point-set-topology

[winged viper]

I think Hatcher gives off the impression that what he’s saying should make sense and that you don’t have to spend lots of time trying to understand each sentence

But that’s just not true

[worthy olive]

Why does homology exist if cohomology does everything homology does and more?

[grave solstice]

I was trying to do this (R^R is given the product topology, with R the standard topology)

I'm thinking that a simple formula won't work, because Q^Q is first countable I think

also I know that f_n-->f means pointwise convergence

[red yoke]

Did someone ask this yesterday

[safe torrent]

I feel like I've seen this question before yeah

Before answering this maybe think about why we care about linear forms

[grave solstice]

what do they mean here? isn't (1/n)_{n natural number} an ultranet in the real numbers?

or do they mean for a given directed set Lambda, you can't construct ultranets in general (i.e., it depends on AC)?

[zinc siren]

it's great compared to A Concise Course for first timers.

the first couple chapters of a Concise Course that I read were the only time something was harder than Baby Rudin for me.

[gritty widget]

what did u just call me

[ebon galleon]

😯

[hot locust]

how do you prove here that the image of an open set, i.e f(V) is open?

[formal jasper]

strong opinions about topo books

enjoying this one now

[visual rock]

I hated reading hatcher, didn't finish it

[empty grove]

I like concise, way more than Dieck

[glacial crane]

i have a few questions about the fundamental group. i was trying to learn some introductory topology/alg top and kind of needed some intuition.

so i know that the fundamental group is defined as the set of equivalence classes of loops under homotopy (w.r.t. some base point x_0), with concatenation.

1. why can we usually omit noting the base point explicitly?
2. when we define the concatenation operation on this set, we're defining it on 2 loops. if the elements of this set are equivalence classes, are we picking an arbitrary representative from each class to apply concatenation to? i'm guessing it does not matter, because concatenating any two representative loops would end up belonging to the same resulting equiv. class?
3. do you have any good resources (books and/or lecture series) for learning topology?

[lime sable]

1. you can do this if you deal with a path-connected space, which nice spaces generally are. (you can just look at one path component at a time if there are multiple)
2. yeah, you have to prove the operation is well defined
3. hatcher is the "standard" but lots of people don't like it, i'm sure other people in this channel have recommendations

munkres for basic topology, but i assume you're talking about algebraic topology

[glacial crane]

okay, thank you! and i'll check out munkres—i probably should learn topology more in-depth before moving onto much algebraic topology

[cerulean oriole]

Going by the definitions, I don't think that sequence is residually in U_n (1/(2n+1), 1/(2n-1)) or its complement.

[cerulean oriole]

In terms of filters, using the Axiom of Choice you can prove that there is a (non-principal/non-“trivial”) ultrafilter on any set X.

As for a non-trivial ultranet indexed by any directed set (without a maximum element, say) in any (say, infinite) space, I'm not sure if that's true or even not false even using Choice. You can show that any net has an “ultra subnet”, but I think only if you define subnet broadly enough that it can have a larger indexing set than the original net.

[west sigil]

bruh topology is so hard

i wanna die

[solemn oar]

You'll get used to it.

[west sigil]

we just got spammed with so many definitions

Show that the intersection of two open dense sets is again open and dense

can we solve the dense part algebraically?

like i was able to solve that its open with algebraic notations

but finding it hard with density

[hidden crag]

Hint: There’s a characterization of density using intersections of arbitrary open sets with your dense set

[west sigil]

wdym in "characterisation"

like an identity?

[hidden crag]

What definition of density are you using

[west sigil]

a set S is dense in X if closure(S) =X

[hidden crag]

By characterization I mean something equivalent to your definition

A set is dense iff the intersection with every non empty open set is non empty

[west sigil]

ye ig thats implied from the definition of boundary thinking about it

[hidden crag]

Now try working with that characterization of density for your problem

[west sigil]

wait sorry another question

if A is in X

(let (X,T) be our topology)

and A is a subeset of X

then is A' boundary of A?

A' is A compliment

[west sigil]

whats the difference between this and boundary as well?

[hidden crag]

Not sure what you mean by that

[west sigil]

the definition of boundary we got given is:
x is a boundary point of a subset S ⊂ X if it is neither an exterior nor interior point; i.e., for all neighbourhoods U of x, U ∩ S ̸= ∅ and U ∩ S ̸= ∅. The boundary of a set S is the collection of all boundary points and denoted by δS.

[west sigil]

how is that defintion of boundary different to this definition of density?

[grave solstice]

ah ok I see, thanks

[worthy olive]

I asked it yesterday yeah

[violet summit]

If f is a homeomorphism and f is a simplicial map, is f^{-1} a simplicial map as well?

[umbral panther]

That’s not a simplicial map

[red yoke]

Now bw's message is wrong

(context: someone deleted messages)

[unreal stratus]

Sorry oop

I messed up

[winter verge]

Anyone know how Sober spaces got their name? Nothing coming up on google.

[gritty widget]

the guy who came up with them happened to stop drinking at the same time

[ebon galleon]

https://ncatlab.org/nlab/show/sober+topological+space
as per there (and i remember seeing this same justification on MSE or MO at some point), it's that you're seeing the "correct" number of points: You're not seeing double of any points (i.e. you have a sort of very weak separation axiom like T_0: you can always tell distinct points apart by at least one open set), but you also aren't not seeing any of the points (there are enough points as described on nlab)

[ebon galleon]

this too

[winter verge]

ah, that actually makes more sense than I anticipated

[hidden crag]

not true, topologists never stop drinking

[empty grove]

Stay drinking stay driving stay winning 💪

[west sigil]

driving is boring

drinking while driving makes it fun

[swift fjord]

Every open set vs every open nbhd of x

[jagged rampart]

hi guys, i picked up the Gamelin book recently

this is the index

https://i.imgur.com/DFvRTv4.png

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

around when do i stop being able to distinguish between a coffee mug and a donut?

[tender halo]

not actually sure if you can just from this book, i guess you will need the classification theorem for surfaces to rigorously convince youself of that

[jagged rampart]

oh wow that sort of thing is pretty much what everyone talks about when they try to introduce someone to the idea of topology, I thought its more central than that

[solemn oar]

You can reach this statement as soon as you learn about homotopy equivalences.

Or even sooner, as soon as you learn about homeomorphisms

[red yoke]

You can already explicitly construct a metric space equivalence between a donut surface and coffee mug surface

Or their interiors

(Not that it would be a particularly fun or useful exercise)

Munkres has a chapter on classifying compact surfaces, involving fundamental groups

[worthy olive]

What are the differences between algebraic and differential topology?

[paper wedge]

differential topology is the study of manifolds ig

algebraic topology is how you attach invariants that are algebraic into topological spaces

[cedar pebble]

usually in differential topology you are making essential use of the extra structure that exists on things like smooth/complex/Kahler/whatever manifolds and using this extra structure to say something about the topology of these spaces

there is a big overlap with algebraic topology, but usually in strictly algebraic topology you're not really making use of extra structure like this, you're usually working more combinatorially and using things like simplicial methods

for example you can study things like characteristic classes for vector bundles/principal bundles on a topological space algebraically in terms of cohomology rings of classifying spaces and prove things about these in an algebraic way. If you're working on a smooth manifold, then Chern-Weil theory tells you that you can compute a lot of information about characteristic classes in terms of connections and curvature in a way that generalizes things like the Gauss-Bonnet formula. The latter is really an analytic thing that makes use of additional structure that exists on smooth manifolds...

[gritty widget]

Where to find works on stone cech compactification of complex numbers

I have come across results like How we can embed C in l_\inft(C)

I would like to know more results on how the stone cech compactification of these spaces and what are its properties

[last topaz]

By page 69 would be my guess.

[jagged rampart]

nice

[gritty widget]

Are there different open set?

[gritty widget]

Your question is not clear can you rephrase it?

Rx{1} and Rx{0} are indeed different open sets of that topological space

one only contains elements of form (x,0) and other only of form (x,1) and they aren't empty so they are different

[gritty widget]

thank you brother

[gritty widget]

i'm not your brother, i'm your sister

[worthy olive]

What is a dual group?

[opaque scroll]

Could depend on context, but for a locally compact abelian group G
Hom(G, R/Z)
is called the Pontryagin dual of G. Or simply the dual group of G.

[prime elbow]

If I = [0, 1] then subspace topology I × I on R × R is strictly finer than ordered topology on I × I.

For strictly finer it gives the argument that [ 0.5 × 0, 0.5 × 1 ] is open in the Subspace topology but it is not open in ordered topology in I × I.

Why?

[unreal stratus]

i think the notation is a bit wrong there , do you mean [0.5,0.5] x [0,1]

though thtat isn't open in subspace top so it can't be what you meant

[prime elbow]

No actually they use the coordinate of X × Y as x × y

[unreal stratus]

Oh is this munkres

Yeah do not use that notation as I've never seen anyone else use that aha

But okay sure

By ordered, do you mean dictionary order?

[prime elbow]

Yes

[unreal stratus]

Hm then I'm confused because isn't that set just like {1/2} x [0,1]

Unless the dictionary order is different to what i imagine

[prime elbow]

Yes it is

[unreal stratus]

Hm that isn't open in the subspace topology

[prime elbow]

Why ? R × R has ordered topology

You can take an open set ( 0.5 × -2, 0.5 × 3) and take the intersection with I × I then it will be ( 0.5 × 0, 0.5 × 1 ), right?

[umbral panther]

That’s backwards. The lex order topology is strictly finer than the product

[vast estuary]

how do we get the last statement?

[unreal stratus]

Okay but you didn't say that

[unreal stratus]

The fibre over b is just F from that diagram

like π-1(b) = proj_1^-1(b) = {b} x F

[vast estuary]

upto the isomorphism yeah

and i see your point, thanks!

[north hedge]

if I have a manifold^2 I know that the open neighbourhoods of each point of M^2 would be homeomorphic to B^2(open ball in R^2), but in order to map for example S^2 on a subset of R^2, I would need to map all the charts and overlap them?

