point-set-topology

[alpine nest]

I recommend letting go of that if you ever want to study measure theory (or indeed general topology)

[silver ridge]

maybe when I'm more comfortable with them

[silver ridge]

welp. guess im never studyign measure theory

[alpine nest]

Also fair

[quartz horizon]

What makes you uncomfy with them?

[silver ridge]

Being wrong about them in the past

lol

[quartz horizon]

Ah, right

[alpine nest]

Then stop being wrong!

[silver ridge]

I'm trying!

[quartz horizon]

Well you can basically never go wrong with preimage

It has a left and right adjoint

[silver ridge]

I don't know what you mean by adjoint

[quartz horizon]

It’s a category-theoretic thing

[silver ridge]

ah okay

sounds like a thing for future me to understand

[alpine nest]

Also we're not really talking function operations on sets; the function is still defined between the original spaces.

$f^{-1}(A) = {x: f(x) \in A}$

[silver ridge]

Yes I know, i wa doing an abuse of language

[quartz horizon]

I mean one part of it is just

[gentle osprey]

Outsider

[quartz horizon]

$A \subseteq f^{-1}(B) \iff f(A) \subseteq B$

[gentle osprey]

Pseudonium

[alpine nest]

Yeah, but I mean ultimately you're still working in terms of the values of functions at individual points, you just consider many points at once.

[quartz horizon]

This expresses that direct image is the left adjoint to preimage

[alpine nest]

So it's not some kind of big qualitative change

[silver ridge]

right

next time I get something like this wrong, you're on the chopping block for making me confident

[quartz horizon]

You can use the adjointness to help prove your preimage stuff

If you know the universal properties of intersection and union

[silver ridge]

that's enough category theory for today pseudo, i appreciate it though

[quartz horizon]

Okkkk

[pulsar junco]

so there is no way to build such surface without building rectangles, gluing them and then twist?

[red yoke]

I meant like, you can take a long rectangular strip and twist it a few times

[alpine nest]

You can do huge swathes of topology without any concern for category theory. I understand it becomes very helpful if/when you get into the algebraic topology kind of thing, but other than that I've never felt the categorical arguments to bring much.

[quartz horizon]

Oh yeah you totally can

I just like category theory

[alpine nest]

That does come across, yes

[silver ridge]

lol

[quartz horizon]

hehe

[silver ridge]

Everything in time

[quartz horizon]

i mean it has also genuinely improved my explanations

[red yoke]

Categories make everything better

[quartz horizon]

Not everything

But lotsa things i care about

[alpine nest]

And none of the things I care about

[red yoke]

Press X to doubt

[alpine nest]

Actually one or two of my papers make use of the notion of inverse limit, but without the general categorical language or machinery.

Just in terms of a specific sequence of topological spaces with continuous maps between them.

[alpine nest]

Fair, but I don't recall seeing any papers/hearing any talks in my field that would mention category theory even obliquely.

Maybe we're just blind to the entire world of possibilities

[quartz horizon]

one of the main things category theory does is make you an expert at manipulating functions

which i think is just a generally good math skill

[alpine nest]

We do draw commutative diagrams a lot

Although again, usually without employing category-theoretical language

[quartz horizon]

i think commutative diagrams are a good way to convey information

you don't need category theory for them

[alpine nest]

This is about as big as they get, though

[quartz horizon]

mhm

[alpine nest]

(from a genuine paper of mine, so I'm almost certainly going to get doxxed now)

[quartz horizon]

my ones are usually a lot smaller!

though i did make one for the implicit function theorem recently

[red yoke]

Your diagrams are cursed

[quartz horizon]

why tho

my diagrams are fine

[silver ridge]

if you didn't say this no one wouldve known

[alpine nest]

Curses!

[silver ridge]

In general, if "such and such condition generates the topology on X" that is referring to the fact that the condition given specifies the sets in the subbasis, right?

[quartz horizon]

do you have an example?

[alpine nest]

Sort of, although ultimately "topology generated by a collection of sets" is just "the smallest topology in which all of those sets are open"

[quartz horizon]

it's an adjunction

[alpine nest]

But I think as it happens. those sets will indeed be a subbasis for the generated topology

[silver ridge]

By "give this the strong order topology" he means "the elements of the subbasis are the one-sided intervals and complements of finite sets"

is what I'm asking

[alpine nest]

Hasn't strong order topology been defined above?

[silver ridge]

It was defined here

I think I asked arki already and he said that he means "these are the elements of the subbasis", but I figure I'll ask again to be really sure

[quartz horizon]

yeah this is what i'd defer to

[alpine nest]

Yep, although it's equivalent to being the smallest topology in which all of the specified sets are open.

And then the basis is formed by their finite intersections.

[silver ridge]

I see

[silver ridge]

Actually, I'm not sure I believe that this space isn't T2. {1,2}, {3, ...} \cup {x,y}, {1} {2, .... } \cup {x,y} are in the subbasis, {1} = {1, 2} \cap {1]. by the same argument, every singleton subset of "omega" is open. Moreover, finite complements are in the subbasis. x^c = omega \cup {y}, {x,y}^c = \omega. then omega \cup {y} \cap omega = {y}, so this should be open too. So, this is the discrete topology, which is hausdorff, as for distinct p,q we can just take {p}, {q} as open and disjoint

[red yoke]

Why is {y} open?

I think this space isn't even T0

[merry geode]

Looking into torsion, I don't understand what we do not understand with torsion.

There should be something missing considering conjectures, but..

[silver ridge]

{y} = (N U {y}) \cap N

And those are both elements of the subbasis

And they're complements of finite sets

[red yoke]

N∪{y} is not open

[silver ridge]

I'm not claiming it is

I'm saying it's in the subbasis

[red yoke]

Oh I missed complements of finite sets

[red yoke]

This is N though?

[silver ridge]

Wait

You're right

Let's ignore the preceding discussion

[merry geode]

@red yoke do you happen to know abt the Reidemaister torsion?

[red yoke]

Nop, sorry

[merry geode]

Ah it's fine

@unreal stratus do you know?

[unreal stratus]

No I am just aware it exists

[merry geode]

Ah I see, sorry for bothering you

[unreal stratus]

I am quite ignorant of most geometric topology stuff

[merry geode]

It's not anything major, but it came up in volume conjecture

[alpine nest]

You seem to have already noticed the flaw in your reasoning, but in any case I can confirm that the exercise statement is correct (this space is T1 but not T2)

[gritty widget]

I know who you are…

[alpine nest]

Fame at last

[merry geode]

What is this

[red yoke]

Exhibit B

[quartz horizon]

Implicit function theorem

[quartz horizon]

There was no exhibit A…

Also I’ve used this diagram when explaining the implicit function theorem

And it genuinely helped

[merry geode]

Which setup?

[quartz horizon]

U is an open subset of R^(n + m)

f is a function from U to R^m

You’re trying to solve f(x, y) = 0

Where x is in R^n, y in R^m

And you have that f is C^1 near a solution (x_0, y_0)

Where the jacobian of f wrt the y variables is invertible at x_0, y_0

You want to be able to find an open subset W of R^n, and a U’ of R^(n + m)

With a function g : W -> U’

Such that for all (x, y) in U’, f(x, y) = 0 iff y = g(x)

Also U’ should contain (x0, y0)

[merry geode]

Hmm, considering it as R^(n+m) seems a bit weird to me

[quartz horizon]

Why

This is how it was stated in my course

[gritty widget]

Ok sorry if that sounded creepy lmao

[merry geode]

Ah, I guess it was taught differently for me.

[quartz horizon]

how was it taught for you?

[alpine nest]

I thought it was supposed to

[merry geode]

Nothing big, just R^n and R^m

[quartz horizon]

but like

what was the statement

oh hmm

did you have n >= m or something

[merry geode]

Maybe my class only did n = m.

[quartz horizon]

isn't that the inverse function theorem?

the implicit one helps when you're going from higher dimension to lower dimension

e.g. f(x, y) = x^2 + y^2 - 25

[merry geode]

Wait

[quartz horizon]

goes from R^2 to R^1

[merry geode]

Yeah, I guess it was my brain reading it all wrong

[quartz horizon]

i see

[merry geode]

Yeah implicit was taught R^(n+m) -> R^m for me as well, my bad

[quartz horizon]

there's also a third version of it technically

though i haven't seen it much

for going from lower dimension to higher dimension

all subsumed by the constant rank theorem, though

[merry geode]

Indeed

[quartz horizon]

I’ve always wondered what the neatest way to prove the inverse function theorem is

[silver ridge]

In an appendix I'd imagine

[quartz horizon]

i mean properly

it’s definitely one of the biggest proofs i did in my undergrad

[red yoke]

• A C¹ function R^n → R^n is locally injective at a nonsingular point by MVT
• It is locally surjective by AT

[quartz horizon]

What’s AT?

Also wait - mean value theorem gives you local injectivity?

[red yoke]

Algebraic topology

[quartz horizon]

huh…

you can use AT for stuff like that?

[red yoke]

If you choose a neighbourhood U of x s.t. |Dy(v) - Dx(v)| < ε for any unit tangent vector at y, then there must be no two points mapped to the same value

By applying MVT on the path connecting them

[prime elbow]

I don't understand how we can prove the second direction? Because now I need to prove that any open set can be written as union of elements in SIGMA, any hint?

Does I need to define how an open set creates by the basis element?

