|
= |x/n - 0|
= |x/n|

So we need

|x/n| < e

for some bound e.

But no matter how large n or e are, we can break this inequality by setting x = n * e.

@glass bison oh, that's exactly what you wrote

[glass bison]

Yeah I think what I did works

seems to be the same thing

I think we're chillin there

the real funky one is the compact convergence

[fossil spindle]

@glass bison I think in the compact case, you just try to prove uniform convergence of the restriction $f_{n | [a, b]}$ of $f_n$ to an arbitrary interval $[a, b]$

[glass bison]

can all the compact sets on R be described by closed intervals

Alright

[gentle osprey]

Mars Industrial

[fossil spindle]

I'm actually not sure if it applies here. Every compact set X in R is the disjoint union of a countable set of intervals X_1, X_2,.... But proving that f_n is uniformly convergent on each interval X_i may not be enough to prove f_n is uniformly convergent on X. I'm not sure, definitely do some study and convince yourself what is right here!

[glass bison]

ah ok

I thhink it might be here

[fossil spindle]

yeah, there is something that makes me feel we should be able to shortcut this way here. I'll get back to you if I figure out why

[glass bison]

alright thanks man

[fossil spindle]

@glass bison I have a parting idea.

If f_n converges uniformly on every interval in R, then it converges uniformly on some interval X* containing our compact X. So it should converge uniformly on X itself as well. Definitely verify if this is reasonable or not!

Take care!

[glass bison]

Sounds good, thank you!

[unreal stratus]

One major thing is that the existence of a fibration can tell you a lot about the (weak) homotopy type of a space e.g. via the Serre spectral sequence, or even just by the associated LES

but also the properties are interesting enough in themselves to warrant interest ig

[little hemlock]

ur talking about applying five lemma to the LES morphism, right? I mean, sure, i think i understand that. I was moreso curious if there was some super conceptually simple application of like, excision or something, that i was missing since Hatcher deemed it clear enough to write down H_n(R^n, R^n - B) \cong H_n(R^n, R^n - {x}) without any explanation

[wise ruin]

I'm having a bit of trouble parsing this Hatcher problem: Prove the Brouwer fixed point theorem for maps f : D^n→D^n by applying degree theory to the map S^n→S^n that sends both the northern and southern hemispheres of S^n to the southern hemisphere via f.

The actual theorem is of course for arbitrary f, so this added condition is confusing. Are we supposed to assume that f maps the northern hemisphere of S^n to the southern hemisphere, are we supposed to assume/prove this can be done without loss of generality, or something else?

[little hemlock]

im about to leave, but as a hint iirc you construct a function g : S^n -> S^n which acts as f on each of the hemispheres, and globally maps both hemispheres to the southern hemisphere

[fossil spindle]

See Topics on Topological Graph Theory, published in 2009. See also Graphs of Groups on Surfaces (2001) and Graphs on surfaces (2001).

[little hemlock]

(something like this)

[fossil spindle]

@gilded oyster "Topics" has a helpful a bibliography for each chapter. Each section links to key papers for the actual "tools". Eg, rotation systems, current graphs, voltage graphs.

[wise ruin]

Thanks kxrider, but I'm not asking for a hint on the solution. I'm fine with tackling it once I can exactly parse the question, which was my original question. Feel free to reread it if you need to

[coarse night]

can you refer to the page number?

[coarse night]

well homotopy then, (Rⁿ, Rⁿ\x) and (Rⁿ, Rⁿ\b) are homotopic

[wise ruin]

Exercise 1 on page 155

[coarse night]

nope no idea what he means

[wise ruin]

Lol thanks, it's nice to have a second pair of eyes for things like these :)

[fossil spindle]

@gilded oyster I think you might like Graphs of Groups on Surfaces. It paces itself and spends time on prerequisites in the first half of the book.

Hrm, sorry for not posting this in #book-recommendations, didn't realize the images would stack vertically

[little hemlock]

yes, i understand your question

hatcher is asking you to construct a new map g : S^n -> S^n using the map f : D^n -> D^n you're given. that's what he means by

applying degree theory to the map S^n→S^n that sends both the northern and southern hemispheres of S^n to the southern hemisphere via f

i.e. you shouldn't have to assume anything about f

[little hemlock]

okay, if im understanding correctly, the inclusion has a deformation retract which is a homotopy equivalence of pairs?

[jagged sage]

would anyone be able to review a presentation on simplicial complexes? it is fairly short and elementary but I learned about them as I did pres so want to make sure I understand

[gritty widget]

i am struggling to prove if a map from S1 to S1 is nullhomotopic it has a fixed point

any ideas?

i want to somehow use browers fpt but cannot see how

[little hemlock]

you're familiar with degree of maps S^n -> S^n, right?

[gritty widget]

no..

i think we briefly defined degree of a map of S^1 to S^1 in proving pi1(S^1) = Z

[little hemlock]

ah okay, thats fine

[gritty widget]

iirc its number of times it loops around

[little hemlock]

if your map is nullhomotopic, what is the induced map on pi1?

[gritty widget]

it is the trivial map

[little hemlock]

yes. Now suppose your map has no fixed point. Try to construct a homotopy of f to some other map such that the induced map on pi1 is nonzero. then you'll have a contradiction

[gritty widget]

i guess my first thought is to define g(x) = (f(x)- x)/|f(x)-x| but i dont see any obvious homotopy from f to g and i also am not certain if/why the pushforward g_* is nonzero

[little hemlock]

Nah, that’s not the right g, but you are on the track by considering functions that could not exist if f had fixed points. Instead, the correct g is a rather simple function and the homotopy you construct between f and g should not be able to exist unless f doesn’t have fixed points

[gritty widget]

ok i will keep thinking

ty

[little hemlock]

np

[odd flame]

is it accurate to say that the product topology on X1 x X2 x X3 x ... is the set of all products of Xi with an open set Uj ⊆ Xj in finitely many indices?

maybe it's phrased weirdly

[gritty widget]

that's fine. phrased a bit strangely, but it's not wrong this is merely the basis for the product topology, but you have described that correctly

tripped me up

it'd be better to say that finitely many of the indices are allowed to vary between the open sets of the X_i's

[odd flame]

idk i just see it better like that, the definition munkres gives in terms of projection mappings is weird to me

i like that more tho

[gritty widget]

what's munkres's definition?