[unreal stratus]

ye

[umbral panther]

Yes

[safe torrent]

Why R/Z is a circle?

[solemn oar]

If you think of it as real numbers modulo 1, does that ring a bell?

[empty grove]

Note that they mean it as a group quotient

The quotient notation for groups and topologies clashes. They're not just collapsing all integers (which is what the topological interpretation of the notation would be), but also collapsing all cosets of the integers.

[safe torrent]

no

[solemn oar]

Try to think of the circle as the unit circle in the complex plane. There's a group structure inherited from multiplication of complex numbers. What does that look like?

[obtuse meteor]

R/Z is made up of Z many intervals [n,n+1]. All of these are identified. So we have an interval [0,1], and we identify the endpoints. So topologically it is a circle

[safe torrent]

Oh now I get that

[obtuse meteor]

Another way to view it

The map R -> S1 by theta Maps to e^(2*pi*theta)

This map is Z-invariant (translating theta by an integer doesn’t change it)

So quotienting by the integer as a group you get R/Z -> S1 as a map

It’s a good exercise to see this map is a homeomorphism

Do it

1. directly, by constructing an inverse/showing continuity of the inverse
2. using the theorem about maps of compact spaces and Hausdorff spaces

Quotients by group actions are incredibly important as topology soups up, so understanding both general quotient topology, how you prove things abt them, and specifics about quotients by a group action are worthwhile

Interesting note: isometrics of R are translations. So if G is a group acting on R by isometries nontrivially, and the action is discrete (say Hausdorff quotient), then R/G is always the circle

This is part of why the circle is so fundamental

It’s lurking in basically group action that leaves invariant a line

Groups of Isometries of hyperbolic space are very important as well in a lot of math. As are their quotient spaces

[safe torrent]

I'm trying to think of it in terms of first iso theorem

it's been a while

[obtuse meteor]

First iso will work in terms of the group isomorphism

But as a space there’s more to keep track of

First iso gives you the result immediately in those terms :)

[safe torrent]

how do you set that up?

I forgot

Oh wait nvm lol

Idk what's wrong with me today

Thanks @obtuse meteor

[visual rock]

reflections are also isometries. Unless you require the action to be free, you can also get the interval from the infinite dihedral group acting on R.

and I guess [0,1) from G being generated by a single reflection

[obtuse meteor]

Sorry I should have specified orientation preserving

I don’t believe in non-oriented things rn in my life 😌

[safe torrent]

The more I do algtop the more I think it's really a based field

[obtuse meteor]

Depends on what counts as algtop imo but ya I agree a lot of the ideas are really cool

[prime elbow]

Yeah that's my mistake

[zinc siren]

...this looks like something my prof would assign.

I like the physics notation of ; for covariant derivative indices.

And ∂_i f is nicer than ∂/∂x^i f

[prime elbow]

How closed intervals in R are connected space. And is this connected subspace of R ?

[prime elbow]

Is that like, let any closed interval A any closed set in A is closed interval and any proper closed set of A can not be open in A, right?

[zinc siren]

The answer is always clopen.

[real granite]

Can someone explain the highlighted bit? I’m not getting how $[x,z] \in I$ and $y \in A$ lead to $[y, y+\epsilon) \in A$.

I think you could say that since we’re working with the standard topology (which is metrizable), we can say there is an open ball centred on $y \in A$: $(y-\epsilon, y+\epsilon) \subseteq A$, and therefore $y+\epsilon/2 \in A$, which contradicts $y$ being the supremum.

So I can see that there is a contradiction, but I’m not sure on the exact reasoning used in the proof.

[gentle osprey]

Douglas

[grim knot]

Does somebody know a good paper where I can find an algorithm for computing homologies with the boundary matrices?

[knotty vine]

Homology over a field or something more general (usually a principal ideal domain)?

https://www.matem.unam.mx/~omar/mathX27/smith-form.html seems good (not a paper tho)

[grim knot]

Over a field k

[knotty vine]

Any textbook about topological data analysis will have a chapter on this

[viral atlas]

The proof is saying the same thing

[grim knot]

Oh okay that's nice thanks a lot, if you have a recommendation they are welcomed

[north hedge]

what is clopen?

[quiet thorn]

closed and open

[knotty vine]

Tbh I was looking and most of what I found was either about general PID's or about persistence. These lecture notes treat computation of homology over a field tho: https://www.few.vu.nl/~botnan/lecture_notes.pdf see chapter 3

[north hedge]

how xd..

I mean a set can only be open to a space X right?

If I take a closed ball on a topological metric space

[north hedge]

it would be closed, but not open

[knotty vine]

We need a bot command to post the "Hitler learns topology" video

[north hedge]

lol

ye where he gets angry about clopen

[knotty vine]

https://youtu.be/SyD4p8_y8Kw?feature=shared

[grim knot]

My professor sent me a paper and I think it would be prop 2. 18 and 2.19,but I don't really see how to apply it https://www.cambridge.org/core/journals/acta-numerica/article/topological-pattern-recognition-for-point-cloud-data/BB0DA0F0EBD79809C563AF80B555A23C

[knotty vine]

The idea for a field is not so difficult: all that we need to know about the boundary matrices is their rank and nullity

[grim knot]

You that is how I would do it, but whz do I need to do these matrix transformations?

I wanted to compute it by hand as you're suggesting, but my professor said that she wants us to learn the algorithm

[knotty vine]

The only reason for the transformation is to find the rank/nullity of the matrix

How else would you find the rank?

[grim knot]

I mean I just wanted to apply Gauss, but without having to deal with the matrix itself

[knotty vine]

Yeah Gaussian elimination is a matrix transformation. wdym without having to deal with the matrix?

[grim knot]

Gaussian transformation is equivalent to multiplying the matrix A by the matrix which gets created by those transformation

And I thought we needed to define that (?)

[north hedge]

(0,1) U (1,2) as a subtopology of R, but the complementary of (0,1) is open, so (0,1) is both closed and open, so it's clopen

[knotty vine]

Yeah, using (augmented) Gaussian elimination of a matrix M, we get two matrices R and V with R=MV (did I get this right?) where R is in row echelon form and V is of full rank/invertible.

Then Prop 2.18 in that paper you linked says that we can basically forget about V for the purposes of homology

And the reason why we want row echelon form is because it's easy to read the rank of the matrix from it

[prime elbow]

The continuous image of the path connected is the path connected.

First they define a path, a path in X from x to y is a continuous map f:[a, b] -> X of some closed interval in the real line such that f(a) = x and f(b)= y. A space is said to be a path connected if every pair of points of X can be joined by a path in X.

Now if I let there is a function h: X-> Y such that h is a continuous function and X is the path connected.

So now let any two points from h(X) say y_1 and y_2. Then there are points x_1 and x_2 in X such that h(x_1) = y_1 and h(x_2) = y_2. Now since X is path connected so there is function f: [a,b] -> X such that f(a)=x_1 and f(b)=x_2 then if I define g = h(f) then g maps from [a,b] to Y and g is continuous because f and h are both continuous and such that g(a)= y_1 and g(b) = y_2.

Hence, h(X) is path connected.
Is it correct?

[kindred cairn]

yes

[prime elbow]

Okay thank you

[lament vortex]

feeling silly about a detail in the TDA package for R, if anyone knows about this and could give some clarification that'd be much appreciated -
I'm theorizing about how much of the geometry of a space can be recovered from a persistence diagram, as opposed to just the topology. I'm starting with a very simple example, in which we have 3 points and are given that they lie on the circle of radius 1. Our PD looks like the attached image, and the matrix that generated it is this:
dimension Birth Death
[1,] 0 0 2.0000000
[2,] 0 0 1.6281593
[3,] 0 0 0.5594266
The birth/death times are the epsilon values at which those events occur, correct? ie, the two closest points in our sample meet when e = 0.5594266, meaning they are 1.1188532 units apart. But then the next event is what's making me scratch my head. It's saying the next two closest points meet at e = 1.6281593. but that would mean they are 3.2563186 units apart... yet they lie on a circle of diameter 2. What am I misinterpreting?

[limpid dragon]

uh yeah if theres space around here id appreciate some ideas here because im blanking. as i said elsewhere, kind of get the sense the structure of the cover should be a "root" at one torus and the surrounding n h-1 torii can cover the h-torus. but this cover woud surely assign the "root" to different places simultaneously right?

oh sorry, didnt mean to interrupt

[formal tide]

Is there anything interesting to say about the -1-th reduced homology of the empty set? It's Z, which is weird

it's tempting to just say it's an artefact of the machinery, but I'm not so sure of this, since one can express this machinery in terms of simplicial objects, which seems very "uniform" to me

[novel acorn]

maybe think about what the euler characteristic of a g-holed torus is and how the characteristic acts under lifting to an n-fold covering

[feral copper]

Hey! Does anyone know of an exhaustive list of all possible fixed-point sets of involutions on non-orientable surfaces?
I know I can obtain it from https://arxiv.org/abs/1612.08489 by hand, but I'm wondering if it's been written down already somewhere. Thanks!

[gentle osprey]

Matplotlib

[worthy olive]

What does a[x,y] mean graphically where [x,y] is an n-chain and a is in an abelian group G?

[shadow charm]

I was wondering if anyone could help me walk through lemma 2 of section 8 here? I think I sort of get the general idea but I'm not quite comfortable enough with alg top to fix the details

in particular the part about using a path from x_i to x_r+1 to conclude linear independence

im guessing there's some poincare duality shenanigans but im not too good on it

(the rest of the proof im relatively okay with)

[signal fox]

Please I really need your help, does anyone know the formula of the homeomorphism between the unit circle and [0,1]×[0,1] ??

