# 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 - Posted by | Differential Topology, Topology

## 1 Comment »

1. […] time, while talking about homotopies as morphisms I said that I didn’t want to get too deeply into the reparameterization thing because it […]

Pingback by Homotopies as 2-Morphisms « The Unapologetic Mathematician | November 30, 2011 | Reply