[red yoke]

Yea you should probably construct the topology first

And show it is indeed a topology, and is generated by ∑

[prime elbow]

I think I need to define a topology T such that the element of SIGMA is open in it

And any non-empty set U is open in T if and only if for any x in U there exists B in SIGMA such that B contains x and B contained in U

[red yoke]

In some some small enough neighbourhood U of x the straight path homotopy from g(y) = f(y-x)-f(x) to Dx(v): U-x → U-x exists

So it must induce an isomorphism on H^n-1

[quartz horizon]

right…

this is where I don’t know any homology or cohomology

[prime elbow]

Is it correct?

If the intersection of two basis elements is empty then we can say that it is an empty union of basis elements?

[red yoke]

Yes

[red yoke]

Yes

[prime elbow]

Okay, thank you

Can a singleton set be linearly order?

Because if X is a singleton set and linearly ordered set then how do these form a base?

[red yoke]

Indeed it fails

But there's only one possible topology for a point anyway

[prime elbow]

Yes

[prime elbow]

So I need to let X have more than 1 element

[prime elbow]

To prove that in a Topological space each point has a smallest neighborhood if and only if the intersection of any collection of open sets is open.

Now let I be an arbitrary indexing set and U_i are open sets in T for all i in I.

If the intersection of all U_i is empty then it is open.
If intersection of all U_i is non-empty then let x in intersection of all U_i.

It implies that x in all U_i, since each point has a smallest neighborhood, let U be the smallest neighborhood of x then U contained in all U_i.

Thus x in U and U contained in intersection of all U_i implies that intersection of all U_i is open set.

Now let the intersection of any collection of open sets is open.

Then for any point x, we let U be the intersection of all open sets containing x.

Thus, U is open by hypothesis and U is smallest neighborhood of x.

Is it correct?

[umbral panther]

Formally, they are equivalent: homology is a coalgebra and cohomology is an algebra

It is already true for of H^0. More fundamentally, the ring of continuous functions on a space is contra variant

[livid escarp]

But one is a group functor and another is a ring functor? So it is strictly more than the other?

[dim perch]

am I mistaken, or are you taking field coefficients for this to be true?

my understanding was that you need to be able to invert the kunneth map to get the coalgebra structure on homology

[lusty trench]

I think the main starting point is that, if A is a commutative ring and X is a set, then the set of formal A-linear combinations of X's elements is just an A-module, but the set of functions X -> A is an A-algebra. In particular, when defining singular chains and cochains, X is the set of singular simplices, but we have other similar situations. For example, if X is a manifold (or an algebraic or analytic variety), then the set of smooth maps X -> R is an R-algebra (there are analogues in the algebraic and analytic cases). But the mapping that assigns to each manifold its R-algebra of smooth functions is a contravariant functor, just like cohomology.

[umbral panther]

Sure, field coefficients or derived interpretation

[dim perch]

makes sense, although I think there is something to be said about the fact that this does give cohomology itself richer structure

we get to do it integrally and discretely, even with torsion

the real heart of the issue being that failure of the Kunneth sequence to be iso/collapse demonstrates that, while H_*:D(R) -> R^{Z} is lax sym mon, it fails to be strong (or oplax)

[lusty trench]

Is there any situation where we would care about coalgebras (comonoids in the category of vector spaces or modules) for their own sake, and not because they arise from dualizing monoids in some other category?

The only case I've seen is the coordinate ring of an affine algebraic group.

[dim perch]

they matter in part because you can compute with them

especially in algebraic topology

[umbral panther]

Right, the original and main issue is bialgebras, where you can’t just dualize

[lusty trench]

I vaguely remember reading that the cohomology ring of an H-space is a Hopf algebra (right?), and I thought “yes, I can see how that could make sense”, but didn't dwelve in the details.

So, intuitively, if we have a contravariant functor from spaces to rings, then it ought to send group objects (or “group-like enough”, in the case of H-spaces) to Hopf algebras, right?

[umbral panther]

Exactly

[unreal stratus]

I think you need some finiteness perhaps but yeah

Basically cause you want Künneth to hold

I guess you don't need that perhaps cause the natural map still exists

[inland laurel]

Every open map is a locally homeomorphism? Right?

[umbral panther]

No. For any space the map to a point is open. For any non discrete space it is not a local homeomorphism

[inland laurel]

If answer yes, we why use this provides on definition of covering space? Because this same thing

[inland laurel]

Oh okay

Thanks

[umbral panther]

Not all local homeomorphisms are covering spaces. For example, the inclusion of an open set is rarely a covering space

[inland laurel]

Thanks you I understand now

[lusty trench]

For a concrete example, any non-constant holomorphic function C -> C is open. In particular f(z) = z^n, for n \in Z^+. But, if n >= 2, this function isn't a local homeomorphism near z = 0.

[prime elbow]

For 2nd bit, if we take f: R->R where discrete topology defined on R and f(x) = x then f is continuous but if we take coarser topology on R (on domain X) such that Euclidean topology which is coarser than discrete topology then f not be continuous, right? Because for singleton open {x} its preimage {x} is not open in X with respect to Euclidean

[lusty trench]

If T1 and T2 are different topologies on the same set X, then the identity map id : (X, T1) -> (X, T2) is continuous iff T1 is finer than T2.

[prime elbow]

Yes

[prime elbow]

Got it, thank you

[lusty trench]

So, let X' be X with a different topology, and similarly Y', and consider the composition X' -> X -> Y -> Y', where X' -> X and Y -> Y' are the identity map on the respective underlying point sets.

[prime elbow]

Make it easy by letting Y = X

[lusty trench]

From what I said earlier, it follows that, if you want to guarantee that the composition is continuous, and have no other information besides what your problem statement gives you, then you're only allowed to make the topology of X' finer than that of X, and the topology of Y' coarser than that of Y.

[prime elbow]

Got it

But is this a necessary condition?

Maybe not

It is sufficient

[lusty trench]

It's sufficient. Whether it's necessary or not, that depends on f. That's why I added “if you (...) have no other information besides what your problem statement gives you”.

[prime elbow]

Yes

[lusty trench]

For example, if f were a constant function, then you would be allowed to freely replace the topologies of both X and Y.

[prime elbow]

Yes

[lusty trench]

For a slightly less trivial example, you could consider a polynomial f(z) with complex coefficients as a function f : C -> C, where C has either the usual Euclidean topology or the cofinite topology, and in both cases f would be continuous.

[unborn dagger]

Hello, I am not sure if this is the right place to ask my question (persistent homology, TDA); if not, please refer me to the right channel and thanks in advance.

Does someone know the proof of $VR(S, \epsilon) \subseteq Čech (S, \epsilon \cdot \sqrt{2})$?

1. $S$ is a finite point cloud (of cardinality $n+1$) embedded in $\mathbb{R}^n$,
2. $\epsilon \geq 0$,
3. $VR(S, \epsilon)$ is the Vietoris-Rips complex such that a set ${v_0, v_1, ..., v_n}$ is a simplex in $VR(S, \epsilon)$ if and only if the Euclidean distance between any two $v_i, v_j$ is $\leq 2\epsilon$, and
4. $Čech (S, \epsilon \cdot \sqrt{2})$ is the Čech complex such that a set ${v_0, v_1, ..., v_n}$ is a simplex in $Čech (S, \epsilon \cdot \sqrt{2})$ if and only if the closed balls centered around each $v_i$ with radius $\epsilon \cdot \sqrt{2}$ have a nonempty intersection.

[gentle osprey]

the_indefinite_integral

[unborn dagger]

The place that I found this inclusion relationship is https://www.researchgate.net/publication/260107617_Statistical_Inference_For_Persistent_Homology_Confidence_Sets_For_Persistence_Diagrams , page 39, but the paper does not give a proof for this.

[prime elbow]

To show that f: R×R -> R such that f(x,y) = x+y.

Now let (x_n , y_n) is a sequence which converges to (x, y) and using that result that if (x_n,y_n) converges to (x,y) then x_n converge to x and y_n converge to y.

Now we need to prove that f(x_n, y_n) converge to f(x,y) means x_n + y_n converge to x + y which can be proven by real analysis.

Is it correct?

[gritty widget]

What do you want to show about f ? That it's cts?

[prime elbow]

Yes

[gritty widget]

What is your definition of cts?

If it's inverse image of open sets is open then in your proof you need to show the equivalence here

[prime elbow]

They defined as but I proved that sequential criteria is equivalent to this definition in metric space

[gritty widget]

Yeah then it's okay

[prime elbow]

Okay, thank you

[gritty widget]

Try to show it without sequential criteria tho

That lemma is unnecessary

Just find inverse image of a standard basis element

[prime elbow]

But how can I find (f+g)^(-1)(a, b) ?

[gritty widget]

That's same as saying for what x,y a < x+y < b

fix a var

If you can then see it's plot

It's the region in between two diagonal lines?

[prime elbow]

Yes

Btw what are diagonal lines?

[gritty widget]

x+y=a

x+y=b

These lines

[prime elbow]

No I mean which lines said to be diagonal lines, is it the same as straight lines?

[gritty widget]

I meant these when I said that

[prime elbow]

Okay

[gritty widget]

Do you know how to show that it's open?

[prime elbow]

No but wait I am trying

[gritty widget]

||a viable option could be that the every point in that region is an interior point||

||given a point in that region try to find a rectangular open set containing it||

[prime elbow]

What if I pick a point between in that region (x,y) and find the perpendicular distance from x+y = a and x+ y = b be p_1 and p_2 respectively then B((x,y), min{p_1,p_2} ) lie in that region, is it correct?