[odd flame]

one sec

at the bottom

[gritty widget]

box topology kekw

[odd flame]

it's just one of those things that might appear on my exam

[gritty widget]

the preimage of U_b by π_b is just the product of all the X_a's, except when a = b you get U_b

when you take the intersection of finitely many of these, you get the usual basis open sets for the product topology

[gritty widget]

which are defined this way

that's how they're related

[gritty widget]

also this is technically wrong, these only form a basis

should have caught that sooner, oops

my brain inserted an "is generated by"

@odd flame

[odd flame]

one sec im on mobile

will read in a min

[gritty widget]

basically what you described is actually only the basis for the product topology

you need to go one step further and take unions to get the real product topology

[odd flame]

i see what you mean

i get that it's a subbasis but are there any elements of the product topology that might not look explicitly like this

[gritty widget]

for a continuous function from X to Y, if Y is T2 will X be T2 only if f is injective? or not necessarily? I think it has to be injective to determine X is T2 but not sure

[odd flame]

i answered my own question but still posted it so i can come back later

but intersections would prolly give an open set like that

[unreal stratus]

Injective is the normal criterion I guess yes

If two points are distinct, find disjoint opens about their images and then take the preimages

[gritty widget]

open ball in R^2

that's not a rectangle

[unreal stratus]

Anything other than a rectangle

[gritty widget]

lol

[unreal stratus]

So anything you can draw by hand

[gritty widget]

ye I did that but can we infer the preimages of the open sets are disjoint without the function being injective?

[unreal stratus]

No

If the function isn't injective there's nothing we can say about Y from X

After all, if Y is nonempty then every space admits a map to Y

[gritty widget]

ye that's what I thought okay

thanks

[unreal stratus]

Np!

[gritty widget]

Also does the same hold if Y is T3 instead of T2?

[gritty widget]

is composing a covering map with a homeomorphism necessarily a covering map?

i cant seem to find anything about it

[coarse night]

Try to use the def of a covering map

[candid hedge]

what does it mean that A is diagonalizable , iff the set of all similar matrices of A is a closed set in Mn(C)

what's it actually saying, in english

intuitively

[gritty widget]

does it help if i tell you that the set of all diagonalizable matrices in M_n(C) is dense in M_n(C)?

[candid hedge]

i dont think so

i am not sure how it fits here

[unreal stratus]

Lies made to sell more topology

[limpid hedge]

yo, if we have a homeomorphism $f:(X,d_X)\to (Y,d_Y)$ do the metric $d_X$ and $d(x,y):=d_Y(f(x),f(y))$ induce the same topology on $X$?

I think yes just using def of continuity of f and its inverse

[shadow charm]

i assume you mean$ d(x,y) = d_Y(f(x),f(y))$. Then $d$ and $d_X$ are topologically equivalent: take a ball $B_{d_X}(x,\epsilon)$ around $x$. Then by continuity of $f^{-1}$, there is an open ball around $f(x)$, $B_{d_Y}(f(x),\delta)$ such that $f^{-1}(B_{d_Y}(f(x),\delta))\subset B_{d_X}(x,\epsilon)$. The former is an open ball with respect to $d$

same thing for the opposite inclusion

[limpid hedge]

ye ok thx i was thinking about the same proof just to be sure

[gentle osprey]

𝓛ittle ℕarwhal ✓

𝔻аniil

[gritty widget]

why do you always add this word to your questions. Not everything can make intuitive sense

[candid hedge]

to make it clear i am not looking for rigorous proof

Ik not everything can make. A lot of things can, so i ask

and usually after i ask for intuition i try to come up with the proof myself

see if they match

[gritty widget]

oh, I see

context

[odd flame]

try detexify maybe

[odd flame]

does cofinite topology = finite complement?

[hidden crag]

the sets with a finite complement are the open sets in the cofinite topology, yes

[gritty widget]

you could say it's the space in which the only closed sets are finite sets and the whole space

[prisma arrow]

the empty set also 🙂

[gritty widget]

empty set is finite

[hollow charm]

noob question -- how does one "read" $T^2 # T^2$

[gentle osprey]

mjachi

[gritty widget]

tee two connect sum tee two

[hollow charm]

thanks

[winged viper]

does anyone know any sources that construct the cup product in this way other than may

[little hemlock]

so, im trying to understand the proof of theorem 3.26

Let $\varphi : H_n(M;R) \to \Gamma_R(M)$ be the map in the proof. If $\alpha \in H_n(M;R)$ is a generator, then $\varphi(\alpha) \in \Gamma_R(M)$ is a generating section, but how do we know that $\varphi(\alpha)(x)$ is a generator of $H_n(M, M-{x}; R)$ for each $x \in M$?

[gentle osprey]

kxrider

[little hemlock]

the goal here being to show part (a) of 3.26, that the natural map $H_n(M; R) \to H_n(M, M-{x}; R)$ is an isomorphism

[gentle osprey]

kxrider

[vast estuary]

Trying to understand what's going on here

if X is connected, and \pi_1(X) is not abelian,....then what?

[true robin]

The point is that if X is path connected given two points x and y, there is a path l between them, and we have an isomorphism of fundamental groups given by p maps to l^{-1}pl. In the nonabelian case this isomorphism is dependant on the choice of l we used, hence is called non canonical.

[little hemlock]

what book is that? looks like Vakil font

[umbral finch]

where can I find the following result: "an injective map between n-dimensional tori is bijective" ?

[odd flame]

bit of a weird question but what's the point of first countability

can i think of it like a countable "subcover" for a basis

im assuming such a space has nice consequences and whatnot im just wondering how to see it in my head best

[hidden crag]

hmm i usually think of it as a space that isn't too messed up

[odd flame]

[hidden crag]

it does have nice consequences, you are allowed to use sequences instead of nets to characterise closure and continuity

[gritty widget]

first countability is the bare minimum a topological space should have to be nice if you don't want to deal with nets

[hidden crag]

maybe think about why you need first countable to replace nets with sequences

good exercise

[gritty widget]

nets are fake anyways

[unreal stratus]

Ngl still never used a net lol

When do they start actually being useful

Well tbh I guess I have used nets without realising it

[gritty widget]

"when will we use this in real life?"

[hidden crag]

i feel like blitz is gonna murder you for that statement

[unreal stratus]

Well yes but I don't do like pure pointset lol

That wasn't even a statement anyway smh

[hidden crag]

yes i haven't used nets either

[gritty widget]

the loss of tone on the internet and its consequences...

[unreal stratus]

But ye come to think of it I have kinda used them

Meh

[hidden crag]

no i was intentionally twisting potatos tone

[unreal stratus]

Lol

[hidden crag]

another thing that i have never seen again are separation axioms

where do people care about those

[odd flame]

me, studying separation axioms for exam tm

what's a "local criterion for normality"

[gritty widget]

a criterion for normality that is local

[shadow charm]

Im so glad we didn’t cover those in my pointset class

What even is the point

[gritty widget]

doesn't munkres explain what he means by this?

