# The Unapologetic Mathematician

## Enriched Naturality Revisited

Let’s look back at the enriched versions of representable functors. If we fix an object $C\in\mathcal{C}$ we have a $\mathcal{V}$-natural transformation $\hom_\mathcal{C}(C,\underline{\hphantom{X}})_{B,C}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$. This corresponds under the closure adjunction to $\circ:\hom_\mathcal{C}(A,B)\otimes\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$. There’s a similar transformation for composition on the other side.

Now remember that the closest thing we have to a “morphism” in an enriched category is an element of the underlying set of a hom-object. That is, we can talk about an arrow $f:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$. We often abuse the language and say that this is a morphism from $A$ to $B$, which in fact it’s a morphism in the underlying category.

Now, even though this $f$ isn’t really a morphism in our enriched category, we can still come up with a morphism sensibly called $\hom_\mathcal{C}(C,f):\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$. Here’s how it goes:

• We start with $\hom_\mathcal{C}(C,A)$ and use the left unit isomotphism to move to $\mathbf{1}\otimes\hom_\mathcal{C}(C,A)$.
• We now hit $\mathbf{1}$ with our morphism $f$ to land in $\hom_\mathcal{C}(A,B)\otimes\hom_\mathcal{C}(C,A)$.
• Finally, we compose to end up in $\hom_\mathcal{C}(C,B)$.

We can similarly take $g:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$ and construct a morphism $\hom_\mathcal{C}(g,C):\hom_\mathcal{C}(B,C)\rightarrow\hom_\mathcal{C}(A,C)$.

These composites should look familiar from the definition of enriched naturality for a transformation $\eta:S\rightarrow T$. In fact, we have a more compact diagram to replace that big hexagon:

Notice here that the right and bottom arrows in this square expand out to become the top and bottom of a hexagon, and we slip the functors $S$ and $T$ into place.