[gritty widget]

yes but avoid too much calculation when you can

Instead of perpendicular distance from lines

Try distances from axes

Given a x_0, y_0 in that region

How far you can slide it vertically and horizontally

This is same as fixing either x_0 or y_0 and seeing the variation in y

[gritty widget]

Using this inequality

[prime elbow]

Oh

[gritty widget]

This is for verifying
||given (x',y') in that region
x' lies in (a-y',b-y') and similarly for x'||

|| Now pic the minimum of these values and half it and draw a circle with centre as x',y' and radius as that halved distance||

[prime elbow]

But when I want to prove that (x,y) ->xy is continuous mapping then I think sequential criteria is good, right?

[gritty widget]

It is good always but not necessary

The more we interact with our objects the better it is imo

[prime elbow]

Minimum of what?

[gritty widget]

You have 4 distances from the point x',y'

Take the minimum of the 4

[prime elbow]

Sorry I don't understand it, we have a-y' < x< b - y' and a-x' < y' < b-x'

[gritty widget]

The consider min{|a-x'-y'| , |b-x'-y'|}

This is just distance of y' in the interval that we found or x' in the same way

[finite lotus]

Hello, can anyone check if this proof is correct? N(x) is the set of open neighborhoods of x

[prime elbow]

If we let f: [0,1 ) -> S^1 such that f(x) = (x,√(1-x^2) ) is homeomorphism mapping, right?

[red yoke]

It's an embedding

i.e. a homeomorphism into its image as a subspace of S¹

But not a homeomorphism to S¹

[prime elbow]

Oh its image is not S^1

But if we take [0,1], then is it correct?

[plucky veldt]

no

what is f(0) and what is f(1)

[prime elbow]

(0,1) and (1,0) respectively

[red yoke]

The image is an arc

Not the entire circle

[plucky veldt]

right

and is there x such that f(x) = (0, -1) for example

[prime elbow]

Got it my mistake

Thank you

What can be continuous bijective mapping from [0,1) to S^1 such that it is not homeomorphism?

[unreal stratus]

Hint: angles

[prime elbow]

Yes t-> (cost, sint) but I am not sure

[tender halo]

why are you not sure

[plucky veldt]

actually (cos 2πx, sin 2πx)

[unreal stratus]

Yes if you do that then this is quite clearly continuous and bijective , nothing to check

[prime elbow]

Because, t->(cost, sin t) is not bijective, right?

[prime elbow]

Yes it works

[plucky veldt]

it's not surjective if it's only defined on [0, 1) to be precise

you'd have to have domain [0, 2π)

[prime elbow]

Yes

[plucky veldt]

you can also represent it by x ↦ e^2πix

[rancid umbra]

what is this actually showing?
something like, there is a function from X x X into the set pi_1(X, -) x pi_1(X, -) such that...

[merry geode]

Function from X x X?

Ah, do you mean "depend only on.." part?

I think you just need to show beta_h is constant if endpoints are fixed

[fading vale]

To clarify beta_h doesn't have to be constant, but for any two paths h from x to y the result should be the same

[rancid umbra]

i can show the result just fine and i understand how to interpret the result at face value, like account said above

i was asking if there was a kind of more general interpretation

for example, this question:

it shows that there is a map from [I,X] to pi_1(X,x_0) for each x_0

taking a homotopy class of paths to the change of base point map of one of its representatives

and the exercise proves that the map is well-defined

[rancid umbra]

oh, i think this one is a map from X^{S^0} to pi_1(X,x0) for any fixed x0

[rancid umbra]

this one is equivalent to 2.

[red yoke]

You mean a map from [(I, 0, 1), (X, p, q)] to Hom(π1(X, p), π1(X, q))

[rancid umbra]

oh yes

what is the extended notion of pointed space

can you specify as many points as you want? or an entire set even?

[red yoke]

I guess you can add as many points or subspaces as you want

If you consider only loops at p, you get a map from [(S¹, ·), (X, p)] to Hom(π1(X, p))

Which is a group action

[red yoke]

In contrast, this is what's known as a groupoid action

The fundamental groupoid Π1(X) is a category whose objects are points in X and morphisms are homotopy classes of paths

And just like group action, a groupoid action is just a functor Π1(X) → Set

[rancid umbra]

ill look into the fundamental groupoid some more later.

[rancid umbra]

what is the group action?

[red yoke]

π1(X, p) acts on π1(X, p) by conjugation

[rancid umbra]

how did you see that it was conjugation from this?

[red yoke]

Change of base point map is conjugation

By defn

[rancid umbra]

right, but uh. this is taking based loops to their change of basepoint homomorphisms

[red yoke]

The loop γ is sent to the conjugation (x → γxγ^-1) ∈ Aut(π1(X, p))

[rancid umbra]

oh oh

cool

thanks

[rancid umbra]

and then the second one, is fixed to

{(S^0, -1, 1) --> (X, p, q)} --> Hom(pi_1(X,p), pi_1(X,q)) for each p,q in X

so you do end up getting maps X x X into this hom set

via X x X --> {(S^0, -1, 1) --> (X, -, -)}

[rancid umbra]

and again here

via a similar map

i wonder if this map is meaningful?

like X x X quotiented by (p1,q1) ~ (p2,q2) iff their corresponding homomorphism maps are the same

[red yoke]

I think you mean there is a unique map in Hom(π1(X, p), π1(X, q)) for each pair (p, q) ∈ X×X

[merry geode]

I don't think the homomorphism map can be the same, but I dunno what you mean

[rancid umbra]

uh. wait. why can't you do this? f : X --> Y can be quotiented by x1 ~ x2 iff f(x1) = f(x2)

[red yoke]

This translates to the groupoid action being trivial

What do you mean the corresponding homomorphism maps are the same

[rancid umbra]

yes

[red yoke]

The domains and codomains are different for each homomorphism

[merry geode]

Can you concretely talk about which map this one is?

[red yoke]

If you restrict to loops at a point p, you get the statement that π1(X, p) acts trivially on π1(X, p) as a group

[rancid umbra]

so, i keep getting lost in the sauce

[red yoke]

But for pairs (p, q) in general this is not a group action, this is a groupoid action (a trivial groupoid action)

[red yoke]

Same thing happens here: the action by loops at p is a group action, but the action by paths in general is a groupoid action

[rancid umbra]

um.

we got that there is a function {(S^0, -1, 1) --> pi1(X, p, q)} to Hom(pi1(X,p), pi1(X,q)) for each p,q in X

so there is a function X x X to U_{(p,q) in X x X} Hom(pi1(X,p), pi1(X,q)) taking (p,q) to beta_h for any path h from p to q

@merry geode

[red yoke]

What's U

[merry geode]

I think how the union is defined is important.

This would be basically like a generalized function, where each image are disjoint.

[rancid umbra]

well, i just mean the union

[rancid umbra]

why?

[red yoke]

Yea that's just this

[merry geode]

$\prod_{(p, q) \in X \times X} Hom(\pi_1(X, p), \pi_1(X, q))$

[rancid umbra]

but whats wrong with union?

[gentle osprey]

Absta

[merry geode]

Generalized function

[red yoke]

What

[rancid umbra]

i don't see why you need this

[merry geode]

How do you identify element of Hom(pi1(X, p1), pi1(X, q1)) and Hom(pi1(X, p2), pi1(X, q2))?

(I am saying that basically the union is going to be disjoint union, and for each (p, q) in X x X there would be one Hom involved)

[red yoke]

Let's stick with this

[rancid umbra]

its the change of basepoint homomorphism resulting from a path taking q1 to p1.

there is only one such map

[rancid umbra]

by this

[merry geode]

Why?

[rancid umbra]

because it only depends on the endpoints p1 and q1

[merry geode]

Ah, are we assuming that Hom(pi1(X, p), pi1(X, q)) has unique element?

And by extension, maps between such homs are unique (if defined)

[rancid umbra]

wait. im just arguing with you to try and see why im wrong, i have no idea about any extensions.

am i misunderstanding this? each pair (p,q) determines a unique point in Hom(pi1(X,p), pi1(X,q)) via the change of basepoint map taking q to p

[red yoke]

Yes

[rancid umbra]

okay, so then whats wrong with unioning over all (p,q)?

[merry geode]

Yeah the Hom set has only one element

[red yoke]

Nothing

[merry geode]

By Hom set, I mean Hom(pi1(X, p), pi1(X, q))

Or do you mean sth else?

[rancid umbra]

no, thats what i mean

[red yoke]

What was your question again

[rancid umbra]

basically, with regards to this question, is there a map X x X into something, based on the result that the homomorphisms only depend on the homotopy class of h

[red yoke]

Yes, the map you described

But it's better to think of it as a groupoid action

[unreal gull]

in the second part of this proof, why does he define $P = N + b$, rather than just saying that $A$ is a subset of $[-N, N]^n$?

[gentle osprey]

Henry

[red yoke]

Since disjoint union is not a very natural construct here

[merry geode]

The part I want clarified is this: Do you understand that the union you wrote earlier is a disjoint union?

[rancid umbra]

yes but why does it matter?

[merry geode]

Ah, yeah that was one of the point of confusion for me

Let's say, G(p, q) := Hom(pi1(X, p), pi1(X, q)).
X x X -> \bigcup_{(p, q)} G(p, q) isn't that natural, considering (p, q) always maps into the unique element of G(p, q).

You can, instead, consider G(p, q) as a Hom set of category, where points like p and q are objects.