"local" in topology usually means something like "for every point (and maybe every neighborhood of the point)..."

it means you're focusing on what happens near points

[hidden crag]

no idea

i'm aware of the consequences and implications among them but uhh

[odd flame]

oh ig i just haven't seen it then, combing ch4 rn

[gritty widget]

"local" is more of a vibe than a precise mathematical thing

but it's usually either "every point has a neighborhood such that..." or "every point and every neighborhood of the point has a smaller neighborhood such that..."

in point-set "local" usually refers to the second one, and an example of the first one is in "semi-local simple-connectedness"

[hidden crag]

[gritty widget]

well, that one's a bit of a misnomer anyways. it's not "simply connected neighborhood of each point"

most decent spaces are going to have neighborhoods like that anyways (earring get fucked)

aaand before the AG nerds hit the chat i have to mention that "local property" has a well established meaning in commutative algebra...

and said local property does indeed reflect some geometric idea of locality

[odd flame]

would this count as local?

sorry i always disappear terra i be walkin around n shit

it seems like one of those words that gets thrown around very vaguely idk

[gritty widget]

it is

mostly

[odd flame]

hence

[gritty widget]

the property of being normal is not really local

but regularity is, sure

[winged viper]

We already know that M is R-orientable, so we already have a section that is a generator at each point. The fact phi is an isomorphism just says that this section corresponds to some class, which is the alpha you want

[odd flame]

isnt that thm redundant

"X is compact iff every collection of closed sets with the FIP has the FIP" no?

[unreal stratus]

how is that redundant

[odd flame]

hhhh wait

[gritty widget]

think about the statement some more

[unreal stratus]

or do you mean like

[odd flame]

the collection neednt be finite

[unreal stratus]

okay yes

lol

[odd flame]

yeah didnt immediately catch that

[unreal stratus]

the point is basically this is what you get by dualising the open covers definition

intersections correspond to unions, being empty corresponds to being the whole space

[odd flame]

wdym by dualising

[unreal stratus]

okay so like

taking complements, roughly, but I used that in a non-technical sense lol

basically the intuition is uh: open covers definition says if $\mathscr U$ is a collection of open subsets of $X$ such that $\bigcup_{U \in \mathscr U} U = X$, then there's finite $\mathscr F \subset \mathscr U$ with $X = \bigcup_{U \in \mathscr F} U$

[gentle osprey]

potato

potato

[unreal stratus]

So yes, they are the same after taking complements everywhere

[gritty widget]

what about taking compliments?

[unreal stratus]

Well if the space compliments you, you should take it

if the space complements you well, maybe ask it out on a date

[odd flame]

the reals are making me blush

will read up in a moment tho

[gritty widget]

[gritty widget]

I am confused. How do we know $f[\omega + 1]$ is compact?

[gentle osprey]

SB

[gritty widget]

image of compact set by a continuous map

@gritty widget Thank you!

[little hemlock]

Ahh ur totally right. Ty frank!!

[odd flame]

what is a common example of a non normal space

trying to do this

[vivid ingot]

I'm not sure about the conclusions to these theorems not being valid, but I know the radial plane is not normal:

It's R2 with the topology generated by the neighborhood basis of stars at each point. Here a star (about a point p) is any set for which the intersection of a ball about p with any line through p is contained in that set.

@odd flame

The Moore plane is also not normal. I really don't want to describe that one.

Soooo here's this instead:

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

[odd flame]

will try and read a bit

thank you for pointers

[vivid ingot]

Okay. X3

Yeah, no prob. Sorry I can't point you toward the "actual" answer.

Maybe when I start studying again for the topology qual.

[odd flame]

my final is in 10 hours this is a practice exam lol

[vivid ingot]

Well, my first thought would actually be to use the Moore plane. It's Hausdorff, so singletons are closed. With that in mind, you could pick the x-axis and some singleton "above" it to be your closed sets which may not (?) be able to be separated with a continuous function.

[unreal stratus]

How about, say, an infinite set with cofinite topology

Singletons are closed but any two (nonempty) open sets intersect

[calm current]

How would I best go about showing that a metric pseudocompact space is also compact? Would it be good to show that a metric pseudocompact space is completely metrizable and totally bounded and then show that that implies sequentially compact and second countable (since I can use that this implies compactness) or is there some easier route directly to compactness?

[gritty widget]

Let x_n have all terms different and no convergent subsequence, define f(x_n) = n and extend (say using Tietze theorem)

[calm current]

I'm sorry but what does that show exactly?

I have not yet heard about Tietze's theorem

[coarse night]

what is the most elementary argument to say homologies of a compact manifold is finitely generated

[astral mango]

what is A here? Chi is the typical euler characteristic

[coarse night]

what's Aⁿ? affine n space?

[cedar pebble]

This is affine n-space. So this is supposed to be the AG analog of homology invariance

In topology the cohomology of X is the same as the cohomology of XxI where I is the interval

Here the affine line A^1 plays the role of the interval in AG

[obtuse meteor]

I would guess Mayer-vietoris. Although perhaps that is a bit painful

To me it doesn’t seem like there would be a simpler way though—the idea is that you should cover with finitely many charts and the homology is thinking about how they overlap, so you kinda just need Mayer-vietoris for that

[coarse night]

looks like that's the way

[rough cedar]

is there any intuition to the definition of a topology

[coarse night]

short answer: metric spaces

[gritty widget]

A different answer: verifiable properties

Intuitively an open set is a set where you can "verify" that some element belongs in it

For example consider the open interval (0,1). Suppose you have some x in (0,1). If you zoom enough near x you will be able to see that it is indeed inside of (0,1)

but if your x was 1 instead, no matter how much you zoomed you'd never be able to tell whether it is inside (0,1) or not

This justifies why finite intersections are open: to verify if x belongs to a finite intersection you can just separately whether it belongs to any of the components of the intersection;

Arbitrary unions of opens are open because if x belongs to the union then theres one open on which x belongs so you can verify whether it belongs to it

(of course this is not very formal, but i still think it's a nice intuition for topological spaces)

[icy schooner]

How do you show the degree of a map f: CP^1 -> CP^1 that raises each coordinate to dth power is d

[lunar yoke]

Following should work: You can view CP1 as the Riemann sphere, i.e. as S^2, say with [1:0] the point at infinity / the north pole of S^2, and the other points being represented by [x:1]. Clearly f fixes [1:0], and sends [x:1] to [x^d:1]. Under this identification of CP^1 with the Riemann sphere this is just the map x -> x^d on C, and under the identification with S^2 this is the same as the suspension of the map x -> x^d on S^1. The latter map has degree d, and suspension preserves degree

[icy schooner]

thanks I think it works
I can generalize it to CP^n using suspension

[bronze ruin]

i have a problem in my topology course which involves proving a space is simply connected

according to my textbook, a space is simply connected if a) it is path-connected, and b) its fundamental group is trivial

the space in question is described as "two distinct 2-spheres, joined by a line segment. imagine a dumbbell, but with a very thin connector"

it's fairly trivial to see that it's path-connected, but to construct the fundamental group, am i allowed to reduce the space to S^2 v S^2?

that is, two spheres connected at a point instead of a line segment?