[gritty widget]

there isn't one, unless you meant "unit disk"

[zinc siren]

r/max(|x|,|y|) [x,y]

and 0 at 0

note that this is 2-norm/\infty-norm

for continuity, we can use either norm.

thus we get easily continuity

using the \infty norm

by reversing the fraction, we get what we can see is the inverse.

basically: we are mapping the ball in the 2-norm to the ball in the uniform norm.

[limpid dragon]

i used this to show the existence of a covering implies the relation, but how would it be useful to build a covering from that relation?

(i went to sleep, sorry)

[stoic eagle]

The idea is that homologous elements have the same intersection numbers. If they are dependent (if one is homologous to a sum of the others, say, h_1=h_2+h_3) since h2,h3 intersect the curve to x1 trivially and h1 doesnt you get a contradiction. @shadow charm

[near sage]

How can I prove that {U_f | f is a polynomial of two variables (x,y)}, where U_f = {(x,y)∈R^2 | f(x,y) != 0}, is a basis for a topology in R^2?

Like I think I proved the first condition for a basis, but Im having difficulties with the second

[viral atlas]

What do you mean by the second condition?

That if U_f and U_g are two such sets, you must have some set in your collection contained in their intersection?

If so, I would suggest thinking about the most obvious candidate the comes to your mind: given polynomials f and g, how do you construct a polynomial that is nonzero only when both f and g are?

[near sage]

wouldnt such polynomial be equal to their intersection?

I mean pointwise

[viral atlas]

Intersection of polynomials?

Think algebraically

[near sage]

I mean wouldnt the points for which the new polynomial is non-0 be the exact same points for which both the first 2 polynomials are non-0

therefore the sets U_F = U_f intersect U_g

edit: changed the notations

[viral atlas]

Yes, that's right

The point is to find such F

[near sage]

so its fine if its that way? By the definition I used I have that U_F ⊂ U_f intersect U_g should be the condition

[viral atlas]

Do you need it to be a proper containment?

[near sage]

from the Munkres' book

[viral atlas]

This doesn't help as it is, there is no universally accepted convention on whether \subset symbol should mean a proper inclusion or not

Let me check what Munkres' convention is

[near sage]

Oh ok yeah so it would be good

again from his book

ok thats what got me confused tbh... so I think such polynomial that is non-0 when both f and g are non-0 would be fg (multiplication) no?

[viral atlas]

Yes, that works

[near sage]

ok I got it thanks

[near sage]

While finding the Zariski closure of I = {(x,0), x ∈ [0,1]} can I just say that there should exist a polynomial f(x,0) = 0 for all x ∈ [0,1] which is a polynomial with infinitely many zeros therefore f(x,0) ≡ 0 therefore the Zariski closure of I is R^2?

[gritty widget]

No. Let f(x,y)=y. Then the locus of f is {(x,0) : x in K}

[near sage]

Hmm

Then how should I proceed?

[gritty widget]

Let f(x,y) be a polynomial s.t. f(x,0)=0 for all x in [0,1]. Let g(x)=f(x,0). Then g is 0 for infinitely many x, so g=0. Therefore f(x,y) is a polynomial in y with no constant terms.

[near sage]

so what would be the closure then? like a set of points (x,y) where f(x,y) ≡ 0 that contains I, right? what would f being a function of only y do towards the goal?

[gritty widget]

when is f(x,y) = a_n y^n + ... + a_1 y = 0?

[near sage]

Yeah

[gritty widget]

It wasn't a yes or no question

[near sage]

oh wait I didnt see the "is"

well when y=0

so Im looking for the x-axis then?

[gritty widget]

yeah

[near sage]

hmm, ok thanks

[prime elbow]

If X is indiscrete topology and there is continuous mapping from X to Y where Y is Hausdorff space so the mapping must be constant mapping, right ?

[gritty widget]

What would happen if there was a non-constant mapping?

[prime elbow]

Let X has more than one element.

If there is non- constant mapping then there is x_1 ≠ x_2 such that f(x_1) ≠ f( x_2 ).
Then by Hausdorff property there is a disjoint open set U and V containing f(x_1) and f(x_2) respectively.

Then since f is continuous, so f^-1 ( U) and f^-1(V) are open set and since they are non empty therefore it will be X, so there is y which belongs to both f^-1(U) and f^-1(V) then f(y) belongs to U and V, it contradicts that U and V are disjoint open set of Y.

[shadow charm]

is this because of poincare duality or smth else?

[gritty widget]

Sounds good

[shadow charm]

also this uses some sort of assumption on the intersection path no? (I could imagine for example a bounded path in R^2: all 1-cycles are homologous but some have intersection nb 1 while others 0)

[gritty widget]

More generally a similar argument should work for T0 spaces

[prime elbow]

T_0 space means ?

[gritty widget]

Given any two points x,y there is either a neighbourhood of x not containing y or a neighbourhood of y not containing x

[prime elbow]

Okay

[prime elbow]

Is {0} connected subsets of R?

[umbral panther]

This is a very simple thing that is a prerequisite for stating duality
But you could say it’s the local property of manifolds that globalizes into pd

[umbral panther]

Huh?
Are you assuming both are cycles?

[shadow charm]

Okay maybe I’m misunderstanding smth: how are we defining intersection number here

I’m sure it’s quite simple but i did Poincaré duality way too fast and not really understanding it a while back so just assume I don’t know what it says

Like I think I see the idea? The path from x_i to x_r+1 determines a function on 1 chains given by intersection numbers, and i guess intuitively this vanishes on boundaries so it should define a well defined map on homology but the “intuitively” part isn’t formally clear to me

[fading vale]

i think its pretty common to prove this using PD

i guess morally its not really that different

[umbral panther]

The (transverse) intersection of a cycle z with a chain x is a chain and d(z.x) = z.dx. Thus if w and y are homologous, you can find x so that dx = w-y, then z.dx gives a homology between z.w and z.y

In particular, if z is the path between the cusps and b is a boundary = dx, then z.x is a bunch of intervals, so z.b is a bunch of paired cancelling points

[stoic eagle]

I don’t think you need that. These facts are in “A Primer on Mapping Class Groups” I believe

Also Margulis is a goat

[shadow charm]

Sorry again for the questions but where is your operator . coming from?

[shadow charm]

Oki ill look into that

[shadow charm]

So true

[umbral panther]

It’s simply intersection. But it is only the right thing if the chains are transverse to each other. Most of the work is reducing to the transverse case

A generic pair of chains are transverse. Given x and y not transverse, x is homologous to a chain x’ transverse to y. If x’ and x’’ are transverse to y and homologous, then x’-x’’’ Is the boundary of a chain transverse to y

[shadow charm]

So when you say d(z.x) = z.dx how are you taking intersections of a 1 chain with a 0 chain on the Rhs to get a 0 chain on the lhs

[umbral panther]

1 and 2

We want to understand the intersection of two 1-chains. But one of them is a boundary y=dx. So we intersect with the 2-chain x to understand the intersection y that we really care about

[shadow charm]

okay so let me just do a drawing to understand here

So this is a shit drawing but is it what you’re saying?

now in what sense is z.x a one-chain with boundary p-p tilde?

(don't mind that i havent specified orientations here you get my point)

[umbral panther]

If z is a cycle, w is not a boundary

If you want to compare z to z’, your 2-chain should have boundary z-z’

[shadow charm]

wait why is z supposed to be a cycle im confused

doesnt z correspond to the path between our cusps here?

[umbral panther]

You could think of z as a cycle in relative homology. Its boundary is the cusps, which doesn’t count

[shadow charm]

hmm is there some kind of les argument under the hood for working with relative homology?

like we're trying to show linear independence in H_1(X/Gamma), and you want to work in H_1(S,S-X/Gamma) so we need to relate these two somehow?

or i guess this is almost excision?

[umbral panther]

We’re working in H_1(S, cusps)

[shadow charm]

right and how do we obtain a statement on H_1(S-cusps) from there?

[cedar pebble]

use the long exact sequence

[umbral panther]

We have an intersection pairing. H_1(S-cusps) tensor H_1(S, cusps) -> Z

[shadow charm]

cause H_1(S,cusps) = H_1(X/Gamma, punctured neighb of cusps) no?

[umbral panther]

PD says it’s a perfect pairing, but here we just want to know it’s well defined

[shadow charm]

ahh okay

so my path between cusps lives in H_1(S,cusps)

[umbral panther]

True by homotopy invariance, er + excision

[shadow charm]

and the loops about cusps live in H_1(S-cusps)

and i guess if i did want an argument by PD i could just use perfectness no?

[umbral panther]

I don’t think PD gives you anything. Perfection says cycles exist with prescribed intersections. But we already have the cycles

[shadow charm]

oh yeah sorry i see

okay at the end of the day I guess I just need to get familiar with the exact definition of the intersection pairing for this to make sense to me, thanks

in terms of the homology of X/Gamma

in the end will we just get H_1(X/Gamma) = Z^(nb cusps)

and to get torsion we'd need elliptic points?

[merry geode]

,ask cusp

[gentle osprey]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by abstia

[merry geode]

Node vs. Cusp is always confusing to me

[umbral panther]

2 problems (1) The complete surface S has homology and that doesn’t go away. (2) The homology of the cusps isn’t the free abelian group on the cusps, but reduced by 1

If you had torsion in Gamma, the quotient stack would have torsion homology, but the quotient space wouldn’t

[paper wedge]

To compute the fundamental group of RP^2

can i say that , as RP^2 = S^2/x~-x , S^2 is a covering of RP^2 with the projection map

but then if you for example have a point in RP^2 say [x1:x2] (the line) then the inverse image would be just {x1} U {x2}

[gritty widget]

what does it mean to compute RP^2

[paper wedge]

the fundamental group

my bad

[paper wedge]

therefore this is a 2-sheeted covering

we know S^2 is simply connected tho so by like the index shit whatever we get that RP^2's fundamental group must have order 2?

is this correct

[empty grove]

Yes, I'm just not sure what you mean by the inverse image of [x1:x2] being {x1} ∪ {x2}

[paper wedge]

i meant like

okay so for example (S^2,x_0) and (RP^2,x_1)

so the preimage of x_1 whatever its homogenous coordinates are would be x_0 and -x_0

correct?