[rancid umbra]

why would that be not natural? (p,q) maps to a unique element of G(p,q)

[merry geode]

I mean, it's not like you can have any map from the domain to codomain

[rancid umbra]

oh

[merry geode]

(p, q) will only map to G(p, q)'s element, hence there being a huge restriction

[rancid umbra]

so you instead look at a category where objects are pairs (x,y) in X x X and morphisms are...?

[merry geode]

This kind of stuff does happen with sections as well;
Sections of bundle E over X are like, s: X -> E where X -> E -> X is identity.

[merry geode]

Objects are points of X.

That is, p is an object.

q is also an object.

[rancid umbra]

oh my b

[merry geode]

G(p, q) is the set of morphisms between p and q

[rancid umbra]

but im specifying only one

oh oh

okay

why is it better to view this as a category

[merry geode]

Idk, category is everywhere

Might as well view this one as category as well

Also, fundamental groupoid is basically what you need from eliminating the basepoint.

Like, pi(X) is no longer a group, so you need more general object; which happens to be a category!

[rancid umbra]

and pi1(X) is the category whose objects are points of X and Mor(x,y) = Hom(pi_1(X,x), pi_1(X,y))?

[red yoke]

The fundamental groupoid Π1 is the category whose objects are points and morphisms are homotopy classes of paths

[merry geode]

Hmm wait, let me think, that is not the pi1.

[red yoke]

There is a functor Π1 → Set sending p to π1(X, p) and sending any path γ: p → q to the basepoint change map (x → γxγ^-1)

The statement that the basepoint change map depends only on the endpoints is the statement that each path in Hom_Π1(p, q) is sent to the same element of Hom(π1(X, p), π1(X, q))

This is called a trivial groupoid action

[rancid umbra]

and the groupoid that i know as a group with a partial function replacing the binary operation is the same as the category theory groupoid?

[red yoke]

I'm not familiar with the partial function definition

A groupoid is a category where all morphisms are iso

And there is a morphism between any two objects (connected)

So unlike a group, composition will require domain and codomain to match in some way

[rancid umbra]

e.g., if X is not path connected, you cannot define any map in Hom(pi1(X,x), p1(X,y)) for x,y in different path components

so its only defined for certain pairs of points in X x X

hence a partial function X x X --> X

[red yoke]

I think it should be a partial function {paths in X} × {paths in X} → {paths in X}

The multiplication is path composition

[glossy talon]

What is square metric

[rancid umbra]

oh okay

[red yoke]

And it's partial because you need endpoints to match to talk about composing paths

[rancid umbra]

because A is not always a subset of [-N,N]^n, it is a subset of [-P,P]^n

[unreal gull]

Max distance of any component

[glossy talon]

Sup norm ok

[unreal gull]

wait yeah so there was just no reason to define P at all right

[rancid umbra]

no

[red yoke]

Have you seen the categorical definition of groups and group actions

[rancid umbra]

the bound you have is that p(x,0) <= N + b for all x in A

so by definition, A is a subset of [-P,P]^n

[merry geode]

A morphism need not be a (partial) function.

[rancid umbra]

thats not what i was saying

[rancid umbra]

and yes

well, only groups really, as a groupoid where every morphism is an isomorphism

and there is a morphism between any two points

i can look into those later. gotta gts now

thank you both for your help/discussion 🙂

[glossy talon]

Is it that A is not necessarily a subset of -N,N?

That sounds wrong

Oh yeah thats true

Its because -N,N is centered at origin

[glossy talon]

You got it?

[glossy talon]

Also isnt this heine Borel?

[merry geode]

Is nlab's treatment of derived functors worthy of learning?

[unreal stratus]

Well they have two pages on derived functors @merry geode which do you mean?

Anyway - both are very standard, so yes it is good to learn them, but for an introduction I'd probably just look at Weibel or another homological algebra text

[merry geode]

Well, I mean, it seems like infinite category stuff is common in nlab.

(Tho I do need to learn model category, I think)

[dim perch]

I will say that the nlab page on derived functors in homological algebra seems quite traditional (i.e. not ∞-categorical)

which is not a bad thing

if you wanted an ∞-categorical treatment, I do think that HA's treatment of this is pretty great, although it is unapologetic enough that it'll probably force you to learn ∞-cats if you try to read it

(see HA.1.3.3.2)

[limpid fern]

The point could be very far off the origin, i.e. say we have [5,7] and x0 was picked to be 6. then [-1,1] is not going to cover it

[umbral panther]

First read Gelfand-Manin and then read Gelfand-Manin

[merry geode]

Ah thanks, I don't know how I could not find this page.

[paper wedge]

what is

infinity-category

theory

why is it everywhere

even on meme pages

[umbral panther]

Here @paper wedge

[dim perch]

I will throw my hat in the ring

∞-cat theory begins with the notion of "homotopy coherent algebraic structure;" you begin by noting that it is ""wrong"" to assert just that (based) loop spaces just have a monoid structure "up to homotopy," as this is quite weak and you sometimes would like something better (e.g. as is used in the definition of Dyer-Lashof operations)

you need to remember the homotopy witnessing x * (y * z) ~ (x * y) * z for based loops, and you better end up with only one homotopy witnessing 4-ary associativity, up to homotopy

then you recognize, ok well you need to remember the 4-fold "coherence" homotopy you just constructed, and you need to verify that you haven't created ambiguity on the 5-fold stage, and so on and so forth, each time adding more homotopies for higher arities

the paranthesizations of an n-ary operation are parameterized by Stasheff's nth associahedron, and filling in homotopies into the facades of these polytopes yields a precise model that allows you to define "homotopy-coherently associative" (i.e. A∞-algebras) in ordinary foundations

this is obviously uh

really complicated

and if you're doing this in traditional categorical frameworks (e.g. model categories or homotopical categories), this is just a massive algebraic theory you need to lug around everywhere, and be really careful about constantly

enter an idea: make a new categorical framework, where this sort of "coherently filling in homotopies instead of making diagrams strictly commute" process is just what a commutative diagram/functor means

hence all of your constructions are immediately and easily compatible with A∞-structures; that is just what the word "monoid" means in ∞-cat theory. This staves off a lot of the difficulty of bookkeeping that otherwise made a lot of homotopy theory leading into the mid aughts just intractable to work out and near-impossible to read

Boardman-Vogt had the idea in the 70s to do everything homotopy-coherently, Joyal envisioned it in the early aughts with some foundational work on a particular model for ∞-cats, and Lurie carried us out of the 2008 financial crisis by finally writing enough things down about this model to make it usable

16 years later we're still finding that a lot of things are clarified if you stuff the bookkeeping into foundations of the ∞-cat framework, so much so that it's become pretty solidly ubiquitous in homotopy theory; now technologies which are only really imaginable in the ∞-cat framework (e.g. synthetic spectra) are ubiquitous along the cutting edge of both theoretical and computational work in homotopy theory, since these ∞-cat setups excel at getting themselves and nasty technicalities out of the way and letting you work intuitively

[ebon galleon]

i like the phrasing that Lurie's work fixed the 2008 financial crisis

[quartz horizon]

wow somehow I understand the sentiment of what you wrote

[unreal gull]

is it true that in a Haussdorff space, you can always find disjoint nbhds U, V of points x, y, respectively, such that the intersection of V and the closure of U is empty

[ebon galleon]

Yes. Hint: ||Choose disjoint open neighborhoods U,V of x,y since it's Hausdorff. Observe that the complement of U and V are both closed||

[umbral panther]

This sounds very close to T_3, so I’d guess that a space that is Hausdorff but not T_3 provides a counter example

[unreal gull]

so the closure of X - U is just X - U? (and similarly for V)

[unreal gull]

but how does this show ^

[alpine nest]

The closure of a closed set is itself

[unreal gull]

right

[ebon galleon]

If A is contained in a closed set C, then the closure of A is also contained in C.

Take A = V and C = X-U

[unreal gull]

ahhhh

thank you

[night heron]

Any suggestions of how you’d go about proving that the homological and homotopic degree of a continuous map between two unit circles are equal

[fading vale]

Do you have any tools relating pi_1 to H_1

[merry geode]

Is it good to learn infinity cat even if I would not do strictly algtop research?

It does seem quite involved, and idk if my usual intuitions of categories would apply

[strange cove]

hey, is there a general graphical interpretation for a topological space?

[tame arrow]

As a vertex and edge kind of graph?

[strange cove]

hm, i don't really know what you mean by that, I'm new to (general) topology. I mean if there is a way to create a mental (3d) image of a topological space. Not questioning a particular one.

[plush folio]

I think most people just visualize a space like R² or R³, and if you're working with something like a Hausdorff space then most of your intuition will carry over. But I don't think there's a nice way to visualize a general topological space

[strange cove]

What is a Hausdorff space? How advanced is it?

[golden gust]

it's a topological space which has the condition that, given distinct points of your space, there exist open sets containing these points which don't intersect

[night heron]

Well that’s what I’m asking

[golden gust]

do you know what the abelianisation of pi_1 is?

[night heron]

Oh. Isomorphy of the abelianization group of pi_1 with H_1?

Thx

[strange cove]

Oh ok, thx

When did you guys start studying topology?

[night heron]

2 months ago for point set topology

[golden gust]

second year of university

almost 3 years ago 😵‍💫

[strange cove]

Oh wow, do you study pure math?

[strange cove]

Do you also study at uni?

[golden gust]

I just finished my master's

[unreal stratus]

Topology was what got me interested in studying maths (instead of physics) in the first place aha so I was exposed to a tiny amount in high school by a friend, which I didn't really understand

[unreal stratus]

Well done

[golden gust]

likewise !

