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Stream: theory: algebraic topology

Topic: homotopies for simplicial sets

Leopold Schlicht (Oct 13 2021 at 16:01):

Let $X$ be a topological space and $S_{\bullet}(X)$ its singular simplicial set. The set $S_0(X)$ can be identified with the set of points of $X$ and $S_1(X)$ can be identified with the set of paths in $X$. Let $\sigma,\sigma'\in S_1(X)$ be two loops based at $x\in X$. Why is a homotopy from $\sigma$ to $\sigma'$ the same thing as a 2-simplex $\tau\in S_2(X)$ with the faces $d_0(\tau)=s_0(x)$, $d_1(\tau)=\sigma$, and $d_2(\tau)=\sigma'$?

Unwinding the definitions, $s_0(x)$ is the map on the line that is constant $x$. Hence such a $\tau$ is given by a function $f\colon \{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\to X$ such that $f(0,-)=\sigma$, $f(-,1)=\sigma'$, and $f(x,x)=x$. Hence (I think) it is a homotopy from $\sigma'\circ \sigma$ to the constant path (is it?). But why is this the same as a homotopy from $\sigma$ to $\sigma'$?

John Baez (Oct 13 2021 at 16:15):

Leopold Schlicht said:

Let $X$ be a topological space and $S_{\bullet}(X)$ its singular simplicial set. The set $S_0(X)$ can be identified with the set of points of $X$ and $S_1(X)$ can be identified with the set of paths in $X$. Let $\sigma,\sigma'\in S_1(X)$ be two loops based at $x\in X$. Why is a homotopy from $\sigma$ to $\sigma'$ the same thing as a 2-simplex $\tau\in S_2(X)$ with the faces $d_0(\tau)=s_0(x)$, $d_1(\tau)=\sigma$, and $d_2(\tau)=\sigma'$?

Calling them "the same thing" is an exaggeration. But you can turn any such 2-simplex into a homotopy from $\sigma$ to $\sigma'$, and you can also turn any such homotopy into such a 2-simplex. (I'm not going to claim there's a unique way, but there's probably some widely used standard way.)

John Baez (Oct 13 2021 at 16:17):

Unwinding the definitions, $s_0(x)$ is the map on the line that is constant $x$. Hence such a $\tau$ is given by a function $f\colon \{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\to X$ such that $f(0,-)=\sigma$, $f(-,1)=\sigma'$, and $f(x,x)=x$. Hence (I think) it is a homotopy from $\sigma'\circ \sigma$ to the constant path (is it?). But why is this the same as a homotopy from $\sigma$ to $\sigma'$?

A homotopy from $\sigma'\circ \sigma$ to the constant path gives you a homotopy from $\sigma$ to $(\sigma')^{-1}$. So, I think you've made some orientation mistake, which is really easy to do in this game.

Reid Barton (Oct 13 2021 at 16:22):

Leopold Schlicht said:

Unwinding the definitions, $s_0(x)$ is the map on the line that is constant $x$. Hence such a $\tau$ is given by a function $f\colon \{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\to X$ such that $f(0,-)=\sigma$, $f(-,1)=\sigma'$, and $f(x,x)=x$. Hence (I think) it is a homotopy from $\sigma'\circ \sigma$ to the constant path (is it?). But why is this the same as a homotopy from $\sigma$ to $\sigma'$?

These should be $f(0,-) = \sigma'$, $f(t, t) = \sigma(t)$, $f(t, 1) = x$. (I renamed your coordinates on $|\Delta^2|$ because you already have something else called $x$.)

John Baez (Oct 13 2021 at 16:22):

Draw a triangle with vertices 0, 1, 2 - that's your 2-simplex $\tau$.

Draw a $\sigma$ on the edge from 0 to 2 - that's the edge not containing 1, so it's $d_1(\tau)$.

Draw a $\sigma'$ on the edge from 0 to 1 - that's the edge not containing 2, so it's $d_2(\tau)$.

For your triangle to describe a homotopy from $\sigma$ to $\sigma'$, the edge from 1 to 2 should be a constant path.