[empty grove]

Ah ok yeah

[paper wedge]

yeah

[empty grove]

Homogeneous coordinates of ℝP² are usually triples though

[paper wedge]

lmfao yeah im stupodi

mb

[empty grove]

stupodi

[prime elbow]

If X is connected then is there a subspace of X which is not connected ?

[ebon galleon]

Not necessarily: consider any indiscrete space, or more simply, the space with one element.

(Excluding the emptyset, which some people consider to be disconnected by definition)

[prime elbow]

Yes, in R if I take a subspace open interval then maybe it is not connected, right?

[ebon galleon]

Well if you take an open interval then it is connected

like (0,1) is connected

[prime elbow]

Yes

I am not sure about the when the statement says U is connected in X , so does it mean U is connected subspace of X ?

[prime elbow]

If X is a topological space and U is an open set of X and C is a component of U. Does it mean that we are taking U as subspace and then find Components of U?

[gritty widget]

"component of U" does indeed mean "component of U"

[prime elbow]

So are we taking U as subspace of X?

[vital pilot]

A subset U of X is connected if U is connected with the relative topology

And every subset U of X is itself a topological space in the relative topology

[empty grove]

Yes you view U as a space by giving it the subspace topology

[prime elbow]

Relative topology?

[prime elbow]

Okay thank you

[empty grove]

Another name for the subspace topology

[vestal herald]

i'm confused about what (b) is asking. the distance from x to A equals the infimum of all d(x,a) so there is some a0 where d(x,A)=d(x,a0) right? i feel like i'm misunderstanding the question or some definition since it's saying to assume that A is compact

i think i get it now nevermind

[gritty widget]

0 is the infimum of (0,1] but 0 is not an element of (0,1]

[worthy olive]

What a cochain complex vs a standard chain complex?

[umbral panther]

In a chain complex the differential lowers degree
C_n -> C_n-1
In a cochain complex it raises degree
C^n -> C^n+1
It is very superficial, but it is a useful convention

[worthy olive]

But are the chains themselves the same (like C_n)

[worthy olive]

Thanks

[umbral panther]

You can turn a chain complex into a cochain complex by just changing the numbering
C^n = C_-n

[gritty widget]

a topology cofinite topology and also cocountable topology is it possible?

[naive trench]

Wdym by countable topology?

[gritty widget]

yes it is another name

[naive trench]

I meant, I dont understand your question

[pastel linden]

is it true that generally that fiber bundles over X with coycles ${g_{\alpha \beta}}$ and {\tilde{g}{\alpha \beta}}$ are isomorphic (as fiber bundles) iff $g{\alpha \beta} = \lambda_\alpha \tilde{g}{\alpha \beta} \lambda\beta^{-1}$ for functions $\lambda_\alpha : U_\alpha \to \mathrm{Homeo}(F)$ and $\lambda_\beta : U_\beta \to \mathrm{Homeo}(F)$?

[gentle osprey]

bacono
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ebon galleon]

Are you asking whether the cofinite topology and the cocountable topology can agree for some set? If so, that happens if and only if the set is finite

[pastel linden]

I know this is true for vector bundles, and I think the proof follows through if you just forget the linear structure on fibers

[gritty widget]

oh, thanks

[shadow charm]

just to get back to this actually: since X/Gamma is a non compact surface its fundamental group must be free so H_1 is free as well right? So we do get Z^(nb cusps -1)

[umbral panther]

Yes, it’s not compact. Yes, the homology is a free abelian group. But it could have genus, so not all the homology comes from the cusps

[shadow charm]

ahhh okay i see

thanks a bunch

(yeah ofc freeness of homology doesnt even need freeness of the fundamental group since it's torsion free anyways)

[outer harness]

I recently learned that (for example) that the knotted torus is homeomorphic to a regular torus. I understand that this “knottedness” property is a property of how the torus is embedded in R^3, but it still feels…wrong. Does the homeomorphism from the knotted torus to the regular torus look different if they were embedded in R^4 (or some other R^n), and is there any other intuition to make this “feel right”?

[umbral panther]

Yes, if you take an embedding T -> R^3 and compose to get an embedding R^4, all the embeddings become isotopic

[oak python]

What does a induced metric topology mean? I don't even understand the information given in this question. Is there any video or pdf I can read understand?
And any hints on the a)?

[unreal stratus]

Every metric space (X,d) canonically gives rise to a topological space

by saying that a set is open in X iff it is open in the metric spaces sense

[oak python]

how is a set open in metric space?

[unreal stratus]

I suggest you look at metric spaces somewhere like Munkres

This question seems to assume such knowledge

[oak python]

ok thanks

[prime elbow]

Okay thank you

[frigid patrol]

Whats the universal property of a universal cover?

[gentle osprey]

kevinhardy2

[steel glen]

are you asking or were you responding to someone

[frigid patrol]

Asking

[novel acorn]

I think it's something like if M is the universal cover and E is another simply connected covering space then M and E are homeomorphic and the diagram commutes

[steel glen]

i thought i knew but im no longer sure. i figured it would be the initial object in Cov(X), provided it exists

im sure there's some obvious counterexample to that that i can't think of

[red yoke]

It should be initial among based covers

[umbral panther]

The initial cover is empty

The universal cover has automorphisms (deck transformations), so it is not initial. Consider R as the cover of R/Z. Translation by Z commutes with the covering map, so it is not initial

But if you choose a lifting from the the base point of the circle to the line and don’t allow that point to move, it becomes rigid and universal

[worthy olive]

How would you solve

[paper wedge]

one of the reasons i started studying math was to understand what type of black magic this shit is

it's been years

im still missing a few words lmfao

[zinc siren]

My favorite one of those was the one that still had nothing more than numbers and basic algebra, but whose solution requires some serious number theory

[gritty widget]

FLT?

[zinc siren]

no no

the question was simply to find the integer solutions to some diophantine equation, i think

someone wrote a solution and put it on quora

This is actually a neat problem for getting some overview of numbers

https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit

[prime elbow]

In any metric space , the collection of €- balls with € <1 forms a basis element for the metric topology, if this statement is true then can is this true if I replace inequality with € < a , where a is real positive number?

[merry geode]

Yeah, this one could serve a nice intro to num theory

[zinc siren]

€

😂

€>0?

€->0!

(brexit amirite)

[merry geode]

€, ah that's a good trick

[zinc siren]

using € implicitly makes some claims about the European economy

[sullen bear]

well if a >= 1, then the result is obviously true because the original basis is contained in the new one. however if a < 1, its a matter of showing that any open ball B of radius < 1 can be covered by open balls of radius < a (this isnt too tricky, just put every point in a ball that is contained in B)

[zinc siren]

what's delta? ¢?

£?

§?

@?

[sullen bear]

im having a nightmare doing hatcher exercise 2.2.30 right now. Im up to part c. If we say f: (a, b) -> (a, -b), how do we determine the induced map in H_2? i know its -1, but how do we show this?

[prime elbow]

I don't know how to use LaTeX so

[zinc siren]

so learn

$\epsilon$

[gentle osprey]

Xela

[zinc siren]

latex is really easy

[prime elbow]

Okay

[zinc siren]

or go and copy paste the symbol, or type "epsilon", but, € is laugh worthy.

[prime elbow]

😂

[zinc siren]

are you laughing at my laughter?

[prime elbow]

No I am laughing at me

Why I used that symbol

[prime elbow]

Oh

[steel glen]

what is your definition for metrizable

assuming it means there's a metric that induces the topology, then i think it's fine. i think a direct proof may be more neat
also note that if you know that all metric spaces are hausdorff, then you can just say that since X is a metric space it is immediately hausdorff

[prime elbow]

Yes, but all metric spaces are Hausdorff and all metrizable spaces are Hausdorff use exact proof, right?

[steel glen]

what do you mean by exact proof

[prime elbow]

Means they imply each other?

[steel glen]

the idea is that whatever proof you use to show that metrizable spaces are hausdorff will surely work to show any metric space is hausdorff

so if you've proved that metric spaces are hausdorff before, you either:

1. cite that theorem
2. copy that proof

but the proof you have above is fine

[prime elbow]

I know when I take B(x, d(x ,y)/2 ) and B( y, d(x,y)/2 ) are two disjoint open balls which contain x and y respectively then it will work

[prime elbow]

Okay thank you

[steel glen]

yeah this is the direct proof i was talking about above

i think its nicer than a contradiction

[prime elbow]

Yes

[prime elbow]

So are they both equivalent?

[steel glen]

are what equivalent

[prime elbow]

Means they imply each other

[steel glen]

but which two things

[prime elbow]

Does it work for both metrizable space and metric space?

[steel glen]

what is your definition of metrizable

[prime elbow]

Metric space and metrizable spaces

[steel glen]

they should be equivalent, as in there is a natural pairing between them

[prime elbow]

If X is a topological space, X is said to be metrizable space if there exists a metric d on the set X that induces the topology of X.

[steel glen]

then they are "equivalent"
a metrizable space is not a metric space since they are different objects, but a metrizable topological space can certainly be seen as a metric space if you just take the metric that induces it

and similarly, a metric space can be seen as a metrizable space if you take the topology it induces

[prime elbow]

Metrizable space and metrizable Topological space are different?

[steel glen]

no im using "metrizable space" instead of "metrizable topological space"

[prime elbow]

Got it

Thank you

[coral pawn]

Sanity check: forgetting "ho" and using intuition from normal direct limits, one can think of an object of Sp(C) as a sequence C_0, C_1,... of spaces such that C_i is homotopy equivalent to omega C_i+1. Thinking in terms of the concrete definition of prespectra, there are objects in that category called omega prespectra that satisfy the property that the structure maps X_i --> omega X_i+1 are weak homotopy equivalences. When forming the category of spectra, we take the omega prespectra replacement of each spectra, which is why the infinity category notion coincides with this one

Can someone verify if my intuition is correct?

