The Unapologetic Mathematician

Mathematics for the interested outsider

Homotopies as 2-Morphisms

Last time, while talking about homotopies as morphisms I said that I didn’t want to get too deeply into the reparameterization thing because it could get too complicated. But since when would I, of all people, shy away from 2-categories? In case it wasn’t obvious then, it’s because we’re actually going to extend in the other direction.

Given any two topological spaces M and N, we now don’t just have a set of continuous maps \hom(M,N), we have a whole category consisting of those maps and homotopies between them. And I say that composition isn’t just a function that takes two (composable) maps and gives another one, it’s actually a functor.

So let’s say that we have maps f_1,f_2:M\to N, maps g_1,g_2:N\to P, and homotopies F:f_1\to f_2 and G:g_1\to g_2. From this we can build a homotopy G\circ F:g_1\circ f_1\to g_2\circ f_2. The procedure is obvious: for any t\in[0,1] and m\in M, we just define

\displaystyle[G\circ F](m,t)=G(F(m,t),t)

That is, the time-t frame of the composed homotopy is the composition of the time-t frames of the original homotopies. It should be straightforward to verify that this composition is (strictly) associative, and that the identity map — along with its identity homotopy — acts as an (also strict) identity.

What we need to show is that this composition is actually functorial. That is, we add maps f_3:M\to N and g_3:N\to P, change F and G to F_1 and G_1, and add homotopies F_2:f_2\to f_3 and G_2:g_2\to g_3. Then we have to check that

(G_2*G_1)\circ(F_2*F_1)=(G_2\circ F_2)*(G_1\circ F_1)

That is, if we stack G_2 onto G_1 and F_2 onto F_1, and then compose them as defined above, we get the same result as if we compose G_2 with F_2 and G_1 with F_1, and then stack the one onto the other.

This is pretty straightforward from a bird’s-eye view, but let’s check it in detail. On the left we have


Meanwhile, on the right we have

\displaystyle\begin{aligned}{}[(G_2\circ F_2)*(G_1\circ F_1)](m,t)&=\left\{\begin{array}{lr}[G_1\circ F_1](m,2t)&0<t<\frac{1}{2}\\{}[G_2\circ F_2](m,2t-1)&\frac{1}{2}<t<1\end{array}\right.\\&=\left\{\begin{array}{lr}G_1(F_1(m,2t),2t)&0<t<\frac{1}{2}\\G_2(F_2(m,2t-1),2t-1)&\frac{1}{2}<t<1\end{array}\right.\end{aligned}

And so we do indeed have a 2-category with topological spaces as objects, continuous maps as 1-morphisms, and continuous homotopies as 2-morphisms. Of course, if we’re in a differential topological context we get a 2-category with differentiable manifolds as objects, smooth maps as 1-morphisms, and smooth homotopies as 2-morphisms.


November 30, 2011 - Posted by | Differential Topology, Topology

1 Comment »

  1. […] we’ve seen that differentiable manifolds, smooth maps, and homotopies form a 2-category, but it’s not […]

    Pingback by The Poincaré Lemma (setup) « The Unapologetic Mathematician | December 2, 2011 | Reply

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