# The Unapologetic Mathematician

## Homotopies as Morphisms

We can think of homotopies between maps as morphisms in a category that has the maps as objects. In terms of the movie analogy, the composition is obvious: run the movie that takes you from map $f$ to map $g$, then run the one that takes you from map $g$ to map $h$.

In practice, the way we make these intuitive concepts explicit tend to get in the way. For one thing, the naïve approach would be to run the first movie from time $0$ to time $1$, and then the second from time $1$ to time $2$. But this gives us a function $H:M\times[0,2]\to N$ instead of $H:M\times[0,1]\to N$. The usual way to handle this is by rescaling — run the first movie twice as fast from $t=0$ to $t=\frac{1}{2}$, then the second movie twice as fast from $t=\frac{1}{2}$ to $t=1$.

The problem with this is that it makes associativity weird. Let’s say we have homotopies $F:f\to g$, $G:g\to h$, and $H:h\to k$ we want to compose. If we write

$\displaystyle [G*F](p,t)=\left\{\begin{array}{lr}F(p,2t)&0

Then we have

$\displaystyle [H*(G*F)](p,t)=\left\{\begin{array}{lr}F(p,4t)&0

and

$\displaystyle [(H*G)*F](p,t)=\left\{\begin{array}{lr}F(p,2t)&0

these two are indeed homotopies from $f$ to $k$, but they’re not the same homotopy! Associativity doesn’t seem to work for this composition.

The easy answer is to wave our hands and say they’re the same “up to reparameterization”. That is, there is some (invertible) function $r:[0,1]\to[0,1]$ so that

$\displaystyle[H*(G*F)](p,t)=[(H*G)*F](p,r(t))$

It’s not hard to figure it out as an exercise.

The fact that we’re talking about two different things being “really the same” is a clue that there’s some higher categorical structure here that we’re “decategorifying” and forgetting about. In particular, we could flesh out the idea of reparameterizations as morphisms between homotopies, but that will quickly become more complicated than I want to get into.

Still, it’s worth pointing out that the reparameterization in the above exercise behaves like an associator, like we talked about in the context of monoidal categories. And, just like in that case, we will find left and right identity reparameterizations.

What’s the obvious homotopy to use as the identity on a map $f$? Clearly it’s just $I_f(p,t)=f(p)$, independent of $t$. I’ll leave the identity reparameterizations as another exercise. The upshot is that we have identity homotopies — “up to reparameterization” — for each map, which completes the definition of our category.

November 29, 2011

## Homotopy

The common layman’s definition of topology generally involves rubber sheets or clay, with the idea that things are “the same” if they can be stretched, squeezed, or bent from one shape into the other. But the notions of topological equivalence we’ve been using up until now don’t really match up to this picture. Homeomorphism — or diffeomorphism, for differentiable manifolds — is about having continuous maps in either direction, but there’s nothing at all to correspond to the whole stretching and squeezing idea.

Instead, we have homotopy. But instead of saying that spaces are homotopic, we say that two maps $f_0,f_1:M\to N$ are homotopic if the one can be “stretched and squeezed” into the other. And since this stretching and squeezing is a process to take place over time, we will view it sort of like a movie.

We say that a continuous function $H:M\times[0,1]\to N$ is a continuous homotopy from $f_0$ to $f_1$ if $H(p,0)=f_0(p)$ and $H(p,1)=f_1(p)$ for all $p\in M$. For any time $t\in[0,1]$, the map $p\mapsto H(p,t)$ is a continuous map from $M$ to $N$, which is sort of like a “frame” in the movie that takes us from $f_0$ to $f_1$. As time passes over the interval, we highlight one frame at a time to watch the one function transform into the other.

To flip this around, imagine starting with a process of stretching and squeezing to turn one shape into another. In this case, when we say “shape” we really mean a subspace or submanifold of some outside space we occupy, like the three-dimensional space that contains our idiomatic doughnuts and coffee mugs. The maps in this case are the inclusions of the subspaces into the larger space.

Anyway, next imagine carrying out this process, but with a camera recording it at each step. Then cut out all the frames from the movie and stack them up. We see in each layer of this flipbook how the shape $M$ at that time is included into the larger space $N$. That is, we have a homotopy.

Now, for an example: we say that a space is “contractible” if its inclusion into itself is homotopic to a map of the whole space to a single point within the space. As a particular example, the unit ball $B^n\subseteq\mathbb{R}^n$ is contractible. Explicitly, we define a homotopy $H:B^n\times[0,1]\to B^n$latex H(p,t)=(1-t)p\$, which is certainly smooth; we can check that $H(p,0)=p$ and $H(p,1)=0$, so at one end we have the identity map of $B^n$ into itself, while at the other we have the constant map sending all of $B^n$ to the single point at the origin.

We should be careful to point out that homotopy only requires that the function $H$ be continuous, and not invertible in any sense. In particular, there’s no guarantee that the frame $p\mapsto H(p,t)$ for some fixed $t$ is a homeomorphism from $M$ onto its image. If it turns out that each frame is a homeomorphism of $M$ onto its image, then we say that $H$ is an “isotopy”.

November 29, 2011