[gritty widget]

what is a Gδ set?

[tender halo]

a countable intersection of open sets

most often, \delta is countable intersection, \sigma is countable union, G is open set, F is closed set

[gritty widget]

alright, what is a Fσ set?

[tender halo]

a set whose complement is a G_\delta set 😛

alternatively, a countable union of closed sets

[gritty widget]

oh so easy thank you

[gritty widget]

hmm X any open set and X countable mean X=Gδ set.

True?

[tender halo]

yes

dont need X to be countable, any open set is trivially a G_\delta set

[gritty widget]

thank you

[obtuse meteor]

these are not equivalent. "The" metric inducing a topology is not well-defined. Many metrics induce the same topology

[gritty widget]

True?

[brave geode]

i think they mean that the full subcategory of metrisable spaces is equivalent to metric spaces with continuous maps between them

[prime elbow]

U subset of X, right

[gritty widget]

thank you

[prime elbow]

In this T contains empty set X and all subsets which contain a not just empty and itself

[drowsy quest]

Got a midterm today on Hatcher chap 1. Wish me luck guys 🫡🫡

[steel glen]

i should not have said "the" you're right, but that's why i put equivalent in quotation marks

[gritty widget]

but not discrete space, true?

[prime elbow]

Yes

Just take an example X= {1,2} and let a=2 then T contains an empty set, X, and {2} not {1}

[gritty widget]

ı understand, really so thank you

[steel glen]

this is what i was getting at, assuming this is indeed correct

[brave geode]

the embedding of metric space with continuous maps between them into Top is faithful, and full because we chose continuous maps

and by definition every metrisable space is isomorphic to something in the image

so they are equivalent

[steel glen]

naturally?

[brave geode]

no

you need choice

[steel glen]

as in, pick one of those metrics that induces the topology

[brave geode]

yeah it's the same way you show the fundamental group is equivalent to the fundamental groupoid

pick isomorphisms and induce commutative squares

you should work it out yourself before you read "equivalence of categories" in riehl

[steel glen]

why before

[brave geode]

because it's easy and more enlightening

[worthy olive]

What does this theorem actually mean?

[brave geode]

,rotate

[steel glen]

what is S^u_*

[obtuse meteor]

which is why equivalence of categories is often the wrong thing. And why cts maps isn't the right thing for metric spaces

(Lipschitz maps or Holder maps are much more reasonable)

[worthy olive]

U is a covering space for X, The other s stuff means a subset of X and all cycles on X. It’s a subset of the general S_n(X)

[brave geode]

equivalence of categories is the wrong thing? then what is the right thing

[fathom steeple]

where does this use the CW pair hypothesis

[umbral panther]

An example that is not a CW pair is two copies of the Hawaiian earrings glued by an interval between the basepoints. This has fundamental group the free product of two copies of the Hawaiian earrings group. If you collapse this interval to a point, you get a much bigger fundamental group, in which you can go back and forth between the two groups infinitely many times. (Isomorphic to the Hawaiian earrings group)

[unreal stratus]

I'm not sure what you mean - it's part of the proof

You have to use the "homotopy extension" property

Indeed there's no obvious way to create a (non-trivial) map X/A -> X in the first place in general

[fathom steeple]

got it

what does interior of A and B mean here?

[abstract saffron]

as in the segment without the endpoint

i.e. the interior wrt subset topology

[fathom steeple]

ahh i see

[abstract saffron]

intuitive a 1-cell is a line segment, and you glue it so that the endpoints are glued with the endpoints of A

[fathom steeple]

yeah

[unreal stratus]

Is the total space of the ("unit") circle bundle associated to the mth tensor product $\gamma_1^{\otimes m}$ of the tautological line bundle over $\mathbf{CP}^{\infty}$ a model for $B\mathbf Z/m$?

[gentle osprey]

Süßkartoffel

[unreal stratus]

I imagine so, essentially because like

γ_1^(x)m is the pullback of γ_1 along "multiplication by m" CP^oo -> CP^oo

[umbral panther]

Yes. Usually people say the total space modulo Z/m, and call it a lens space, your description works, too

[unreal stratus]

Yes, that's the description/model of BZ/m I'm most familiar with too

Thank you

[unreal stratus]

How would you properly show this is the case?

I guess actually shouldn't be too hard to do it by hand

I guess I'm just not used to thinking of the total space gamma_1^(x)m as a space in its own right

I guess it may be easier just to show that gamma_1 mod Z/m is gamma_1^(x)m

Which i guess holds because you can view S^1 -m-> S^1 as S^1 -> S^1/(Z/m)

Or is there any slick way to see this

[heady skiff]

why exactly would each isngleton set be an open set of A? i'm assuming that we're imposing the subspace topology on A, if we consider a broader topological space X and consider A to be a subset of that space

[paper wedge]

what does points of A being isolated

mean

@heady skiff

[sterile fossil]

Once they are isolated, for every x there is an open set S with $A \cap S = {x}$

[heady skiff]

oh nvm i'm dumb

lol

thanks

[gentle osprey]

Leu

[heady skiff]

i literally forgot the definition of an isolatd point lmfao

[sterile fossil]

I gosto of your writing

I misread it, nvm

[heady skiff]

why do we need to take the limit of both sides for part b? because I know that we have $c_{-1} = \frac{1}{(n - 1)!} \frac{d^{n - 1}}{dz^{n - 1}} \bigl((z - z_0)^n f(z)\bigl)$ but isn't this formula valid as it is?

[gentle osprey]

okeyokay

[heady skiff]

because by definition c_{-1} is the residue

[paper wedge]

ask in #real-complex-analysis

[heady skiff]

oh yeah oops wrong channle

[unreal stratus]

that doesn't look dodgy at all

[gritty widget]

okay

true?

[zinc siren]

there's an open neighborhood of a that contains no other points of A. thus under the subspace topology {a} is open.

ah, I am too late.

[rugged canyon]

guys, are basic questions abt topology ok here?

[unreal stratus]

Yes

Indeed most of the questions here are from first courses or equivalent

[rugged canyon]

like, what is the motivation behind some of the basic general topology definitions?

for example, if we consider metric spaces, it makes intuitive sense to define a set to be closed if it contains its limit points.
but when we define a closed set iff complement is open my brain goes brr – like, yeah, this is true in metric spaces, but why did we choose this particular generalization, for example?

same for closures – why did we choose definition «the smallest closed set that contains A»?
sure, we cant use limit points since we dont have distance – but I just have trouble visualizing it in more abstract context (without a metric)

or mb i just need to see more examples of topological spaces that aren't metric spaces

[zinc siren]

closure seems intuitive tome

the general construction of "smallest [blank] containing S"

it's everywhere in mathematics

[rugged canyon]

yeah. like in «construction» of naturals for example as «the smallest inductive set»
I just don't like this «construction» since it ain't constructive

[zinc siren]

what?

wdym

I'm thinking of, say, the linear span.

Or a subgroup generated by S

Or the normal closure (smallest normal subgroup containing S)

or the subring/subfield/subalgebra generated by S

or the sigma algebra generated by S

or the topology generated by a subbasis.

[rugged canyon]

defining smth as «the smallest» is far less constructive to me that specifying the object by saying its a union with its limit points

[zinc siren]

It's not constructive. But it's everywhere.

It's very useful, too.

Also, union with limit points begs the question of "what about limit points of those limit points?"

To me, the closure is first and foremost the smallest closed set containing S, it's the closed set generated by S, a special case of this general construction, insert abstract nonsense about universal properties.

[rugged canyon]

yeah, that's a technicallity we have to prove, sure. but it at least gives some mental pic

[sterile fossil]

One thing to keep in mind is that when things get more general they tend to become weird as well. I guess the idea of using those definitions is more related to being able to use the same arguments in different contexts

[rugged canyon]

(or a better mental pic)

[rugged canyon]

hmmm, interesting

[zinc siren]

The intuition behind smallest closed set containing blank is that you use the same reasoning you did everywhere else in mathematics

Honestly for me, the harder justification is the axioms defining a topology in the first place!

To this day they feel strange. I am unable to succinctly explain what topology is to someone without starting at the definition and working out some of the results

[rugged canyon]

hm, but they are just heine borel classifications of open sets in metric spaces, no?

[zinc siren]

wat

[rugged canyon]

that union of arb open sets is open and int of fin many open sets is open

[zinc siren]

I don't know this classification, though I probably know the result.

[zinc siren]

Oh, yes.

Didn't realize that was Heinr Borel. I hear "Heine Borel" and think "in R^n, compact iff closed and bounded"

[rugged canyon]

my bad
the wrong name lol

[zinc siren]

Topological spaces are significantly worse than metric spaces.

[patent pulsar]

Hello there, I'm looking for a "math buddy" because my phd is a bit lonely and I miss vocalizing my questions and ideas.

We could talk about literally any topic. Though I'm currently trying to learn (at a low level for a good bunch of them) about the following :
algebraic topology in general, (co)homology, category theory, Lie groups and algebras, representation theory, differential geometry, bundles, projective geometry, quantum mechanics.

We could also choose a paper to read each week and try to go through it together. Also I don't mind starting a whole book if you're motivated.

About me : I'm super chill, I like elegant math, I value pedagogy and taking my time on the basics. Also any level is cool. Let me know if you're interested ! 🙂

[tender halo]

nothing preventing us from using limit points

you can define closure as the set with all of its limit points

[pastel linden]

Hit me up! We have a lot of overlapping interests

[zinc siren]

Sure? I worry I may not be at the required level.

I'm currently reading Fulton and Harris.

Honestly, I'd even be happy just sending random math stuff to you and receiving random math stuff.

[visual rock]

Here are some good things to think about:

1. open sets (in metric spaces) contain a lot of the information of the metric space. For instance, you can characterize continuity as the inverse image of every open set is open, and convergence of sequences as x_n -> x if for every open set containing x, eventually x_n is inside that open set.