[icy schooner]

I think this is just van kampen

[bronze ruin]

doesn't van kampen require the sets u and v be open?

[icy schooner]

take S^2 attach a bit of stuff at the end

it's the same with S^1 V S^1

[bronze ruin]

oh wait

yeah

duh

thanks

i just needed to see it written out i think

[summer sequoia]

someone please correct my proof of: if c is a cluster point for S then there exists a sequence $(s_n)$ in S such that $(s_n) \rightarrow c$: Assume c is a cluster point for S, then let $(s_n) $in S by the definition of a cluster point and let $\epsilon > 0$ such that $|s_n-c| < \epsilon$. Suppose $(s_n) \rightarrow c$ then for any $\delta > 0$ there exists $N$ such that $|s_n-c| < \delta$ for all $n \geq N$ proving the original statement true

[gentle osprey]

imnotrachel

[unreal stratus]

What definition of cluster point are you using

It also looks like you are proving the wrong direction sort of

[maiden pilot]

Why is the following statement true? I feel like im missing something really obvious.

Suppose that $f \colon D^n \to D^n$ is a map with no fixed points. Then we can treat $f$ as a map from the northern hemisphere $D^n_+$ of $S^n$ to itself.

[gentle osprey]

OmnipresentCoffee

[maiden pilot]

In what way does this map of the hemisphere arise?

[maiden pilot]

nevermind, i thought it was inducing some map on the boundary of Dn but that is S{n-1} not Sn. I didnt realize hemisphere of Sn is just homeomorphic to Dn

[summer sequoia]

oh haha whats the correct proof?

[gritty widget]

Limit point is enough

Or well, point from the closure

If is in closure of S, then for any neighbourhood of c we can find a point of S in it

In particular choose s_n in open ball around c of radius 1/n

Then s_n converges to c

This proof translates well if we have a countable basis at c

If not, we can choose a point for every neighbourhood of c and thus we would create a net converging to c

[summer sequoia]

so is this a good way of starting the proof if i say this: Let $s_n$ be a sequence in S that converges to c. The sequence exists as c is a cluster point of s so the neighborhood is non-empty

[gentle osprey]

imnotrachel

[gritty widget]

Not really

If you have claims you should justify them

People in books/articles often don't but you should

[summer sequoia]

im still kind of confused but this is my second attempt, Assume c is a cluster point for S then by the definition of a cluster point, there exists an element in the nieghborhood such that s_n belongs in S. Assume s_n belongs in S for all n in N that converges to c and choose an epsilon > 0, then there exists a N in the natural numbers such that |s_n - c| < epsilon. meaning that s_N belongs in the neighborhood and is a sequence that converges to c

[pseudo coral]

small question for showing closed subset of normal is again normal.

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

oh neevermind I just answered my own question.

[gritty widget]

[gentle osprey]

MyMathYourMath

[gritty widget]

that, you should be confident about

[pseudo coral]

ok cool neevermind then 🙂

ok question in showing a compact metric space is totally bounded. Suppose the space is NOT totally bounded then there is an epsilon such that X has no finite epsilon covering. Then choose a sequence in X as follows:

pick any x_1 \in X then as X \neq B_\epsilon(x_1) there exists some x_2 \in X such that d(x_1,x_2) \geq \epsilon

then repeating this we see that

given some {x_1,...,x_n} there exists some x_{n+1} such that d(x_k,d_{n+1}) \geq \epsilon for 1 \leq k \leq n.

[gritty widget]

backslash your stuff

[pseudo coral]

So any Cauchy sequence cannot converge thus the space is not sequentially compact and thus not compact

sorry lol

i didnt know if i show put it in TeX or not

[gritty widget]

you are massively overthinking this

suppose X is compact. you want to show that for each epsilon > 0 there is a finite covering of X by epsilon balls

what's a super easy way to use covering compactness?

unless you want to use sequential compactness for some reason

[pseudo coral]

any opeen covering has a finite subcover

[gritty widget]

yes

how?

consider the cover consisting of...

[pseudo coral]

epsilon balls?

[gritty widget]

yes

[pseudo coral]

ahh

[gritty widget]

there is a finite subcover so you're done

[pseudo coral]

youre right! I was overcomplicating things

how to show X is compact if it is a totally bounded metric space that is complete? this is the tough direction i wanna say

I wanna say to choose balls of radius 1/n but unsure what to do with the completeness in the argument

[gritty widget]

for this sequential compactness is probably easier to use

[pseudo coral]

ah

[gritty widget]

(and you're right about it being the tough direction)

[pseudo coral]

since in a metric space compact iff sequentially compact

[gritty widget]

(also, you can use sequential compactness for the other direction like you were doing above, and i think your idea is right)

(it's just a lot nicer with covering compactness)

[pseudo coral]

youre right it is lol

[gritty widget]

so you take a sequence in X

and you want a convergent subsequence

what are our assumptions? totally bounded doesn't give us anything that looks immediately useful, but completeness? completeness tells you that cauchy sequences converge, so what if you could find a cauchy subsequence?

[pseudo coral]

using the totally boundedness?

[gritty widget]

somehow, yeah

[pseudo coral]

cause you can bound it by balls of radius 1/n

[gritty widget]

well, i'll let you figure out the details

but i think i've given a good starting point

[pseudo coral]

thanks, you have!

i appreciate it!!!

[gritty widget]

it's how i'd start to prove it at least

do tell if it works out

[pseudo coral]

will do!

[wise ruin]

We are trying to recover a 2-dimensionsal closed TQFT from a monoidal k-linear pivotal bicategory B with duals such that B(O,O)=Vect_k, where O is the unit object of B.
That is, for each object a in B, we would like to construct a commutative Frobenius Algebra A_a from a.
We declare the underlying vector space of A_a to be End(1_a).
The dual a* of an object a in B is interpreted as a label for the bordism patch with opposite orientation.
Now, in order to endow A_a with a bilinear form, we begin by labelling a sphere as a, then project it onto some plane, producing a disc labelled as the tensor product a x a*. Supposedly this is because "the rear part of the sphere has opposite orientation with respect to the plane".
I'm having trouble understanding what's going on here. I can start to understand the intuition being that when you tensor product a and its dual, the parts on the same side of the sphere cancel out, but I don't know why the same wouldn't apply for the plane its projected onto.
Furthermore, why are we even projecting in the first place, what does that have to do with this construction?
Lastly, why not just declare that a x a* will be the disk because that is precisely the unit of the 2-dimensional bordism category?

[rugged swan]

can you define what's a closed TQFT ? Are 2d closed TQFT the same as the monoidal thing ? Like 2d oriented TQFT are the same as Frobenius algebras ?

is your category of cobordism here extended or just a genuine 1-category ?

[tough zephyr]

A rather elementary question but I can't be the only one who finds this proof confusing to read?

