# The Unapologetic Mathematician

## Weak 2-Categories

I’d like to step aside from homology because I’m on the road and I can throw off a post about weak 2-categories in my sleep by now.

When we were talking about enriched categories we mentioned the case of 2-categories, where between each pair of objects we have a hom-category and so on. We also mentioned that if a 2-category has only one object it’s the same thing as a strict monoidal category. But monoidal categories aren’t generally strict! That is, the monoidal product isn’t associative “on the nose”, but only up to a natural isomorphism. And so we should look for the same sort of thing to happen in the case of more general 2-categories.

So, a (weak) 2-category $\mathcal{C}$ will have a collection of objects. Between each pair of objects $C,D\in\mathcal{C}$ there will be a category $\hom_\mathcal{C}(C,D)$ whose objects we call 1-morphisms from $C$ to $D$. Between two such 1-morphisms $f,g:C\rightarrow D$ we have a set of “2-morphisms” $\hom_{\hom_\mathcal{C}(C,D)}(f,g)$.

We can “vertically” compose 2-morphisms using the composition from a given hom-category. That is, given $\phi:f\rightarrow g$ and $\psi:g\rightarrow h$ we have $\psi\bullet\phi:f\rightarrow h$. Of course, there is an “identity” 2-morphism $1_f$ on any 1-morphism $f$, and the composition is associative. At this top level, everything is just like any other category.

Down at the level of 1-morphisms, things get hairier. For each triple of objects $A,B,C\in\mathcal{C}$ we have a functor $\circ:\hom_\mathcal{C}(B,C)\times\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$. This functor is not required to be associative in the usual sense, nor to have identities. Instead, for every triple of 1-morphisms $f:A\rightarrow B$, $g:B\rightarrow C$, and $h:C\rightarrow D$, there is a 2-isomorphism $\alpha_{f,g,h}\in\hom_{\hom_\mathcal{C}(A,D)}((h\circ g)\circ f,h\circ(g\circ f))$ which replaces the associative law. Similarly, for each object we have a 1-morphism $1_C\in\hom_\mathcal{C}(C,C)$ and for each 1-morphism $f:C\rightarrow D$ we have 2-morphisms $\lambda_f\in\hom_{\hom_\mathcal{C}(C,D)}(1_D\circ f,f)$ and $\rho_f\in\hom_{\hom_\mathcal{C}(C,D)}(f\circ1_C,f)$ to replace the left and right unit laws.

Now because $\circ$ is a functor, it also acts on 2-morphisms. That is, if we have 1-morphisms $f_1,g_1:A\rightarrow B$ and $f_2,g_2:B\rightarrow C$, and 2-morphisms $\phi:f_1\rightarrow g_1$ and $\psi:f_2\rightarrow g_2$, then we have a “horizontal” composition $\psi\circ\phi:f_2\circ f_1\rightarrow g_2\circ g_1$.

Functoriality says that this horizontal composition has to preserve the vertical composition inside each hom-category. So let’s take 1-morphisms $f_1,g_1,h_1\in\hom_\mathcal{C}(A,B)$ and $f_2,g_2,h_2\in\hom_\mathcal{C}(B,C)$. Then take 2-morphisms $\phi_1:f_1\rightarrow g_1$, $\psi_1:g_1\rightarrow h_1$, $\phi_2:f_2\rightarrow g_2$, and $\psi_2:g_2\rightarrow h_2$. We can vertically compose to get $\psi_1\bullet\phi_1:f_1\rightarrow h_1$ and $\psi_2\bullet\phi_2:f_2\rightarrow h_2$. These can then be horizontally composed to get $(\psi_2\bullet\phi_2)\circ(\psi_1\bullet\phi_1):f_2\circ f_1\rightarrow h_2\circ h_1$. On the other hand we could have composed horizontally before vertically and obtained $(\psi_2\circ\psi_1)\bullet(\phi_2\circ\phi_1):f_2\circ f_1\rightarrow h_2\circ h_1$. Functoriality tells us that these two 2-morphisms must be the same. We call this equation the “exchange identity”.

As an exercise, write all this out in the case where we just have one object. Then there’s only one hom-category to worry about. Verify that this restates the definition of a (weak) monoidal category, and show what the exchange identity means in this case.