2. Given the above definition of convergence for a sequence, you can show that every closed set in a topological space contains all limit points of sequences in it (however the converse is false).

3. this applies for more than just closures, generally when we define something to be 'the smallest X satisfying P', then if an arbitrary collection X_i satisfy P, we can intersect them to obtain something new satisfying P. Therefore the smallest thing satisfying P is just the intersection of all things that satisfy P.

4. In a metric space, you can define the closure by just adding all limit points of sequences. You can also do this in a general topological space, but you need to replace sequences with the more general nets.

[zinc siren]

I do like the idea that a topology only has a notion of closeness, and not one of how close.

But that's too close to a uniformity

But, with uniformities I believe you have some notion of closer?

Because you can half the closenesses.

[tender halo]

my favorite fact about that is that the closure of an open r-ball is not the closed r-ball

cuz closeness is a local property that cant be "accurately seen" from far away

[rugged canyon]

1. [...] For instance, you can characterize continuity as the inverse image of every open set is open
   we can do that even without a metric, right?

2. [...] that every closed set in a topological space contains all limit points of sequences in it
   but here u mean metric spaces, no?

3. [...]
   yeah, I just rephrased the intersection with «smallest» bc I was too lazy to type it all up

4. [...] You can also do this in a general topological space, but you need to replace sequences with the more general nets.
   oh, that's cool! Thanks!

[visual rock]

for 1, that is typically taken to be the definition of continuity outside of metric spaces. For 2, this applies to all topological spaces, with a suitable definition of limit points (such as the one in part 1). There are several intricacies here that don't apply to metric spaces, such as potential non-uniqueness of limits, and that the converse doesn't generally hold.

[tribal palm]

anyone have some nice resource explaining the big deal with covering maps? (why are they of interest)

my prof sort of just defined them out of the blue without much explanation, except giving exp(iθ) as an example

[umbral panther]

The quotient by a proper free action of a discrete group is a covering map
The universal cover is a covering map. Nice spaces have universal covers

Two application:
Picard’s theorem on essential singularities
a subgroup of a free group is free

[plain raven]

they show up everywhere in complex analysis, a simple way of thinking about them is a formalism for defining multi-valued functions. For example there's no continuous square root function on the complex numbers, but given the map z^2 : C -> C (which is a covering map except at z=0) we can understand the "square root" function to refer to any/all valid continuous sections of this bundle defined on an open subset of C. There are no global sections of this bundle, but there are global sections on any simply connected open subset of C not containing zero.

Similarly, exp(i\theta) doesn't have any global sections, but we can still talk about the "natural logarithm" on the unit circle S^1 as long as we understand that it refers to a choice of section on some simply connected subset of the unit circle.

[tribal palm]

ooooh this sounds very promising, i admit i was rather frustrated by the treatment i’ve seen so far of multivalued functions

maybe these covering maps are just what i need

thank you

[plain raven]

A good book on complex analysis that treats covering spaces early on is "Lectures on Riemann Surfaces" by Otto Forster.

Sheaves are a different formalism than covering spaces for talking about multivalued functions, but they're equivalent in a sense. Certain kinds of sheaves correspond to covering spaces and vice versa.

[tribal palm]

wonderful, will have a look

[patent pulsar]

Just to add for a good heuristic I heard for covering spaces : imagine that you're looking at a loop inside a circle. This is already... Kind of messy to visualize, at least for me. So we would like to "untangle" that loop, that is, instead of considering the loop in the circle, instead consider a path in a "multi-level car park". It's way easier to talk about that because you know at which level you're at

Each level corresponds to a copy of the circle

Well, not sure this is 100% correct, but I'm sure they had the "let's untangle this crap" in mind when they started working on this

Once you have your path in that space, which we call "covering space", you just have to project it to get back to the circle. In others word, to tangle it again

Of course this untanglement is not unique because why choose one level instead of the one right above ? However, you can prove that it's unique up to an integer

[tribal palm]

[zinc siren]

a covering space locally looks like a stack of pancakes

[civic verge]

I hope one day to reach that level!

[paper wedge]

pretty stupid question incoming no lonely assholes please:

yeah nvm

or wait

if i have a space that it is a quotient of some other space

and if the quotient space is contractib le

is the big space contractible

cuz u lift the homotopy ?

[ebon galleon]

why would you be able to lift the homotopy

the singleton is a quotient of any nonempty space

[paper wedge]

if you are asking this then i definitely do not understand the homotopy lifting property

[ebon galleon]

this would imply every space is contractible

[paper wedge]

can you please explain the homotopy lifting property

or anyone

[paper wedge]

i thought you could always do that ?

[ebon galleon]

idk what kind of quotient you are considering so maybe it has the HLP
but the actual problem i guess is that even if you can lift the homotopy, you need a particular type of homotopy to conclude the total space is contractible, right? You'd need a homotopy from the identity to constant at a point, which you can't guarantee you'll get with just this data

Like the endpoints of the lifted homotopy might not be constant or identity

[paper wedge]

yeah

i only know where it starts yeah

the lifted homotopy would have F(_,0) = p^-1(x_0)

but thats it

is that what you mean

[ebon galleon]

it would have to lie in the subspace p^{-1}(x_0), yeah. which may be very large for a general quotient

[unreal stratus]

Isn't this the case tho

[ebon galleon]

[quiet thorn]

Just squeeze hard enough

[rugged canyon]

I guess the idea of using those definitions is more related to being able to use the same arguments in different contexts
Now, I see what u meant

those looks almost exactly like the ones for metric spaces(!)

is there a notation for a nbhd of a point in a topological space? \
don't like writing $\exists , U\in\tau \text{ s.t } x \in U$

[gentle osprey]

Sweet Tea 🧋

[rugged canyon]

how do I show the reverse? \
like, that if

\begin{align*}
\forall B_\varepsilon (x) , \exists N \text{s.t.} \
x_n \in B_\varepsilon(x), \text{ if } n \ge N
\end{align*}

then

\begin{align*}
\forall \text{open subsets } U \text{ of } M , \exists N \text{s.t.} \
x_n \in U, \text{ if } n \ge N
\end{align*}

(the other direction is trivial)

[gentle osprey]

Sweet Tea 🧋

[rugged canyon]

Assume we are given a metric space $(M, d)$ and for any $\mathbb{R} \ni r > 0$ and $x_0 \in M$ we define
$$
B_r (x_0) := {x \in M , | , d(x, x_0) < r}
$$
as usual

[gentle osprey]

Sweet Tea 🧋

[sterile fossil]

What It means for U to be open in M?

[rugged canyon]

I only now noticed I had to include $x \in U$ (i.e. $U$ is a nbhd of $x$)

[gentle osprey]

Sweet Tea 🧋

[rugged canyon]

ahhhhhhhhhhhh. it was that simple

thank you soooo much

btw, I'm struggling to see what's the point of defining convergence (i.e. being eventually arbitrary close to a point) somewhere where we can't measure closeness?

just doesn't make any sense

this assumes injectivity and surjectivity, no? (asking bc there is no specific mention of it)

[gritty widget]

this does not assume that f is bijective

f^{-1}(U) means the preimage of U (as it says in the text)

[unreal stratus]

ye

[gritty widget]

the set of all things in X which f maps into U

[rugged canyon]

oh wow. so not only we don't assume injectivity, we also don't need surjectivity(!)\
like, we are fine with $f^{-1} (U) = \varnothing$, ye?

[gentle osprey]

Sweet Tea 🧋

[unreal stratus]

yes

empty set is always open anyway

also like

1. this characterisation works for metric spaces already

2. you will later prove that f: X -> Y is continuous iff the restriction X -> f(X) is continuous anyway

but that requires giving f(X) a natural topology (though for metric spaces it isobvious what to do)

[rugged canyon]

am I even on the right track?

[unreal stratus]

ye

Chad move is to just say "by basic set theory" and then write down the answer you need

without checking

[rugged canyon]

how do we define $f^{-1}[U]$? Like, with
$$
f^{-1}[U] = {x \in X , | , \exists y \in U \text{ s.t. } f(x)=y}
$$?

Then what if for some $y_0 \in U$ there is no $x$ s.t. $f(x)=y_0$? Do we add the emptyset into the output of $f^{-1}[U]$? \

For example, if $U={y_1, y_2}$ and $f^{-1}(y_1)=x_1$ and $f^{-1}(y_2)$ does not exist \
Do we write $f^{-1}[U]={\varnothing, x_1}$ or just ${x_1}$?

[gentle osprey]

Sweet Tea 🧋

[gritty widget]

f^{-1}(U) = {x in X : f(x) in U}

[sterile fossil]

You can look from your definition, you used existencial quantifier, so you just ignore that y_0

[jagged rampart]

any lectures to follow along for Munkres general topology?

[worthy olive]

What exactly is S_n(X)?

Where X is some K-simplex

[cedar pebble]

I'm assuming this is the group of singular n-simplices in X?

this is like, pretty insanely large, I don't know how satisfying of a description you can really hope for

[viral atlas]

Look up for MAT327 lecture notes from University of Toronto

[worthy olive]

Yeah, but what does it actually mean for a simplex to be in X?

[cedar pebble]

You just consider continuous maps from a simplex

[prisma garnet]

this looks like lee

[rugged canyon]

ya

[prisma garnet]

basedddd

reject munkres' tyranny, accept lee's gospel, join the anarchy!!

[gritty widget]

i like Willard for point set

[gritty widget]

LET'S FUCKING GOOOOOO!!!! IVAN!!!!!!

[stuck geyser]

I have done very little topology but my dumbass had an idea to try to prove the smooth, euclidean case of Urysohn's lemma, that is for any two closed sets A, B in R^n, that there is a function from R^n to [0,1] where the fiber of 0 is A, and the fiber of 1 is B.