I'm not even sure how to find it as a quoted result

[gritty widget]

i hate this already

ike

like

they're too close

[gritty widget]

just look for different characterizations of closure in a topological space

sometimes people define the closure as the set on the right

and some, like you, must have defined it another way

[rancid umbra]

what’s the difference?

[little hemlock]

why is this true?

hmm okay, one way to see it is (M, partial M) is a good pair so the long exact sequence for the triple (M, partial M x [0, 1), partial M) gives you an isomorphism Hi(M, partial M) --> Hi(M, partial M x [0,1)), and then you can apply excision

[gritty widget]

Is $A$ : $\bigoplus_{p} H_n(X_p)$, $\bigoplus_{n} H_n(X_p)$, or $\bigoplus_{n,p} H_n(X_p)$,

[gentle osprey]

lime_soup

[gritty widget]

thanks and

d=jk

is it really jk

or is it jik

[gritty widget]

ah thank you

[gritty widget]

they're too close

my eyes just skip over it

[chrome ridge]

I prob missing something obvious but why is U a Basis for X ? This is from Hatcher's Alg topo page 64

[plain raven]

you want to go back and carefully look up the definitions of locally path connected and semilocally simply connected

and apply them

[little hemlock]

is there any sense in which the cross product in homology is dual to the cross product in cohomology? Say, a commutative diagram of isomorphisms
$$
\begin{tikzcd}
{H^i(M; k)\otimes H^i(N; k)} & {\hom(H_i(M; k)\otimes H_i(N; k), k)} \
{H^{2i}(M\times N; k)} & {\hom(H_{2i}(M\times N; k), k)}
\arrow[from=1-1, to=1-2]
\arrow["\times"', from=1-1, to=2-1]
\arrow["{(\times)^*}", from=1-2, to=2-2]
\arrow[from=2-1, to=2-2]
\end{tikzcd}
$$

[gentle osprey]

kxrider

[little hemlock]

where the horizontal maps would, i suppose, be the same ones that come from the universal coefficient theorem for cohomology

[little hemlock]

oh yea, and the indices probably shouldnt have to be like this either. i imagine you could replace i and 2i with i, j and i + j

[little hemlock]

i noticed that the way I've written this is suggestive that this should follow immediately from naturality of the SES in the universal coefficient theorem, but im going off the definitions in hatcher. The cross product in cohomology is defined in the kunneth formula section of hatcher, and the cross product in homology is defined in the chapter 3 appendix on generalizing the kunneth formula

[hidden crag]

do you have questions about something i said in the other channel?

[glass bison]

Sorry if my brain is acting really fried rn, I'm on like barely any sleep lol, but how would I define it uniquely?

Like the smallest dimension such that there is a homeomorphism between a neighborhood of x and a subset of R^ that dimension?

[hidden crag]

what would your initial definition be

[glass bison]

Like the general definition of a topology?

[hidden crag]

no the dimension at a point

[glass bison]

sorry not topology

I meant manifold

[hidden crag]

think about why you don't need "smallest" here

for every point on your mfd there exists a neighbourhood s.t. this nbhd is homeomorphic to R^n for some n

why is this n unique

[glass bison]

If I don't say smallest wouldn't it no longer be unique?

because it has the same dimension as the neighborhood?

To be quite frank, no, and this is assignment is due in a couple of hours, had to miss a couple lectures to catch up on other exams/hw. Usually I get to reading everything a couple days before the assignment is due, but I've been solely submitting grad school apps for the past 36 hours, and I had an exam before that

[hidden crag]

using the result nobody mentioned you can prove that if x has a nbhd. homeomorphic to R^n then x has no nbhd. homeomorphic to R^m for m not euqal to n

[glass bison]

ah ok

[hidden crag]

the proof for said result requires invariance of domain tho, i recommend checking your lecture notes you are probably allowed to assume invariance of domain

[glass bison]

theres a discussion on that right after

but not before

[hidden crag]

so you think you're not allowed to use it?

[glass bison]

I guess not, but I'm allowed to use this one, which is the same thing you and Nobody were mentioning

[hidden crag]

yes you can use that to prove the result nobody mentioned

[hidden crag]

and then prove this

and then you're done

[glass bison]

Alright I'll give that a go

[hidden crag]

both proofs are fairly short

[glass bison]

Thank you so much!

[hidden crag]

you're welcome

[glass bison]

I may have one more question later if you're free, I understand if you're busy tho

[hidden crag]

just send it here

[glass bison]

Alright

I hadn't looked at this one yet but seems to be on an entirely different topic

I'll work on the other ones before getting back to this though

[glass bison]

Sorry for sneaking another one, just so braindead rn

Aren't manifolds defined by the properties of their points

so how would I give an example of a manifold without using a choice of point

[hidden crag]

i think this is saying that they want you to give an example of a manifold that does not have the same dimension at every point

[glass bison]

ohhh

[rancid umbra]

this really depends on how u define a manifold tho. some authors don’t allow “multidimensional” manifolds

[coral pivot]

"Let M^n be a manifold"

[gritty widget]

i actually love writing the dimension of a manifold as a superscript

saves so much space

ppl should go back to doing that

[little hemlock]

can anyone tell what the "obvious isomorphism" is supposed to be here?

[cedar pebble]

[gritty widget]

why sully

[cedar pebble]

Bad notation

[gritty widget]

why

[cedar pebble]

It overloads products/powers and it's awfully redundant a lot of the time

[gritty widget]

writing skill problem

let me have my fun

it's not like i write for an audience

[coral pivot]

I usually just say, let M^n be a manifold and never the superscript again unless like

its needed.

[gritty widget]

i find this is common whenever a book denotes dimension like that

[cedar pebble]

Wait so you change notation of the manifold willy nilly

[gritty widget]

come on

why do you care?

[coral pivot]

yeah its called being fluid with notation lol

u do this for many things too, where you specify something in the notation only when needed

its nothing different here.

[gritty widget]

it's not like you're switching between different letters

[cedar pebble]

Idk this is deranged

Maybe not confused, but definitely displeased

[gritty widget]

cry about it

[lunar yoke]

yeah im not a huge fan either but its super common, and for lectures quite useful when you dont want to write "let M be n-dimensional" every time

[cedar pebble]

"M: n-mfd" is usually what people will write in lecture

If they are avoiding this notation

[coral pivot]

I have seen way more M^n than that lol

[cedar pebble]

Somehow I don't remember seeing this in any of the low dimensional topology courses I took

[gritty widget]

please

[cedar pebble]

Maybe low dimensional topology people are smart enough they don't need constant reminders of what dimension things are

Who knows

[gritty widget]

this is kind of cringy

[cedar pebble]

They only need to remember numbers between 1 and 3

[coral pivot]

Maybe low dimensional topology people are smart enough they don't need constant reminders of what dimension things are

[cedar pebble]

Lol

[little hemlock]

bumping my question

[lunar yoke]

isnt the inclusion a homotopy equivalence via the collar nbhd? I imagine its like the inclusion [0,1) -> [0,1]

[little hemlock]

yeah, im having a moment

[feral copper]

Hi! Is anyone there a pro of Kirby calculus? I'm willing to show that the following two 3-manifolds are homeomorphic. What Kirby moves should I use? I'm not sure how to blow down the +2-framed unknot... I know that a +1-framed or a -1-framed unknot can be removed at the cost of a -1 or +1 in framing, respectively, but here?