Reid Barton (Oct 13 2021 at 16:23):

and in these coordinates 0 is at (0, 0), 1 is at (1, 0), 2 is at (1, 1)

Daniel Teixeira (Oct 13 2021 at 18:47):

I think a sequence of homeomorphisms $|\Delta^2| \cong D^2\cong [0,1]\times [0,1]$ like in the picture below should do the trick

image.png

Daniel Teixeira (Oct 13 2021 at 18:48):

You can also roundabout this through basic quasicategorical babble. As John was saying, a homotopy from f to g corresponds to a 2-simplex H that looks like this:
image.png

Your $\tau$ however witnesses that $s_0(x)$ serves as a composite for $\sigma'\circ\sigma$. Another possibility is the path concatenation $\sigma'\star\sigma$, but any such composites are homotopic (in the quasicategorical sense). Then they're homotopic, but quasicategorical homotopy <-> topological homotopy in $Sing(X)$.

Leopold Schlicht (Oct 14 2021 at 15:33):

Thanks all! But I still don't get it.

Daniel Teixeira said:

As John was saying, a homotopy from f to g corresponds to a 2-simplex H that looks like this:
image.png

That's precisely the question I'm asking. Why is a homotopy from $\sigma$ to $\sigma'$ the same as a triangle with edges $\sigma$, $\sigma'$, and $s_0(x)$?

Your $\tau$ however witnesses that $s_0(x)$ serves as a composite for $\sigma'\circ\sigma$. Another possibility is the path concatenation $\sigma'\star\sigma$, but any such composites are homotopic (in the quasicategorical sense). Then they're homotopic, but quasicategorical homotopy <-> topological homotopy in $Sing(X)$.

Unfortunately that's too handwavy for me. :sweat_smile:

Leopold Schlicht (Oct 14 2021 at 15:50):

Daniel Teixeira said:

I think a sequence of homeomorphisms $|\Delta^2| \cong D^2\cong [0,1]\times [0,1]$ like in the picture below should do the trick

image.png

Of course intuitively that makes sense, but I don't know how to translate this into mathematics: how to formally write down the bijection between the "triangle homotopies" $\{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\to X$ and "square homotopies" $[0,1]\times[0,1]\to X$? I'm pretty sure it isn't enough to just precompose with any homeomorphism $\{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\cong [0,1]\times [0,1]$, as you seem to suggest.

Reid Barton (Oct 14 2021 at 15:56):

You should precompose with a map $F : [0,1] \times [0,1] \to \{\,(x,y) \mid 0 \le y \le x \le 1\,\}$ that satisfies the right boundary conditions: $F(0,u) = (0, 0)$ for all $u$, $F(1, u)$ lies on the edge $(1, *)$ for all $u$, $F(t, 0) = (t, 0)$, $F(t, 1) = (t, t)$. For example, $F(t, u) = (t, tu)$.

Leopold Schlicht (Oct 14 2021 at 16:45):

This isn't even a homeomorphism! :grinning_face_with_smiling_eyes: Thanks!
Does this yield a bijection between "triangle homotopies" and "square homotopies" or something a bit weaker?

John Baez (Oct 14 2021 at 19:19):

By the way, when you're a topologist you look at this:

John Baez (Oct 14 2021 at 19:21):

and you see the top left edge is the path $\sigma$, while the bottom edge is the path $\sigma'$, and the right edge is a constant path...

... so you know that if you slide the top left edge down to the bottom edge - keeping its left endpoint fixed while sliding its right endpoint down the constant path $s_0(x)$ - you get a homotopy from $\sigma$ to $\sigma'$.

John Baez (Oct 14 2021 at 19:22):

You don't think consciously about formulas for maps from the square to the triangle. You just think "here is a way to morph the path $\sigma$ to the path $\sigma'$ while keeping both its endpoints fixed - so, a homotopy!"

John Baez (Oct 14 2021 at 19:24):

When you're getting started in topology you write down a bunch of formulas for maps, but this eventually becomes too slow: it's enough to know you could.