Namely using the fact that in a euclidean space, every open cover admits a countable subcover, and applying this to a covering of open balls of a fixed radius

I know if any open covering of E / (A \cup B) must have a countable subcover. The problem is that the union over the subcover might overflow and intersect into A or B

[unreal stratus]

I think this is easier by other methods

[stuck geyser]

I was going to construct, for each sequence of balls that cover E / (A \cup B), a geometric-like series function

[unreal stratus]

i may be wrong lol

[stuck geyser]

I'm avoiding convolution by mollifiers

[unreal stratus]

Oh lol

Why

[stuck geyser]

I was wondering if there was a way that doesn't appeal to measure theory

I might try to prove that for each set X in R^N there is a locally-finite covering of balls for it, probably through Lindelof Property

[worthy olive]

Maps from a simplex to ?
Sorry if I’m just stupid

[gritty widget]

to X

[honest shore]

Guys, is borsuk ulam's theorem excluding the point (0,0) ? Like the theorem is useless if its not cause f(0)=f(0)

And if it's then that's kinda weird to get

[clear jackal]

The origin is not a point of a sphere

It does not have norm 1

[honest shore]

Oh right

[honest shore]

U got an eg of pi(X) being isomorphic to pi(Y) without X and Y being homeomorphic ?

[umbral panther]

X=R^0, Y=R^1

[honest shore]

Tf is R^0

[umbral panther]

Or X=R^2

[honest shore]

Ty

[unreal stratus]

A point

[worthy olive]

Is there such things as non positive integer simplexes?

[knotty vine]

Sometimes it makes sense to call the empty space a (-1)-simplex

[unreal stratus]

and a 0-simplex is a point

[plain raven]

Beyond that the (-n) simplex is

{ 0 > -1 > -2 > ... > -n }

|| This is a shitpost. ||

[gritty widget]

Wasnt there something about some kind of negative dimentional objects using spectra or something

[solemn oar]

The notion of "dimension" is not really appropriate in the category of spectra. You can have homotopy groups indexed by any integer, but it is better to think of these as analogous to the homology groups of an unbounded chain complex.

[grim knot]

can sombody help me find the homology group of del 1 and del 2 using this method:

I'm trying to solve this exercise and I got del 2 = (1, 1, 0, 0, 1) and
del 1 = [(1 0 0 1 1), (1 1 0 0 0), (0 1 1 0 1), (0 0 1 1 0)]

[opaque scroll]

The only relevant information is the dimension of the image and kernel, which you can find by row reducing the matrices.

Also, are you working in characteristic 2? Otherwise I think you're missing some signs

[willow viper]

hello, can I prove this with a chain of equations? $\bar A^c=A^{c\circ}$

[gentle osprey]

based

[paper wedge]

chain of definition unfolding yeah

[unreal stratus]

Yes, stare at the definition of interior and closure

[rotund halo]

Are nets typically assumed to be nonempty (or the directed set used for their domain)? I'm trying to prove that $x\in X$ is an accumulation point of a net ${x_a }{a\in J}$ iff some subnet of ${x_a}{a\in J}$ converges to $x$, I can build up a subnet that I think works but I keep getting stuck proving convergence because I don't know why $J$ can't just be empty?

[gentle osprey]

DootDooter

[paper wedge]

arent like

subnets defined on like

cofinal subsets

of directed sets

which are nonempty?

[rotund halo]

Yeah the definitions involve cofinal and directed sets, I just don't see why those have to be nonempty.

[paper wedge]

i might be wrong but irrc a subnet isjnt defined as just a map from a subset of the directed set

[rotund halo]

I could be missing it

[paper wedge]

and u say your proof doesn't work if its empty?

why is that

[rotund halo]

Because convergence reqs showing a particular index exists

If the set I need to produce an index from is empty, then I can't do that.

[paper wedge]

okay well

new definitoin just drofpped

a directed set is a nonemtpy set such that...

😄

[rotund halo]

Lol yah, my suspicion here is that somewhere in my chain of definitions I'm missing a "blah blah such and such is nonempty"

[paper wedge]

yea your right

i just googled it

directed sets are nonemtpy

[rotund halo]

Imma double check. That's a relief then.

[opaque scroll]

I guess one definition of a directed set would be a poset in which every finite subset has an upper bound.

Then it would follow that the set is nonempty, since the empty set needs an upper bound.

[rotund halo]

Mine is along the lines of a partially ordered set where for any pair of elements you can find an upper bound for the pair.

This is from munkres. I'm paraphrasing it a bit but the nonemptyness is dropped afaict. I looked back at the definition of a partial order and don't see nonemptyness there either.

Gonna assume they just forgot it or something lol.

Thanks for the help!

[willow viper]

is it okay to write it like this $\bar A^c= (X \setminus \bigcap \set{C:A \subseteq C,C~closed})=(\bigcup\set{X\setminus C: A \subseteq C, C~closed})=\bigcap\set{(X\setminus C)^\complement: A \subseteq C, C~closed} = A^{c\circ}$

[unreal stratus]

Yes nice

[willow viper]

Would it also be okay to write it as C1 ∪ ... ∪ Cn?

[unreal stratus]

No because there many be infinitely many things

[willow viper]

but isn't the union of set countable?

[unreal stratus]

Wdym

Nothing here needs to be countable

[willow viper]

Isn't that the word? A must consist of a finite amount of sets no?

[unreal stratus]

No

Idk what you mean really

[willow viper]

this seems easier to use when using set theory operations that's why

[ebon galleon]

Oh I see, the bigcups. Yeah, writing it like this would imply that there are finitely many sets your taking a union over, but really this will be infinite in general

So you shouldn't write unions or intersections this way unless there are only finitely many

[willow viper]

But why infinite I don't quite understand that ^^

[ebon galleon]

In general, there might be infinitely many closed sets containing A

For example: A = (0,1) and X = R. Then any [-n,n] contains A, for n a positive integer; there are infinity many positive integers

[unreal stratus]

I think the question is also more - why should it be finite or countable

There is nothing in the definitions to that effect

[willow viper]

Ok yes that makes sense thx

[willow viper]

I think I mixed some terms up sry ^^

[merry geode]

,rotate

[zinc siren]

... isn't a long exact sequence defined as exact?

[merry geode]

Maybe it is asking why the long exact sequence exists

[zinc siren]

that's seriously left as an exercise?

point is you have a short exact sequence of chain complexes

using my own letters,
0->A->B->C->0
0->H1(A)->H1(B)->H1(C)->0
0->H2(A)->H2(B)->H2(C)->0

err with chain complexes pretend I put C_i instead of H_i

point is, snake lemma

err, zigzag lemma

if above image

then below image

there's this annoying diagram chase thing you can do to construct \delta

now anyways the snake lemma is a generalization

and this is an application of it

coker is A'/im a

so note that the map between kernels is great and the map to the coker is fine since after passing to homology im a becomes trivial.

at least, I think that's the explanation

[old plover]

Tf is this😭🙏💀

[zinc siren]

arrow theory

[willow viper]

hello, how can I start proving that every subset of a separable metric space is a separable metric space again? How can I find a countable subset which is dense?

[zinc siren]

just take the original one

[willow viper]

in what sense?

[zinc siren]

you have a separable space

take a countable dense subset

restrict to the subspace

QED

[willow viper]

what the heck

[zinc siren]

what?

[willow viper]

I don't quite understand ^^

[zinc siren]

Let X be a separable space.

Let A be a subset, give A the subspace topologym

X has a countable dense subset S

Now S \cap A is a countable subset of A dense in A.

[umbral panther]

Not if A is the complement of S

[zinc siren]

...shit

[willow viper]

I read about creating a base but I don't quite get it

[ebon galleon]

Is this statement even true? I thought it was only open subsets

No, it's not. Consider R U {infty} where R is given the discrete topology, and then infty is a generic point. Then R with the subspace topology is not separable, but R U {infty} is, with {infty} being dense only for topological spaces

[zinc siren]

that is, one point compactification of an uncountable discrete space

[willow viper]

But it's only about metric spaces, not general spaces

[umbral panther]

If S is the countable dense subspace, I suggest attempting to find a countable dense subspace of an arbitrary subset A by starting with the countable set SxN and for each s,n, finding a point of A within 1/n of s, if any such point exists in A

[ebon galleon]

oh is this actually true for metric spaces

my bad

[spiral lagoon]

<@&286206848099549185> I'm asked to define a homotopy, with a homotopy square/diagram/rectangle, to show that f(t) = x0 is the identity on the fundamental group π1(X,x0) for f a path in X. How can a single homotopy show this?

[gritty widget]

you're probably not going to get much mileage out of pinging helpers for questions in the "advanced mathematics" channels

[spiral lagoon]

Sorry, who should I ping?

[steel glen]

do you have the exact question in hand?

it sounds like they are asking you to show the path product with x_0 induces the identity

[spiral lagoon]

"Write out the formula for a homotopy (and draw the rectangle) to show that f(t) = x0 is the identity for the fundamental group"

[steel glen]

yeah im still not sure what they're looking for. where is the question coming from

[spiral lagoon]

Just a Canvas assignment

[swift fjord]

They want you to show that for any other loop concatenating with the identity on either side is homotopic to the loop

And write out the homotopy explicitly

[spiral lagoon]

Ohhh

[sonic hill]

Does anyone happen to know if the generator (of order 12) of $\pi_6(S^3)$ is possibly the quaternionic multiplication $S^3\times S^3\to S^3$ (after a suitable modification as in the J-homomorphism)?

[gentle osprey]

Icy0

[rough arch]

I have a question in Topology iam currently reading Topology by James Munkres and he claims that the one point set is a Basis for the discrete Topology of any set X.

[alpine nest]

The collection of all one-point sets, you probably mean?

Because just one such set isn't going to be a basis

[rough arch]

But i do not understand why that should be the case for example the set X=(1,2,3) the first condition is satisfied but the second is not definable because every element of the OPS is disjoint from one another so every construcable intersection is empty

[tender halo]

the intresections are empty so the condition is vacuously true

i assume the condition you mean is something like for ever point inside the intersection there is a basis set contained inside it

[alpine nest]

The intersection of any two distinct basis sets in this case is empty. So vacuously, every element of the intersection belongs to some basis set that's a subset of the intersection.