Maybe it's better to make the drawings in terms of plumbing diagrams: what's the rule to reduce the plumbing graph in terms of framing?

[gritty widget]

@tardy urchin ?

[tardy urchin]

Me calculus???

[pseudo coral]

can i do a proof walk through of separable metric spaces imply second countable and someone crituque my proof

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

thus

[hidden crag]

I cant check all the details rn but it looks good

The basis is chosen correctly

[gentle osprey]

MyMathYourMath

[pseudo coral]

ok sweet

[gentle osprey]

MyMathYourMath

MyMathYourMath

[old oak]

what is the example for b? the only space I'm familiar with that is not first countable is $\mathbb{R}$ with finite complement topology but I dont think that satisfies this

[gentle osprey]

Bilboswaggins

[old oak]

oh actually nvm, I guess discrete topology on R works

[gritty widget]

No it doesn't

[old oak]

why not?

Oh its first countable 💀

[gritty widget]

@old oak did you figure it out yet

[old oak]

nope

[gritty widget]

Box topology

[junior coral]

hi, i'm trying to understand this proof of dehn's lemma, but i'm not too familiar with covering space theory

the last page is really the main relevant one, the first two are kinda supplementary to my question

one setup involved in the proof is

you have a map from the disc to some surface in a 3-manifold

and they have this map whose image is a surface that intersects itself 10 times

but by considering the double cover of the surface

the double cover only intersects itself 5 times instead of 10

i don't really understand how lifting the map to the double cover reduces the number of times it goes around, so if someone could explain that, that would be great

sorry for the long text

[ocean narwhal]

sps X is metrizable and ~ is an equivalence relation such that X/~ is hausdorff (w/ the quotient top). under what conditions can we conclude that X/~ is metrizable?

[paper wedge]

X being compact

[ocean narwhal]

compact is too narrow

[wise ruin]

Could someone possibly help me understand these grading corrections? The “counterexample” they gave is just not clicking

[plain raven]

what is the question

[wise ruin]

Oh yes, sorry

[plain raven]

In your original argument you are correct up to and including the comment that $[\iota c] = [\iota d]$ but this refers to a different implicit equivalence relation than in the equation $[c]=[d]$. In the first one, it means that the two cycles $\iota c, \iota d$ differ by a boundary of a chain in $C(X){n+1}$; in the second one it means that the two cycles $c,d$ differ by a boundary of a chain in $C(A){n+1}$. The chain implicitly referred to in the first equation may not live in the subgroup $C(A){n+1}\subset C(X){n+1}$.

[gentle osprey]

diligentClerk

[wise ruin]

That makes sense - thanks!

[plain raven]

do you know the split exact sequence lemma?

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

ignore the shit about abelian categories lol you just want it in the case of abelian groups

That might be overkill actually

My point is you need to use that homology is a functor.

Not just that it sends maps to maps, but also the other laws regarding composition and identity

[crisp helm]

Not sure the appropriate channel for this:
Is there a study of "discrete" manifolds, I guess they're just simplicial complexes. But then having somehow equivalent definitions of vector fields on the complex (this seams easy) and then having k-forms on the complex (gut says this would be a bit more complicated to define properly)

[gritty widget]

If $X$ is separable metric, $q:X\to X/\sim$ is closed with compact fibers, then $X/\sim$ is separable metric

[gentle osprey]

Blitz

[gritty widget]

Also note that R/N is not metrizable so this theorem is pretty tight

It uses Urysohn metrization theorem, for general metrizable spaces all conditions for metrizability are beautiful (well except for the star-refinement ones) but not so easy to check, I wouldn't expect anything in the general metric space setting

[gentle osprey]

Blitz

[junior coral]

why is the thing in parentheses about the no connected 2-sheeted covering space true

[gaunt linden]

Sorry, why would R/N not be metrizable?

[gritty widget]

It's not first countable at N

[gaunt linden]

Ah, I see.

[void gazelle]

Hi, guys, i am reading the proof that a topological manifold is paracompact. it says for any open cover U, and any basis B, there is a countable open refinement of U consisting of elements of B. But What if the open cover U is uncountably infinite? or do we just assume U is countable by second- countability?

[hollow harbor]

you shouldn't need to assume U is countable

if U is uncountably infinite, then you'll have sets in your refinement which are refining uncountably many sets in U at a time

[void gazelle]

Thank you!

[unreal stratus]

I'm not sure what the question is - if paracompactness just said countable covers had countable refinements, that wouldn't really make sense

[void gazelle]

It says if B is any basis for the manifold M, U is any open cover, then there exists a countable, locally finite open refinement consisting of elements in B

I guess it is saying for any U, i must miss something in the proof

[pseudo coral]

question

for showing a separable metric space is second countable so I take the basis

[gentle osprey]

MyMathYourMath

[pseudo coral]

then show that these basis elements are contained in an arbitrary open set of my ambient space

[gentle osprey]

MyMathYourMath

[pseudo coral]

does that suffice to show the space is second countable? if every arbitrary open set of ambient space contains these basis elements

did i choose the correct basis

i missed this question on my final and im worried itll show up on the qualifying exam for top

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

anybody?

[gritty widget]

any open cover of a second countable space has a countable refinement

simply write every set in the cover as a union of some elements of the basis

[void gazelle]

Thank you! Yes! I messed up the definition of refinement. I was thinking other way around : If U has a refinement V, then for all U_a\in U, there exists V_b such that V_b\subset U_a, which is not the definition

[gritty widget]

how do you know it's a basis? That's what you got to prove

[pseudo coral]

isnt it enoigh to show each element is contained in an arbitrary open set of my ambeint space

therefore each open set of ambient space can be written as a union of them

[gritty widget]

no

that doesn't sound true at all

[pseudo coral]

i thought a basis it

a basis if for each open set

[gritty widget]

Here: Let $\mathcal{B}_n$ be a family of open covers, each consisting of sets with diameters $< \frac{1}{n}$. Then $\mathcal{B} = \bigcup_n \mathcal{B}_n$ is a basis.

[gentle osprey]

Blitz

[pseudo coral]

and point contined in the open set

you can put a basis element around that point and inside of the open set

so given an arbitrary y \in U \in \tau I have to show B_1/n(x) \subset U

with y \in B_1/n(x)

[pseudo coral]

centered about points of the countable dense subset right?

[gritty widget]

no

we're not talking about balls

[pseudo coral]

but were in a metric space

[gentle osprey]

AlexSchopbarteld

AlexSchopbarteld

[supple sable]

Struggling in general because i dont see how preserving the equivalence relation I am going to find a bijection at 0,0

[prisma arrow]

o lol

am i missing smth

y=y’ and norm y are equal

?

[supple sable]

ye

Ehm

Abs y is 1

So only the edges of the y

[prisma arrow]

i guess lol

[supple sable]

Ahhhh wait

Yeah you are right

Im missing something

[prisma arrow]

i mean its fine

ik what u mean

If you were to write it out its norm y,y’=1

like u want a cylinder?

[supple sable]

Yeah

I dont see how this could possibly be homeomorphic to a disk

[lunar yoke]

you are not gluing the two edges together

you are collapsing each of the top and bottom edges to a point

also you dont wanna use cos and sin since your input is already in cartesian form, not in radian

so bijection should be more like (x,y) -> (x/sqrt(1-y^2),y)

[supple sable]

Well if this is the case then yeah, you dont want that haha

[lunar yoke]

this probably doesnt work entirely, but i would expect it to look similar

like y shouldnt change too much, but the range x can vary needs to be scaled down depending on y

[supple sable]

Should probably first find one that works for [0,1]^2 and then translate and scale

I now see where I went wrong, collapsing makes more senes

[prisma arrow]

oh good question

what ways besides simply connected is there to show disc and cylinder not homeomorphic?

closed cylinder and closed disc

[lunar yoke]

unless you mean the boundary of a cylinder, the cylinder is 3 dimensional and disk 2 dimensional so there is a variety of reasons this cant work

[prisma arrow]

yeah boudary lol

[lunar yoke]

wait you say simply connected so i guess you mean boundary

[prisma arrow]

i didnt mean solid

[lunar yoke]

ok so then its just the sphere

and you may use any technique from algebraic topology

[prisma arrow]

i think looking at the boundary makes sense

[lunar yoke]

homology, euler characterisitc, homotopy groups, etc

[prisma arrow]

without at techniques

im guessing looking at boundaries

and removing points

and connectedness

nah

oh yeah lol

boundary of cylinder isnt connected

boundary of disc is

its that

[lunar yoke]

you have to be careful with these things, since boundary depends on surrounding space

apriori i dont know what you mean by boundary of cylinder

[prisma arrow]

yeah treat them as manifolds with boundary

[lunar yoke]

then cylinder has no boundary

cause its S^2

[prisma arrow]

closed cylinder

[prisma arrow]

?

[lunar yoke]

the cylinder is homeomorphic to S^2 im pretty sure

just radial projection

like square is homeomorphic to circle

[prisma arrow]

nah

like

[0,1]^2

and glue one side to the opposite

[lunar yoke]

bruh

[prisma arrow]

like not S1 x R

[lunar yoke]

you're missing top and bottom lids

[prisma arrow]

but S1 x I

[lunar yoke]

whatever

[prisma arrow]

yeah its details

but

ig this is ambiguous

i thought ppl never talk about lids

[lunar yoke]

idk

[prisma arrow]

same here

[lunar yoke]

i only do abstract nonsense

[prisma arrow]

?

[cedar pebble]

[prisma arrow]

yeah ng be kind

[rain ether]

general abstract nonsense

[lunar yoke]

wait im dumb i basically already had it, you do (x,y) -> (x*sqrt(1-y^2),y), not dividing but multiplying

[lunar yoke]

like (infinity) categories and abstract homotopy theory and so on

stuff that these guys do

[prisma arrow]

🤟

[supple sable]

Does anyone have an equivalence relation so P2/~ is homeomorphic to S2?

I know one so that S2/~ is homeomorphic to P2

But not the other way around

[gaunt linden]

P2 is the real projective plane?

In that case I think it ought to work to pick a line L and define a ~ b iff a = b or {a,b} ⊆ L.

[junior coral]

does anyone know of a reference on the computational complexity of calculating the first Betti number of a triangulation of a 3-manifold?

[old oak]

n would belong to the intersection right?

I guess something like {R-(-n,n)} would work

Think of it as a set of open sets that require infinitely many elements to cover the entire space

[neat current]

"A space X is locally connected if and only if for every open set U of X, each component of U is open in X."

what exactly does "component of U" mean? a component of U meaning U as a subspace of X?

[hidden crag]

connected component maybe?

yeah that should be it

[gentle osprey]

Shivvaghel

[vivid ingot]

My work so far is to suppose per contra that A and B form a separation of this union, namely as open sets in the subspace topology on this big union (which I’ve called Y). My strat at the moment is to assume that they don’t restrict to form a separation of any of the X_i, but then that means they form a separation of X_0 — a contradiction.

Is that a good plan of attack, and how do I enact it?

[unreal stratus]

The easiest imo is to use the characterisation of connected spaces in terms of continuous maps into {0,1} (with discrete topology)

But yours should wind up basically the same

[neat current]

couldn't you also show it's path connected pretty easy by induction

and then path connected implies connected

[unreal stratus]

Definitely can't show it is path connected

Otherwise you'd be showing connected implies path connected

[neat current]

o nvm

[unreal stratus]

By taking all sets to be the same

Or I empty or whatever

[vivid ingot]

The continuity argument is interesting, and I definitely saw that one online too. The problem is that nothing I found online matches the phrasing here exactly. (Undergrad problem to complain about, I know.)

((No hate to undergrads either.))

[unreal stratus]

Hm I think you may need X0 cap Xi to be connected

[vivid ingot]

Did I not type that?

[unreal stratus]

Unless I'm being silly

No you did not

But dw

But yes suppose A and B separate this set and then consider intersections with each of the Xi and X0 cap Xi and X0

You'll wind up showing one of A or B is empty

[vivid ingot]

Though I realize a picture is not a proof, if you stick this into R2 then it shows that the intersections need not be connected. The problem is something I saw a while ago before grad school; it shows that all you need is this X_0 “hub” for the subspace to be connected.

[vivid ingot]

I think this is the hint I needed.

[unreal stratus]

Actually yeah the proof I'd give doesn't require intersection to be connected so it'll be okay

[neat current]

so, I have a feeling this proof is quite wrong

since the hint in the book has nothing to do with my approach

and as far as I can tell, it seems like it should work for any surjective continuous map, not just a quotient map

anyone care to enlighten me?

actually think I just realized lol

we would need p to be an open map so that p(J) is open

[chrome ridge]

Any intuitive proof of the fact that two homotopic maps induce the same homomorphism of Homology groups? Hatcher constructs this map called prism operator which feels very not motivated.