[night heron]

Yeah but I do mathematical physics

Doing a PhD program

[golden gust]

I want to learn more math physics

[night heron]

Lmao. I wish it were easy to switch to pure math for that same reason

[unreal stratus]

Oh nice !

[golden gust]

trying to do higher gauge theory this summer but the category theory background is a lot

[unreal stratus]

I swapped from physics undergrad to maff lol

[unreal stratus]

Super cool

[night heron]

Thankfully did both so I can switch with some effort

[night heron]

You got it, m8

[strange cove]

Wow thats crazy. I want to study math myself at university full time after school. I took part in a program, where you can study at uni before graduating school but I got really burned out. Then I stopped and now I'm trying to get back in.

Can you use topology in physics?

[night heron]

Yeah

You kinda use it in QFT and condensed matter

String theory if you’re going there too

But you’ll probably not need it

[gentle osprey]

vernatron

vernatron

[fierce lily]

Can someone give me an example of a set D?

[opaque zodiac]

R^2 minus the origin as a subset of R^3

[umbral panther]

No, you should not automatically learn such an involved thing, unless it is directly applicable. The difficult question is how much you should know to recognize a new discovery that it is applicable

[merry geode]

Ah, I see. Thanks a lot!

I will judge based on where my interest goes.

[rancid umbra]

for maps $f,g:(S^n,s0)\longmapsto (X,x_0)$, define
$f\vee g : S^n\vee S^n\to X$ as $f$ on the first sphere and $g$ on the second sphere. $\newline$

Let $\psi : S^n\to S^n\vee S^n$ be the quotient map resulting from collapsing the equator of $S^n$ to a point, taking the upper hemisphere to the first sphere and the lower hemisphere to the second sphere. $\newline$

Define the concatenation of two maps $f$ and $g$ as $f + g = (f\vee g)\circ\psi : S^n\to X$ $\newline$

How do you go about showing that this is equivalent to the definition of the $n$th homotopy group defined for maps fixing the boundary of $I^n$?

[gentle osprey]

c squared

[scenic ridge]

What is the typical/most common topology on the tensor product of multiple hilbert spaces?

[hearty gazelle]

Can someone explain hopf invariants?

I know about hopf fibrations and homotopy, but I'm not super amazing at topology

[coral pawn]

Is there a nice construction of $\mathbb{G}_m\wedge \mathbb{G}_m$ in the category of (Zariski/Etale/Nisnevich) sheaves valued in spectra (or spaces)?

[gentle osprey]

Finitely Many Bananas

[fading vale]

The one induced by the norm, where the value of the norm on a simple tensor is the product of the norms of each component

[coral pawn]

By nice I mean a construction that can be formed using schemes and categorical constructions (pushouts/pullbacks,fibrations,cofibs)

[golden gust]

do people ever use sheaves valued in (a subcategory of) topological spaces?

[quick bough]

has anyone read donald yau’s book on colored operads? i’m trying to understand the details of a proof when he showed that the category of algebras over the ass-operad is isomorphic to the category of monoids

[umbral panther]

There are lots of places where sheaves should be valued in topological vector spaces, but it is difficult to exploit the structure. The program of condensed sets is intended to attack this

For example, on a complex manifold, coherent sheaves are naturally valued in topological vector spaces. You can use this to prove that on a compact space the cohomology is finitely generated. More generally, there is the Grauert-Remmert theorem that the push forward of a coherent sheaf across a proper map is coherent. This is a difficult theorem, whereas in algebraic geometry it is just the relative version of the previous theorem. I think if you had a good handle on the topological vector spaces, it wouldn’t be any harder

[fierce lily]

I am wondering if it is possible that lim x->0 f(x) does not exist?

[lethal oxide]

a topological pair (X,A), let (X,A) be a HEP then any map from A go to X is a cofibration right?

[red yoke]

It's the inclusion map A → X specifically

[fierce lily]

is there anyone who can answer this question?

[red yoke]

Yes, 1/|x| blows up near 0

[plucky veldt]

so we're treating infinite limit as "limit doesn't exist"

[alpine nest]

Yep, although it's a bit of a gray area since we're allowed to write $\lim_{x\to x_0} f(x) = \infty$. So it sort of exists, but not really

[gentle osprey]

Outsider

[alpine nest]

It definitely doesn't exist in a topological sense.

(Unless something something compactification)

(but even with something something compactification it won't be a limit in R / R^n)

[red yoke]

Well the codomain of f is R so conventionally ∞ and -∞ wouldn't count

[alpine nest]

Yep, I agree, I'm just pointing out that in areas such as calculus and measure theory it's perfectly normal to consider ∞ as valid limits of sequences/functions.

With varying degrees of rigor involved

[red yoke]

Is every such function guaranteed to converge in stone cech compactification of R
Idts

[obtuse meteor]

The best example for a question like this imo is |e^(1/z)| on the complex plane

[alpine nest]

Well that one certainly doesn't have a limit at 0

[lethal oxide]

my question on about question b, how that is possıble? we are saying X contractible if identify a map should be null homotopic, but question just explain a map if identify a function null homotopic then we need two map ?

[red yoke]

I don't follow

Contractible means there is some continuous X × [0, 1] → X that restricts to id: X × {0} → X and X × {1} → {p} ⊂ X

Can you show that such a map exists for Y?

[lethal oxide]

r a retraction then Y x [0,1]--->Y, Y x {0}---->Y and Y x {1}--->{p} and this is subset of X and Y

[red yoke]

Yes, this is what you need to show does exist

[lethal oxide]

thank you

[proud tulip]

The usual topology on Sn is the subspace one right

[opaque zodiac]

Sn being the sphere?

If so, yes

[ebon galleon]

arki honourable

[left eagle]

I am having some trouble with proving the backwards ocnditional (<=) [the stuff circled in red]. I have a full proof however it seems kinda awkward and not very solid. Is there either an easier way to do this or a way to improve my current proof?

[lusty trench]

Let X be an arbitrary set, and let B be a basis for a topology T on X. Then T is the discrete topology on X iff B contains all the singletons in X.

But your basis for the product topology on $\prod \nolimits_{i \in \Lambda} { 0, 1 }$ contains the singletons precisely when $\Lambda$ is finite.

[gentle osprey]

Eduardo León

[left eagle]

wait why is that precisely when lamba is finite?

*why

[lusty trench]

The elements of your given basis are of the form $\prod \nolimits_{i \in \Lambda} U_i$, where each $U_i$ is an open in ${ 0, 1 }$, and only finitely many of the $U_i$'s are proper subsets of ${ 0, 1 }$.

[gentle osprey]

Eduardo León

[left eagle]

yes

how would u prove that if lamda not finite then B doesnt contain the singletons?? that part kinda tripped me up

[lusty trench]

First a trivial remark. The expression $\prod \nolimits_{i \in \Lambda} U_i$ denotes a singleton set if and only if each $U_i$ is a singleton.

[gentle osprey]

Eduardo León

[left eagle]

yep makes sense now ty

damn i didnt think of just doing it that way instead of proivng => and <= individually

[lusty trench]

I hate double implications and double inclusions so much.

I mean, if there's no other way, then I will use them, of course.

But I prefer proving equalities directly, whenever possible.

[left eagle]

Yes i need to start being more lazy. Cuz most of the time im like ugh and then i just do them anyways instead of spending some time trying to avoid it

which ends up with a page long proof and half hour down the drain

😅

damn wtf is my writing...

[lusty trench]

When I took topology, all my take-home test answer sheets were full of correction fluid.

Actually, not just topology.

[left eagle]

I just had to google correction fluid...

🤦‍♂️

[radiant pollen]

I am confused by some exercise I am doing and would like some clarification.
Assume X is an infinite CW complex (i. e. has an infinite number of cells) show that the cone CX cannot be embedded in R^n for any n.

The source of my confusion. Can't we take X to simply be a discrete set? (Infinite number of 0-cells). Then for the cone we simply connect all of these to a new point and easily embed it in R^n?

[red yoke]

How would you embed that in R^n

[lusty trench]

The cone over Z isn't embeddable in R^2?

[red yoke]

There is a continuous injective map into R^n indeed

It's not

[lusty trench]

Oh, I was thinking of using Z x {0} as the image of Z and (0,1) as the vertex of the cone. But then I realized that the segments connecting each (n,0) to (0,1) get very close to each other near (0,1).

Anyway, that cone is itself a CW complex, right? Because we can take first X x I and then collapse the subcomplex X x {1}.

[red yoke]

CZ is a CW complex

[lusty trench]

Then the vertex has infinitely many adjacent cells, in the scenario of the original question.

[lusty trench]

Okay, suppose we have a CW complex Y and an embedding Y -> R^n. Does there necessarily exist a CW decomposition of R^n that makes Y a subcomplex of R^n?

[red yoke]

No

Consider the embedding Z → R, n → 1/n

[lusty trench]

Oh, duh, right.

Okay, let me rephrase then.

Suppose Y is embeddable in R^n. (The embedding isn't given a priori.) Do there necessarily exist an embedding Y -> R^n and a CW decomposition of R^n that makes Y a subcomplex of R^n?

[languid patrol]

To make this question more approachable: I suspect the answer is no. But it comes down to finding a CW complex Y which one can show has an embedding into R^n which is not closed, perhaps it helps to know that one way to show this would be to show that no embedding could be a configuration

*cofibration

[languid patrol]

I will go further and say I bet there is an example which is a graph that embeds in R^2 but has no closed embedding

[lusty trench]

That sounds wild.

What kind of graph could be that?