It works for tricksy logical reasons, but it does work

[rough arch]

The condition says that if x is element of the intersection of two basis elements then there must be another base element B3 that contains x and B3 ⊂ B1intersectB2 which is for my given example X(1,2,3) not the case

[alpine nest]

Right, so if the interesection is empty, it has no element.

[rough arch]

Exactly

[alpine nest]

So in the "if x is element..." implication, the first part is false

And thus the implication is true

[brittle rapids]

it's not even false

it just doesn't apply there

[alpine nest]

It is

For every x, x is not an element of the intersection

"x is an element of the emptyset" has a well-defined logical value

[brittle rapids]

you can take two singletons that are the same, and their intersection will be nonempty

[alpine nest]

Which is always false

[brittle rapids]

it's exactly like say open intervals on R

[rough arch]

But why does he then say for every set when it’s clearly not the case for every set given my example

[alpine nest]

Yes, and then the one element of that intersection will belong to that intersection, which is a basis set and a subset of the itnersection

Still works

[brittle rapids]

the point was that you can avoid the tricksy logic because really this is what we're used to

given a point contained in the intersection of two basis sets we can find a smaller one

[alpine nest]

It is the case for every set, it's just that it's often vacuous and not very practical/interesting

But it is true

[brittle rapids]

consider two disconnected open intervals in R

[rough arch]

So the justification that its true even for my case is that our premise is wrong because the empty set is not an element but an subset and 0->1 so we base true on logic but not really mathematical truth

[brittle rapids]

this is mathematical truth

you dispose of the cases where the assumption is false

[alpine nest]

If the intersection is empty, it's true there's no condition you could/would actually want to check, and that's why the logic is defined so that an implication with false predecessor is always true.

Precisely so that the only cases that are interesting/merit investigation are ones where there's something to actually check

It is sometimes unintuitive, a frequent stumble, and a lot of seminar talks have been ruined by the innocuous equestion of "but what about the empty set?", but it's still overall the aproach we've come to prefer.

[brittle rapids]

(again this isn't even a case of vacuous truth. the axiom the basis needs to satisfy is supplied by pairs of identical basis sets)

[alpine nest]

Yeah, but the vacuous truth comes in when you consider a pair of disjoint basis sets.

[brittle rapids]

that's the argument for why vacuous truth is natural

[alpine nest]

Yes

[rough arch]

Ok can u say that then if all base elements are disjoint from one another it is enough to just show the first condition to show that it is a base

[brittle rapids]

ya

in particular any partition forms a base of a topology

[rough arch]

Ok thank you guys so much i highly appreciate your help this dispute was keeping me up all night

[brittle rapids]

one of the motivations for vacuous truth is that true implications should stay true when you restrict your universe of discourse

so by thinking, "wlog i need only consider pairwise distinct base sets" you immediately imply vacuous truth

[gritty widget]

In the proof of Whitney-Graustein it is stated that if there's a regular homotopy as t goes from 0 to 1, then the turning number depends on t continuously. Can someone explain why is it obvious that the dependence of turning number is continuous?

[umbral panther]

What is your definition of the turning number?

[gritty widget]

if $\gamma(s)$ is a closed smooth curve, then the turning number is defined to be the number of rotations of the $\frac{d\gamma(s)}{ds}$ vector, where clockwise rotations are counted with a minus sign

[gentle osprey]

topology student

[umbral panther]

Express it as an integral. Integrals are continuous

[gritty widget]

thanks, will have to think about it but I get the idea

[tender halo]

can there be an uncountable closed set \subseteq R \ Q?

[naive trench]

Mmmm Cantor set(?

[tender halo]

cantor set has rational points right?

can you wiggle the set so that it no longer has them?

[naive trench]

Yeah I did not read it well

[ebon galleon]

I want to say no nevermind

[umbral panther]

Isn’t there a positive measure closed subset of [0,1] containing no rationals?

[tender halo]

oh yeah true

[umbral panther]

Eg, enumerate the rationals and delete a neighborhood of each one. Let the n-th neighborhood have size 3^-n, so the total area removed is less than 1

[alpine nest]

yeah, the complement of a small open set containing all the rationals will be closed, uncountable and contain no rationals

that said, I wonder if you can prove this without invoking measure

is there a purely topological argument, probably involving Baire somehow

[tender halo]

small as in

of a non full measure?

[alpine nest]

yeah, as bw wrote above me

union of open intervals around each rational, with lengths summable to an arbitrarily small total

[umbral panther]

The standard cantor set is the set of numbers that when expressed in a base 3 have only 0 and 2, no 1

Consider the cantor set of numbers such the n-th digit is either the n-th digit of sqrt(2) or the n-th digit of sqrt(3), independent choice for each n. Easily closed and uncountable. Probably possible to prove no rationals

Easy to prove example:
Numbers of the form sum of a_n / n! Where a_n is 1 or 2

[alpine nest]

neat

[tender halo]

neat

[unreal stratus]

R\Q

[tender halo]

not very closed

[unreal stratus]

I joke though as I assume you mean closed in R

But the way you've written it makes it seem like it should be closed in R \ Q

[limpid sandal]

I’m a bit confused on the claim B(a,delta) is a subset of f^-1(V). Based on the definition of continuity it seems the x they are using must also be in f^-1(V) so their assumption that x is in B(a, delta) should really be x is in the intersection of B(a, delta) and f^-1(V) which would not lead to B being a subset of f^-1(V). Every proof I’ve looked at seems to ignore this so I’m not sure

[wispy veldt]

You have a homeomorphism between the irrationals and N^N, and you can take the inverse image by any homeomorphism, of {0,1}^N which is compact by tychonoff and clearly uncountable

[alpine nest]

Thanks, I hate it

[uneven bronze]

I'm new to this. I was wondering how to show $$\phi(x)=\begin{cases} 1 &\text{if } x=\infty \ \frac{x}{\sqrt{1+x^2}}&\text{if } -\infty<x<\infty \ -1 &\text{if } x=-\infty\end{cases}$$ is a homeomorphism between $\overline{\mathbb R}$ and $[-1,1]$. Does it suffices to simply show that $\phi(x)$ has an inverse and the function+inverse is continuous? Or do I need to do something more?

[gentle osprey]

Philip

[wispy veldt]

yes you can do this directly

[worthy olive]

What sets do you use for this? I tried to do homotopy for two joined toruses, but the Mayervictories sequence is still not known

[uneven bronze]

ok, when should I be cautious about this? I've heard a continuous bijection need not always be a homeomorphism...

[worthy olive]

Exercise 6 plz btw

[wispy veldt]

if the domain is compact and range is hausdorff then any continuous bijection is a homeomorphism, so be careful when these conditions are not available

There is certain situations where this is true too, for example see the bounded inverse theorem.

[uneven bronze]

Do we know though that the extended real line is compact a priori? Isn't that what the homeomorphism establishes? Feels like we are using things we need to show...

[wispy veldt]

you dont need to use this result, you have a explicit map and you can check the continuity of the inverse directly.

but if you want to, you can just show f^-1 is a continous bijection

extended reals are hausdorff

[uneven bronze]

alright, thanks

[sterile fossil]

Do you agree that the set which satisfez $|x-a| < \delta$ is precisely $B(a, \delta)$

[gentle osprey]

Leu

[limpid sandal]

Yeah

[sterile fossil]

The proof says that if x satisfies the inequation then f(x) satisfies the second inequation

And the f(x) satisfying the second inequation means that $f(x) \in B(f(a), \epsilon)$

[gentle osprey]

Leu

[sterile fossil]

Thus $x \in f^{-1}(B(f(a),\epsilon) \subset f^{-1}(V)$, but x was arbitrarily chosen in $B(a, \delta)$, so in fact $B(a, \delta) \subset f^{-1}(V)$

[gentle osprey]

Leu

[limpid sandal]

It just seems we have to require that x is in f^-1(V) as well to conclude f(x) is in B(f(a),epsilon)

[sterile fossil]

Look

You agree that the first inequation implies the second one?

[limpid sandal]

Like I’m thinking of a case that a is on the boundary of the set f^-1(V) so B(a,delta) has points not in f^-1(V)

Yeah but only if x is also in f^-1(V) as well

[sterile fossil]

x is in U

[limpid sandal]

So that f(x) is defined

[sterile fossil]

x in is the domain, U

So f(x) is defined already

Is in*

[sterile fossil]

And U is open

So ultimately you could say there is an open ball of radius r_1 contained in U where x is inside

And you have that open ball with that \delta, you can then construct an open ball inside those two

Do you agree with that?

[sterile fossil]

The intersection is open, x is element of the intersection then there is an open ball inside the intersection with center at a

[limpid sandal]

I’m not sure what you mean. But I didn’t mean f(x) isn’t defined outside of f^-1(V). I meant to say it won’t map to f(V) so f(x) will not be in V and hence not in B(f(a),epsilon) if x is not in both B(a) and f^-1(V)

[sterile fossil]

You think that the first inequality doesn't imply the second inequality?

[limpid sandal]

Yeah I think it only implies the second if x is also in f^-1(V) which they never state

This is the case I’m trying to disprove idk if this helps

Like it seems like we have to show no such y could exist

[sterile fossil]

1 - f is continuous in U

2 - U is open

3 - V is open

4 - $a \in f^{-1}(V)$

5 - By (4) $f(a) \in V$

6 - By (3) there exists $\epsilon > 0$, such that $B(f(a), \epsilon) \subset V$

7 - By (1) there is $\delta > 0$, such that $x \in B(a, \delta) \subset f^{-1}(V) \implies |f(x) - a| < \epsilon$ this follow from the definition of continuous function

[sterile fossil]

$f(y) \in B(f(a), \epsilon)$