[plain raven]

you need to understand how to visualize the prism operator geometrically

it's a triangulation of $\Delta^n\times I$

[gentle osprey]

diligentClerk

[plain raven]

Whenever you have a homotopy between two maps $f, g : X\to Y$

[gentle osprey]

diligentClerk

[plain raven]

then you of course are going to have, for each simplex $\sigma: \Delta^n\to X$

[gentle osprey]

diligentClerk

[plain raven]

an induced homotopy of sigma $h_\sigma : \Delta^n\times I\to Y$ induced by $h : f\Rightarrow g$

[gentle osprey]

diligentClerk

[plain raven]

i.e., $h_\sigma = h\circ (\sigma\times \operatorname{id}I)$. This is essentially the only relationship you have between the simplices $f\ast\sigma = f\circ\sigma$ and $g_\ast\sigma = g\circ \sigma$ so i feel like it should be like, somewhat intuitive a priori that the proof \textbf{has} to somehow make use of this homotopy $h_\sigma :\Delta^n\times I\to Y$

[gentle osprey]

diligentClerk

[plain raven]

and like, because we have a theory based around simplices, again i feel like it's a fairly straightforward conceptual step to go, ok is there a way we can write $\Delta^n\times I$ itself as being built out of simplices/ triangulated?

[gentle osprey]

diligentClerk

[plain raven]

And the answer is, yes, there is a canonical way to equip $\Delta^n\times I$ with the structure of a simplicial complex, which tbh kinda falls out of the definition of the cartesian product if you ever end up studying simplicial sets

[gentle osprey]

diligentClerk

[plain raven]

I really suggest trying to draw the triangulation of $\Delta^n\times I$ in the cases $n=1,2$ and trying to get a feel for it geometrically

[gentle osprey]

diligentClerk

[plain raven]

the proof is then just like, turning the triangulation into homological algebra, but the "prism" itself, it's inevitable imo that you have to use this

[chrome ridge]

Thanks a lot I will try to work that out

[pseudo ocean]

guys i'm a little confused

it is said in the definition of an open set that it is any set part of the topology of the original set we took the topology of.

but the entire set is inside the topology

so it's impossible to construct any closed set because every subset of the original set and itself IS part of the topology

where did this reasoning go wrong? it sounds fine ^

[prisma arrow]

not every subset of your base space is an open set unless you choose the discrete topology

[pseudo ocean]

ohhhhhh

[prisma arrow]

A topology is a specific subset of the power set that has some properties. You can also think of a topology as a collection of subsets of your base space. Base space is the original set.

[pseudo ocean]

yep so ig it depends on which topology we choose

[prisma arrow]

So those three properties quickly summarized are that your topology T contains the base space and the empty set, it is closed under arbitrary unions and closed under finite intersections

[prisma arrow]

Yes

[pseudo ocean]

oh also

following that logic, does a discrete topology only contain open sets?

or can they have closed sets too?

[prisma arrow]

you missing a word?

[pseudo ocean]

there fixed it heh

[prisma arrow]

oh

So the elements of the topology are called ‘open sets’

[pseudo ocean]

well yeah

[prisma arrow]

if you choose the discrete topology then any subset of your base space is considered an open set relative to the discrete topology

open sets are a relative concept

[pseudo ocean]

ahhh

[prisma arrow]

they depend on the topology you choose/ is implicitly understood

have you heard of the standard topology on R?

[pseudo ocean]

yeah

ig?

[prisma arrow]

its the topology where open sets are unions of open intervals

[pseudo ocean]

yep

[prisma arrow]

however there are many different topologies on R other than this

[pseudo ocean]

wait

what are those other topologies on R

[prisma arrow]

finite complement topology is another example

So the open sets are complements of finite collections of objects

[pseudo ocean]

ah

[prisma arrow]

So if you choose a set X={1,4.32,pi/2,sqrt3} then R\X is open in the finite complement topology

there are many ways you can construct topologies from existing ones but you should take it slowly if you are just starting

[pseudo ocean]

fair enough

thanks for the help man

[prisma arrow]

hakuna matata

[gritty widget]

well, R on its own is a set

we can give it different topologies, there's a lot flexibility about that, and it only really depends on cardinality of R

[zinc siren]

(as an example, take the discrete topology, where every set is an open set)

[gritty widget]

oh wait. I just realized that I'm wrong, deleted

[drowsy quest]

I am trying to understand spinors (in their broad mathematical scope rather than just pure QM) and I keep reading how they are the result of SU(2) being the double cover of SO(3) (at least in the case of 3 dimensions), what does double cover mean?

My knowledge in topology is limited, so any good reads is appreciated too

[gaunt linden]

Basically there a continuous homomorphism from SU(2) to SO(3) that hits each element of SO(3) exactly twice.

[drowsy quest]

So is double covering just mean there is a 2 to one 1 function that maps elements from SU(2) to SO(3)?

[prisma arrow]

yes

or you can phrase it that there are two elements in the preimage of any point under projection map

[drowsy quest]

Okay, idk why I had so much trouble with this before, now it's making a lot more sense

[tough zephyr]

Is it just me but I've managed to convince myself that all the propositions hold for topological spaces not necessarily Hausdorff

N.B. $A’$ is defined as the set of accumulation points of $A$.

[gentle osprey]

Philka

[rancid umbra]

what r your counterexamples/reasoning?

[gritty widget]

that's correct

you don't need Hausdorff anywhere

[tough zephyr]

Strange exercise…

[unreal stratus]

This is what we all said when we did this xd

I think 3) needs T1 though

Some people incorrectly deduced 3) or 4) from 2) but that doesn't work

[gritty widget]

is it Munkres?

[unreal stratus]

I did this homework at my uni

evidently same uni

I think that works at least

@ blitz

[gritty widget]

ah yeah. This one needs Hausdorffness

[unreal stratus]

ye

[shy phoenix]

anyone have a pdf copy of "the topology of fiber bundles" by steenrod?

[coarse night]

you can find online

I have but it's better to not share here

[prisma arrow]

How did you convince yourself for 4?

2 is closed in your example?

[unreal stratus]

Lmfao

Yeah it is

[prisma arrow]

Ig we rely on theorem that a closed set is limit points union isolated points

[unreal stratus]

Deleted, but yeah you do need T1

I'll try to come up with an actual counterexample

[prisma arrow]

same example but choose 0?

limit points are 0,1 and this isnt closed

[unreal stratus]

i don't think 0 is a limit point of {0} in that example

whilst 2 is a limit point

But yeah

[prisma arrow]

yeah it cant be right

[unreal stratus]

Okay, sure I found this perturbation online: X = {0,1,2} with the only non-trivial open set being {0,1}

Then the set of limit points of {0} is {1,2}, which isn't closed

[prisma arrow]

isnt your example fine though?

[unreal stratus]

I don't think so

Eh i've deleted and this new thing works so it's fine owo