[languid patrol]

probably something along these lines

https://researchspace.auckland.ac.nz/bitstream/handle/2292/5156/472.pdf?sequence=2&isAllowed=y

I am not going to read the paper carefully enough to discover if by embedding they mean "injective continuous map" or "injective homeomorphism onto the image"

but I hope that at least one of their examples admits the latter

anyway if it doesn't work for a 1-dimensional CW complex I firmly believe there will be a higher dimensional counterexample

[red yoke]

Take the tetrahedron graph and add a ray to each vertex

[languid patrol]

that works

well at least for an injective continuous map

[red yoke]

It embeds

[languid patrol]

ah yes it does

so there is an example

[languid patrol]

^^ there you go

[lusty trench]

Ah, thanks!

[rocky wave]

Say I have a topological space $X$ and a sequence $(x_n)_{n \in \mathbb N}$ in it.
Put $E = { x_n : n \in \mathbb N }$. Say $y \in \overline E \setminus E$. Can I say, then, that $(x_n)$ has a subsequence converging to $y$?

[gentle osprey]

phao

[rocky wave]

For metric spaces, I believe this can be proved with a diagonal argument. Is it true for a general topological space?

[quartz horizon]

I think no

N with cofinite topology maybe?

[paper wedge]

probably true with nets

[quartz horizon]

Yeah

[tender halo]

it is true in Frechet spaces

well its what they are really

not the ones from functional analysis

https://en.wikipedia.org/wiki/Fréchet–Urysohn_space

in particular, all first-countable spaces are frechet

[rocky wave]

Thank you.

[quartz horizon]

I don’t think this works

But im pretty sure this is false in general topological spaces

[tender halo]

there are topological spaces in which no sequences converge except eventually constant ones

(nondiscrete ones, of course, otherwise that would be boring)

for example, SC compactification of N is such a space, in it N is dense but you cant get anywhere with just a sequence of elements of N

[quartz horizon]

Ok nice

[rocky wave]

Thank you all

[fair tusk]

Can 2 loops (or, more generally, any 2 paths) be homotopic if they have "double points?" For instance, below are 2 loops - f0 at basepoint x0 and f1 at x1, except f1 crosses itself at a single point. If f0 and f1 shared the same basepoint, could they be homotopic?
Similarly (and this may answer my question), can loops that intersect themselves be used in the calculation of fundamental groups? I'm almost positive they must be, since calculating the fundamental group of a circle requires that we consider loops that retrace themselves as distinct.

[unreal stratus]

Yes, if you move x0 to x1

[obtuse meteor]

Yes these two things are freely homotopic

[unreal stratus]

Indeed both are homotopic to the constant loop

[fair tusk]

oh ok

[unreal stratus]

if you want loops within D^2

Well, D^2 is contractible

So this will always be the case

[fair tusk]

yeah i kind of figured since the fundamental group of a circle required it

[unreal stratus]

If you want fundamental group of the circle, then uh these aren't even loops in the circle

[fair tusk]

yeah lol just the loops that constitute it would cross themselves

[uncut stratus]

Does anyone know a good book that discuss persistent homology and persistence diagrams?

[languid patrol]

there is no diagonal argument needed, it's true that if $\bar{E} - E$ is nonempty then by definition there is a subsequence converging to $y$ (as long as you allow constant subsequences). Here $E$ converges to $y$ is basically defined as $y$ is a limit point of $E$ which is not in $E$ .

[gentle osprey]

Math_Discord_Final_Girl

[languid patrol]

but even in metric spaces you can have sequences where no subsequence which is not eventually constant converges

[chilly mica]

(you probably understand already, but) the fundamental issue you were grappling with is that, when we say "loop", we mean the function f: [0,1] -> X, with f(0) = f(1) = x_0, we don't mean the image of the function (which a layperson may call a "loop"). nonhomotopic loops may have the same image (you don't expect, in general, for some loop run in reverse to be homotopic to the original, nor the loop run twice over) and when we're manipulating loops around, we're manipulating the function not the image, so cross over points dont restrict anything.

[fair tusk]

Oh, I guess I never thought of it that way, that makes a lot of sense. Does that mean we can compare double points to the function being non-injective?

[quartz horizon]

Yes exactly

Well

[quartz horizon]

Strictly speaking, as a function from [0, 1] it is already not injective

After all $f(0) = f(1)$

[gentle osprey]

Pseudonium

[quartz horizon]

But uh

[fair tusk]

Oh true

[quartz horizon]

$0 \neq 1$

[gentle osprey]

Pseudonium

[fair tusk]

But excluding the basepoint

[quartz horizon]

So maybe a better way to phrase this is

One can use the universal property of the quotient

To convert to a function $\tilde f : S^1 \to X$

[gentle osprey]

Pseudonium

[quartz horizon]

And double points will mean this function is not injective

[fair tusk]

Oh yeah the circle kinda makes that compute

Is that true of the interval if we just exclude the basepoint, though?

[quartz horizon]

Sure, so you can restrict the function to $(0, 1]$ or something

[gentle osprey]

Pseudonium

[fair tusk]

Ah neat

[quartz horizon]

But yeah in general

Curves it makes more sense to think of as functions rather than as just subsets

[lethal oxide]

Hi, how I identify (a,b)? If c is a curvature of a manifold c:(a,b)---->M (M is a manifold).

[gentle osprey]

ТТерра

[lethal oxide]

Okay thank you

Then how can be c a closed curvature?

[low knot]

I have a doubt. We know that the countable union of countable sets is countable. Does this result still hold if we take the arbitrary union of countable sets?

[ebon galleon]

No.

Every set is the union of its singletons, or the union of its finite subsets, or etc.... Now pick your favorite uncountable set like real numbers

[low knot]

Okay. Thank you!

[lusty trench]

Are Stone-Čech compactifications actually useful for anything, or are they just a logical curiosity?

[languid patrol]

Condensed mathematics for instance

But also Stone Cech compactifications are just a natural thing that pop up

[lusty trench]

Where do they pop up?

[languid patrol]

For instance Spec(\times_n F) is the stone Cech compactification of N

if you let the product be over positive integers

Generally the Stone Cech compactification shows up inside of the space of maximal ideals of all kinds of rings of functions (even just entire functions on C)

[tender halo]

well first of all even categorically its an interesting object because all maps into compact spaces factor through it

[lusty trench]

Ah, understandable. Because it's compact and it contains a dense subspace that is the countable disjoint union of Spec(F)'s...

[languid patrol]

Yes also you can explicitly show that ultra filters are the same as maximal ideals

[tender halo]

they are useful when studying how badly something fails to be compact

N fails spectacularly so its SC is a monster

[lusty trench]

Does R fail that badly too?

[tender halo]

dont actually know much about SC of R, sorry

one way to view SC is all the "virtual" points your space has

[languid patrol]

anyway there are lots of examples. I don’t think it’s fair to say they’re “just a logical curiousity” because they’re a very natural part of mathematics. They show up in functional analysis as well due to the natural iso $C_0^{bd}(X) \cong C_0(SC(X))$

[gentle osprey]

Math_Discord_Final_Girl

[tender halo]

all the limits you can take in your space

[languid patrol]

In particular SC(R) contains at least one copy of SC(N) by picking any discrete infinite subset of R

So yeah it’s also a mess

[lusty trench]

Ah, I know no functional analysis. But when a friend who took functional analysis gave me his notes, there was nothing in them that would've depended on the axiom of choice.

[tender halo]

the compactifications that are smaller than SC, then, are basically you taking some of the "virtual" points and gluing them together

[languid patrol]

that’s not possible

Hahn Banach is one of the most fundamental tools in functional analysis and is equivalent to some version of choice

Or no what’s it called

Hahn Banach

[lusty trench]

Makes sense.

[alpine nest]

Personally I've never had any use for the Stone-Cech compactification ||because my research only involved spaces which were compact to begin with||

[tender halo]

note that it is also true for all non-pseudocompact hausdorff spaces

uhh maybe you need like being Tychonoff instead of hausdorff

a (Tychonoff) space is not pseudocompact iff it has a C-embedded copy of N

(C embedded meaning that you can find a continuous extension of any continous function defined on the subset)

[tender halo]

a good exercise to see the "naturalness" of SC is

pick a dense countable subset in your favorite compact space (works funnier if the space is not first-countable), fix an enumeration of it

then for every ultrafilter on N there is a prime filter corresponding to it in your space, i.e. there is a unique limit for that filter

for easy visualisation you can do it with [0; 1], you divide the space in half each time, only one half of the dense subset lies in the ultrafilter

so ultrafilters on N are kinda like your maps to find limits in a compact space

[ocean narwhal]

when talking about compactifications, i always assume the space is tychonoff anyways

[tender halo]

oh yeah for sure

i meant this part

I think being hausdorff is enough

actually

yea it should be enough

[umbral panther]

It’s not standard, but it is possible. Hahn-Banach is a crutch, not fundamental. No useful application of functional analysis depends on choice

[lusty trench]

Okay, no, my bad. My friend's notes do mention Hahn-Banach.

[lusty trench]

How is it a crutch?

[umbral panther]

How is it useful?

[lusty trench]

No idea. I don't know functional analysis, as I said before.

[quartz horizon]

wait how do you do Hahn-banach without choice

[alpine nest]

I think they're saying Hahn-Banach is unnecessary

[umbral panther]

It’s equivalent to choice

[quartz horizon]

oh

[quartz horizon]

huh…

[opaque scroll]

It's weaker than choice

[quartz horizon]

i remember using it a bunch

it’s very useful for getting elements of the dual space

[alpine nest]

Well, the statement was that it has no "useful" applications, and I am not going to tread on the extremely thin ice of arguing which parts of mathematics are useful

[opaque scroll]

So I guess the way to do it without choice is to take it as your axiom instead 😛

[alpine nest]

(the answer is "almost none of it")

[quartz horizon]

i mean…

[red yoke]

Fun is useful and usefulness is transitive

[quartz horizon]

apparently functional analysis is what you need to make QM rigorous or smth

[alpine nest]

Absolutely, I'm not saying mathematics shouldn't be done, I just prefer to think of it as a form of abstract art.

With the value being mostly aesthetic

[quartz horizon]

yeah ig we have different perspectives on it

for me maths is a tool

so i like it when it’s useful

even if that use is in other maths

that’s still fine

[umbral panther]

Maybe you need Hahn-Banach to prove the existence of dual vector spaces. But in practice you can just write down the duals

[quartz horizon]

wdym

[tender halo]

yeah i like my vector spaces to have bases sorry

[alpine nest]

I don't really need Hahn-Banach or Axiom of Choice, and I happily reject them in favour of the Zorn Lemma

[lusty trench]

I guess Hahn-Banach ensures that the continuous dual of a nonzero vector space is again nonzero.

[quartz horizon]

yeah that seems pretty important lol

[lusty trench]

Anyway, I'm much more willing to admit Stone-Čech compactifications (which I originally called "a logical curiosity") are useful than this.

[rain mason]

I thought Hahn-Banach was weaker than choice?

[umbral panther]

Apparently HB is equivalent to Stone-Czech, which is weaker than AC

[rain mason]

Yeah

The only reason why I know this is that we have a question "what is yellow and equivalent to the axiom of choice?", for which the answer is "zorn's lemon", but me and my friend were trying to get "Hahn-Bananach" to work lol

[red yoke]

@topaz plover

[alpine nest]

At my uni Hahn-Banach was known as the Hanna-Barbera theorem

(makes more sense in Polish because it's "Twierdzenie Hahna-Banacha", so the rhythm fits)

[languid patrol]

So you assume both not AOC and Zorn’s lemma before proving anything? You must be a very successful mathematician

[languid patrol]

for hilbert spaces you don’t need Hahn Banach, you can actually construct a separating hyperplane by hand

[quartz horizon]

Oh huh

[alpine nest]

like you wouldn't believe

lets me deduce anything

[alpine nest]

Even in non-separable ones?

(I'm not questioning you, just curious)

[opaque zodiac]

This is probably going to depend on the precise statement

What are you separating? Convex sets or strictly convex sets?

Are they compact?

[alpine nest]

With the "extending continuous functionals from subspaces" version you can probably just explicitly construct the extension by referring to the orthogonal complement.

Unless that relies on AOC in a subtle way

[opaque zodiac]

It only does if you define 'Hilbert space' in a slightly dumb way

Specifically completions of metric spaces are slightly subtle without countable choice. It's not good enough to use sequences

You have to use something like Cauchy filters

But if you use Cauchy filters in the definition then orthocomplements work fine

[lusty trench]

Cauchy filters?

[opaque zodiac]

Uh yeah so recall that a filter on a space $X$ is a family $F$ of (let's say non-empty) subsets of $X$ that is closed upwards and closed under intersections. Given a metric space $(X,d)$, a \emph{Cauchy filter} is a filter $F$ such that for every $\varepsilon > 0$, there is an $A \in F$ with diameter less than $\varepsilon$.

[gentle osprey]

Exomnium (MSC2020 03C66)

[opaque zodiac]

If you're trying to do metric space stuff without countable choice for some reason, then the right way to define completions of spaces is in terms of Cauchy filters instead of Cauchy sequences

Constructively they're also a lot better behaved

[lusty trench]

You'll take dependent choice from my cold dead hands. And dependent choice implies countable choice.

[opaque zodiac]

sure

Yeah live without dependent choice is rough

[lusty trench]

Dependent choice says “nondeterministic algorithms either terminate or don't”, which is so obviously intuitive to me.

[opaque zodiac]

(although I would also say life without BPI is rough)

[ebon galleon]

What's BPI?

[opaque zodiac]

Boolean prime ideal theorem

equivalent to Hahn-Banach iirc

[ebon galleon]

Oh yeah that's tough

[lusty trench]

What does BPI buy you?

[opaque zodiac]

Compactness theorem in model theory

Hahn-Banach

existence of prime ideals in rings

ultrafilters

[lusty trench]

Is BPI stronger than dependent choice?

[opaque zodiac]

they're incomparable

[unreal gull]

cant you take the set of basis of the form [a, b), for rationals a and b to be a countable basis of R_l

also [a, infty) and (-infty, b)

[opaque zodiac]

How do you write $[\pi,\pi+1)$ as a union of open sets of that form?

[gentle osprey]

Exomnium (MSC2020 03C66)

[plucky veldt]

no because then there won't be a basis member contained in [x, x + 1) for irrational points

[heady bolt]

does this proof of thm 17.8.1 really hold for the case where the base point is fixed?

some assumption might be missed, like what wikipedia tells us for Leray-Hirsch the restriction on each fiber to the basis for a cohomology

[lethal oxide]

Let D¹ be a disk and D¹={1} then d)D¹={1}. My question let X be topological space and p:d)D¹---->X a attacking map right? And d)D¹=1 meaning 1~p(1) a cell right?

[ebon galleon]

I absolutely cannot understand what you are trying to ask.

Asking in your original language or asking a more carefully translated question would be more helpful.

[ebon galleon]

(arki what does this mean?)

(Oh okay, that clarifies a lot )

[gentle girder]

they’re learning about CW complexes/decompositions and attaching maps

[lethal oxide]

This question arose due to definition I dont think i understand to attacking map i need a disk and boundary of disk also this map from boundary of disk go to any topological space this is definition of attacking map right? And we saying image on topological space of boundary of disk with boundary of disk relation called a cell the definition and my understanding the same thing?

[gentle girder]

is the definition that of a CW complex?

[lethal oxide]

I have only seen attacking map and attacking n-cell at the moment and is the definition i understand the same as the correct definition?

[gentle girder]

when you attach an $n$-cell to $X$, you take a $D^n$, you take a map $\phi$ from $S^{n-1}=\partial D^n\to X$ and you get a new space by identifying each point $x$ in $\partial D$ with $\phi(x)$ in $X$. $\phi$ is the attaching map here

[gentle osprey]

australian pizzeria

[lethal oxide]

thank you I understand now @gentle girder

[ebon galleon]

Ohhh I just realized it's a p:(an)d(a) attacking map

[unreal gull]

in the first part of the response to this question https://math.stackexchange.com/questions/3652459/prob-11-sec-30-in-munkres-topology-2nd-ed-a-continuous-image-of-a-lindelo, $f(X) = f(\overline{D}) \subseteq \overline{f(D)}$, but don't we have to show containment in the other direction? dont we need to show $\overline{f(D)} \subseteq f(X)$?

[gentle osprey]

Henry

[gentle osprey]

(Т*Терра, dqⁱ ∧ dpᵢ)

[unreal gull]

uhhhh

but can't Cl_Y(Y) not be in Y?

[full merlin]

From my understanding, it is generally easy to determine the homology of a finite cw-complex.
Is it also generally easy to determine the homotopy type?

Also, is there such a thing as a classification of homotopy types for, say, manifolds?

[languid patrol]

Here is a finite CW complex that is very difficult to understand precisely in the homotopy category:

let \Delta^n be an n simplex

Now remove the inside so you’re only left with the n-1 dimensional faces

[languid patrol]

no not really, being homotopy equivalent to a compact manifold puts some mild restrictions on the homotopy type but not many.

[full merlin]

thanks

[umbral panther]

https://en.m.wikipedia.org/wiki/Surgery_theory

[gaunt linden]

That's another way to say (n-1)-sphere, right?

[unreal gull]

what does it mean for RP^n to be "covered" by n + 1 affine charts?

[red yoke]

It is a union of affine subspaces

[unreal gull]

uh

i dont know what those are either

all ive seen in the text so far are principle affine charts

[lusty trench]

When you construct $\mathbb{RP}^n$, you use $(n+1)$ copies of $\mathbb R^n$, where the $i$-th copy $U_i$ has coordinates $x_0, \dots, \widehat{x_i}, \dots, x_n$. Letting $U_{ij}$ be the open subset of $U_i$ where $x_j \ne 0$, you glue $U_i$ and $U_j$ with the transition functions $\varphi_{ij} : U_{ij} \to U_{ji}$, $\varphi_{ij}(x_0, \dots, \widehat{x_i}, \dots, x_n) = (x_0 / x_j, \dots, x_{i-1} / x_j, 1 / x_j, x_{i+1} / x_j, \dots, \widehat{x_j / x_j}, \dots, x_n /x_j)$ for each $i < j$.
But this construction works perfectly fine if you consider the $U_i$'s as a copies of $\mathbb C^n$ rather than $\mathbb R^n$. And then you get $\mathbb {CP}^n$.

[gentle osprey]

Eduardo León

[unreal gull]

what does the hat mean?

[full merlin]

Question: Let X be a (finite) regular CW-complex and f: X -> X be cellular such that f fixes the centers of each cell. Is f necessarily homotopic to id ?

[umbral panther]

Degree -1 map of the circle to itself

[full merlin]

But if you use 1 1-cell, its not regular. If you use 2, its not fixing centers.

(I think)

[umbral panther]

What is the definition of regular? The map on the boundary of a cell is injective?

[full merlin]

yes

[umbral panther]

If you don’t require the boundary to go into the lower skeleton then you can repeat the S^1 example where the two ends go into the interior of D^1

[full merlin]

hmm okay ^^. What if we require the ends to go into the lower skeleton? Do you expect the statement to be true?

[umbral panther]

Have you tried proving it by induction?
Have you tried proving it for simplicial sets?
Chain complexes?

[candid wyvern]

is there a way to uniquely define each knot up to ambient isotopy

eg assign each knot to something such that any two knots assigned to the same thing are the same knot up to isotopy

[umbral panther]

Yes, I believe it follows formally from the decidability of the knot equivalence problem that there is a complete invariant. A standard open problem is whether the collection of all finite type variants distinguishes knots. Better, a well-defined truncation so that it can be finitely specified

[candid wyvern]

so, there exist a complete invariant? but we do not knot of it yet?

sorry i didnt really understand your response

[umbral panther]

From an algorithm, you can extract an invariant. But it is ugly, just a statement of what the algorithm does to that particular knot

[candid wyvern]

oh so there exist a way but no way to nicely and simply classify knots?

that we know of

[umbral panther]

Write some description of knots. Say, knot diagrams. Order by simplicity and within that find some order, some kind of lex order on descriptions. Then a complete invariant of a knot is the first description of that knot in the list. No guarantee of computation, but it’s a complete invariant. If you can compute knot equivalence, you can compute this invariant

[candid wyvern]

lex order?

[umbral panther]

Lexicographic

[candid wyvern]

does this include knots that are made of multiply components? or is there a nice way to classify knots of only 1 component?

[left eagle]

Im stuck, not really sure what to do with an infinite intersection..?

[lusty trench]

Your notation is a bit inconvenient, so please allow me to change it a bit.

Let $X = \prod \nolimits_{i \in \Lambda} X_i$, let $\pi_i : X \to X_i$ be the $i$-th canonical projection, and let $f : Y \to X$ be an arbitrary function. By definition, $X$ has the smallest topology such that every $\pi_i$ is continuous. Thus, in particular, if $f$ is continuous, so is every $f_i = \pi_i \circ f : Y \to X_i$.

Now, conversely, suppose every $f_i$ is continuous. To show that $f$ is continuous, it suffices to check that, when $U$ is a \underline{basic} open subset of $X$, its preimage $V = f^{-1}(U)$ is open in $Y$. Write $U = \prod \nolimits_{i \in \Lambda} U_i$, where finitely many of the $U_i$'s are proper open subsets of the respective $X_i$'s, and the rest are equal to the respective $X_i$'s. Then $V = \bigcap_{i \in \Lambda} V_i$, where $V_i = f^{-1}(U_i)$. Notice that finitely many of the $V_i$'s are proper open subsets of $Y$, and the rest are equal to $Y$. Therefore, $V$ is open in $Y$.

[gentle osprey]

Eduardo León

[lusty trench]

ERRR... I'll rewrite the second paragraph.

Now, conversely, suppose every $f_i$ is continuous. To show that $f$ is continuous, it suffices to check that, when $U$ is a \underline{basic} open subset of $X$, its preimage $f^{-1}(U)$ is open in $Y$. Write $U = \prod \nolimits_{i \in \Gamma} U_i \times \prod \nolimits_{i \in \Lambda \setminus \Gamma} X_i$, where $\Gamma$ is a finite subset of $\Lambda$, and each $U_i$ is an open subset of $X_i$. Then $f^{-1}(U) = \bigcap \nolimits_{i \in \Gamma} f^{-1}(U_i)$ is a finite intersection of open subsets of $V$, and therefore is itself open in $V$.

[gentle osprey]

Eduardo León

[lusty trench]

Argh, typo. Should've been $f^{-1}(U) = \bigcap_{i \in \Gamma} f_i^{-1}(U_i)$.

[gentle osprey]

Eduardo León

[night heron]

Why are you adding a subscript for the inverse function

[scenic ridge]

All norm induced topologys on vector spaces are a Topological Vector space right? So is the tensor product of multiple Hilbert spaces a Topological Vector space under the norm induced topology? And does this apply when some of those hilbert spaces are not finite dimensional

[lusty trench]

Strictly speaking, what you said isn't correct. The topology of a topological vector space isn't the same thing as the topological vector space itself.

But it's true that a norm on a vector space induces a topology. So every normed vector space is a topological vector space.

The tensor product of Hilbert spaces is again a Hilbert space, and in particular a normed vector space, so the previous remark applies.

[candid wyvern]

could someone explain to me about how the slice knot of unknot bounds a disk in S3

[scenic ridge]

wait im confused? wdym

[scenic ridge]

and this applies even if they are infinite dimensional correct?

[lusty trench]

Sure. But be mindful that, in the infinite-dimensional case, you can't just take the tensor product of the underlying abstract vector spaces.

[gritty widget]

I'm going to get inside an algebraic topology course, any advice?

[gentle girder]

try the door, or maybe just homotope a window to be way too big for them to stop you

[gritty widget]

Im confused about this description for a set X with topology tau. The author is describing X as the union of sets belonging to tau. But we could have X={a,b,c} with topology {a},{b},{a,b} the union of which would give us {a,b}. Of course X itself is a part of the topology but then we're describing X by itself

[red yoke]

I think they're defining X to be the union

[hidden crag]

it is kind of weird i agree

[gritty widget]

I'll just stick to Munkres

[lusty trench]

Have fun, of course. What point is there to studying math if you aren't having fun?

[gritty widget]

The price you are gonna pay in Spain if you repeat that course

[limber ravine]

The hemisphere of a sphere and a cylinder are homemorphic right?

making the cylinder compact

[unreal stratus]

What do you mean by cylinder

[limber ravine]

fuck

I meant

cone

[unreal stratus]

Oh

[limber ravine]

sorry

[unreal stratus]

Like an infinite cone or what?

[limber ravine]

make it compact..

[unreal stratus]

Or like S^1 x I / S^1 x {1} ig

[limber ravine]

like an ice cream cone

[unreal stratus]

Sure yes

Then yes, that is homeomorphic to a hemisphere

Both are homeomorphic to a closed disc

(By projection)

[limber ravine]

indeed

thanks

[unreal stratus]

Np

[unreal gull]

does a space X which has a countable basis mean that it is first or second countable?

i see

i was confused cuz munkres defines countable basis at a point as first countable

[unreal gull]

why isnt a compact space always second countable?

isnt compactness stronger than second countability?

[opaque zodiac]

So 2^kappa is compact for any kappa, but it's second countable iff kappa is countable or finite

[umbral panther]

If compact doesn’t imply finite, why should it imply second countable?

[unreal gull]

what is 2^kappa?

[opaque zodiac]

Product of kappa many two-element spaces

Each with the discrete topology

[unreal gull]

oh wait compact means every open covering has a finite subcovering but second countable has to do with a countable basis...

i kinda confused an open covering to be a basis mb

[unreal gull]

wait if u take an open covering of all possible 1 element open sets, then isn't it not compact?

[opaque zodiac]

There are no one element open sets in that space

[unreal gull]

oh but in the product topology only finitely many terms can be != 2

[opaque zodiac]

No this is the full product

[unreal gull]

but i thought it was the discrete topology?

[opaque zodiac]

The factors are discrete but the product space isn't

[unreal gull]

yeah so if 2 = {1, 2} then the open sets of 2 are 0, {1}, {2}, {1, 2} right

so open sets of 2^kappa can look like {1} x {1} x {2} x {1} x ....

[opaque zodiac]

No so in the product a basic open set has to have all but finitely many factors be {1,2}

What you're describing is called the 'box topology'

[unreal gull]

oh yeah

[opaque zodiac]

It's finer than the product topology

[unreal gull]

thats what i meant

what do u mean by the "full product"? Is that different than the product topology?

[opaque zodiac]

I mean that we don't have this restriction

[unreal gull]

doesnt that mean the same thing?

[unreal gull]

this just means there are finitely many parts in the product that are \emptyset, {1}, or {2}

[opaque zodiac]

Yes

Although note not all open sets are of this form. This is just the basic opens

[unreal gull]

ty

[rancid umbra]

if you attach a disk to the torus so that it fills in the hole in the middle, the fundamental group of that space should be Z^2/(Z,0) = Z, correct?

[red yoke]

Yea

[rancid umbra]

sweet

[unreal gull]

https://math.stackexchange.com/questions/3619691/show-that-a-regular-lindelöf-space-is-metrizable-if-it-is-locally-metrizable in this question, why does \overline{V} have a countable basis? that doesnt follow directly from \overline{V} being Lindelof I believe...

[rancid umbra]

and if i attach a line across the gap , then it should be (Z^2 * Z * Z) / Z = Z * Z * Z?

[red yoke]

What's a line across the gap

[rancid umbra]

like a pretzel bridge across the donut hole

a cross section would look like a theta

[red yoke]

Any line attachment should be Z² ⊔ Z

[rancid umbra]

that has a group structure?

[red yoke]

Coproduct

[rancid umbra]

oh

hmm

[quartz horizon]

Coproduct mentioned👀

[rancid umbra]

okay, my * was free product

[red yoke]

||Cuz it's homotopy equivalent to wedge sum with S¹||

[rancid umbra]

hmm

oh

okay

[rancid umbra]

i canceled the wrong factor of Z here

[rancid umbra]

wedge of a torus with S^1?

[red yoke]

Yea

You can take a path on the torus connecting endpoints of the line