Leopold Schlicht (Oct 16 2021 at 17:24):

How would one formulate precisely the statement that there is a "bijection" or "correspondence" between "triangle homotopies" $\{(x,y)\in\mathbb R^1\mid 0\leq x\leq 1\text{ and }y\leq x\}\to X$ and "square homotopies" $[0,1]\times[0,1]\to X$? The map that sends each triangle homotopy $\tau$ to $F\circ \tau$ is not surjective I guess, since $F$ is not injective.

Reid Barton (Oct 17 2021 at 11:35):

The "triangle homotopies" and "square homotopies" aren't just maps of those types, they have to satisfy certain boundary conditions. In particular, they have to be constant on certain subsets of the triangle/square (one edge of the triangle, and two opposite edges of the square); that means they correspond uniquely to maps out of the quotients by those subsets. The map $F$ induces a homeomorphism of those quotients, so the original "triangle homotopies" and "square homotopies" are in fact in bijection.

Reid Barton (Oct 17 2021 at 11:36):

However, it's not necessarily a great idea to attach too much significance to this specific bijection because the map $F$ was kind of arbitrarily chosen. The main point is that two paths are "triangularly homotopic" if and only if they are "squarely homotopic".

Leopold Schlicht (Oct 18 2021 at 14:23):

Thanks, that's very helpful!

Leopold Schlicht (Oct 18 2021 at 18:10):

Reid Barton said:

The "triangle homotopies" and "square homotopies" aren't just maps of those types, they have to satisfy certain boundary conditions. In particular, they have to be constant on certain subsets of the triangle/square (one edge of the triangle, and two opposite edges of the square) [...]

Actually, why does a homotopy $H\colon [0,1]\times [0,1]\to X$ from a loop $\sigma$ to a loop $\sigma'$ need to be constant on two opposite edges of the square? The only conditions I am aware of are $H(-,0)=\sigma$ and $H(-,1)=\sigma'$. There seem to be no conditions on the edges $(0,-)$ and $(1,-)$.

Reid Barton (Oct 18 2021 at 18:14):

Good question!

Reid Barton (Oct 18 2021 at 18:14):

This is what people mean when they talk about a homotopy between paths, even if sometimes they forget to say it.

Patrick Nicodemus (Oct 18 2021 at 18:14):

Leopold Schlicht said:

Actually, why does a homotopy $H\colon [0,1]\times [0,1]\to X$ from a loop $\sigma$ to a loop $\sigma'$ need to be constant on two opposite edges of the square? The only considtions I am aware of are $H(-,0)=\sigma$ and $H(-,1)=\sigma'$. There seem to be no conditions on the edges $(0,-)$ and $(1,-)$.

The definition you're considering is that of a general homotopy. There is a notion of path homotopy with fixed endpoints. The fundamental group would be totally uninteresting if we used the standard definition of homotopy rather than the notion of path homotopy with fixed endpoints.

James Deikun (Oct 18 2021 at 18:15):

I don't think they really have to be constant, but they definitely have to be equal to each other all the way down, otherwise you could take apart one loop, drag it along itself and the other, and join it, making all loops homotopic to each other!

Reid Barton (Oct 18 2021 at 18:18):

For the purpose of defining the fundamental group you really do want them to be constant--if you just ask for a homotopy through loops, the basepoint could trace out a nontrivial loop $\gamma$ and then you end up identifying two loops that are conjugates by $\gamma$.

Reid Barton (Oct 18 2021 at 18:19):

Because if you go around the square one way you have $\sigma$ then $\gamma$, but the other way you have $\gamma$ then $\sigma'$.

James Deikun (Oct 18 2021 at 18:21):

... and in any case fixed endpoints is what you need for the fundamental groupoid, and if you obtain the fundamental group from the fundamental groupoid it's what you get.

Leopold Schlicht (Oct 19 2021 at 10:13):

Ah, I see, thanks so much! Hence in this case the right wikipedia page to look at is Path (topology) and not Homotopy. :grinning_face_with_smiling_eyes: