Keith wrote:

> I believe Puzzle 162 can be answered the same way using the initial objects and terminal objects of \\(\mathbf{Cat}\\) and \\(\mathbf{Set}\\).

I'm not sure how to use this idea, so my attempt takes the long and tedious route.

> **Puzzle 162.** Show that this bijection is natural.

The two functors in question are:

\[
\begin{align}
\mathbf{Cat}(\mathrm{Disc}(-), =) &: \mathbf{Set}^{\mathrm{op}} \times \mathbf{Cat} \to \mathbf{Set} \\\\
\mathbf{Set}(-, \mathrm{Ob}(=)) &: \mathbf{Set}^{\mathrm{op}} \times \mathbf{Cat} \to \mathbf{Set}.
\end{align}
\]

We will check the naturality of the two functors in both arguments, which are denoted above by \\(-\\) and \\(=\\).

First, we show that for any morphism in \\(\mathbf{Cat}\\), \\(F: \mathcal{C} \to \mathcal{D}\\) (functor), the following diagram commutes:

\[
\begin{matrix}
\mathbf{Cat}(\mathrm{Disc}(S), \mathcal{C}) & \rightarrow & \mathbf{Cat}(\mathrm{Disc}(S), \mathcal{D}) \\\\
\downarrow & & \downarrow \\\\
\mathbf{Set}(S, \mathrm{Ob}(\mathcal{C})) & \rightarrow & \mathbf{Set}(S, \mathrm{Ob}(\mathcal{D}))
\end{matrix}
\]

Written as an equation, this means:

\[
\mathbf{Set}(S, \mathrm{Ob}(F)) \circ \alpha_{S,\mathcal{C}} = \alpha_{S,D} \circ \mathbf{Cat}(\mathrm{Disc}(S), F).
\]

For any functor \\(G : \mathrm{Disc}(S) \to \mathcal{C}\\), we have:

\[
\begin{align}
& (\mathbf{Set}(S, \mathrm{Ob}(F)) \circ \alpha_{S,\mathcal{C}})(G) \\\\
& = \text{{ composition }} \\\\
& \mathbf{Set}(S, \mathrm{Ob}(F))(\alpha_{S,\mathcal{C}}(G)) \\\\
& = \text{{ definition of the hom functor }} \\\\
& \alpha_{S,\mathcal{C}}(G) \circ \mathrm{Ob}(F) \\\\
& = \left\\{ \; \alpha \text{ behaves like Ob on functors } \right\\} \\\\
& \mathrm{Ob}(G) \circ \mathrm{Ob}(F) \\\\
& = \text{{ Ob preserves composition }} \\\\
& \mathrm{Ob}(G \circ F) \\\\
& = \left\\{ \; \alpha \text{ behaves like Ob on functors } \right\\} \\\\
& \alpha_{S,\mathcal{D}}(G \circ F)) \\\\
& = \text{{ definition of the hom functor }} \\\\
& \alpha_{S,\mathcal{D}}(\mathbf{Cat}(\mathrm{Disc}(S), F)(G))) \\\\
& = \text{{ composition }} \\\\
& (\alpha_{S,\mathcal{D}} \circ \mathbf{Cat}(\mathrm{Disc}(S), F))(G).
\end{align}
\]

Second, we show that for any morphism in \\(\mathbf{Set}\\), \\(f: T \to S\\) (function), the following diagram commutes:

\[
\begin{matrix}
\mathbf{Cat}(\mathrm{Disc}(S), \mathcal{C}) & \rightarrow & \mathbf{Cat}(\mathrm{Disc}(T), \mathcal{C}) \\\\
\downarrow & & \downarrow \\\\
\mathbf{Set}(S, \mathrm{Ob}(\mathcal{C})) & \rightarrow & \mathbf{Set}(T, \mathrm{Ob}(\mathcal{C}))
\end{matrix}
\]

Written as an equation, this means:

\[
\mathbf{Set}(f, \mathrm{Ob}(\mathcal{C})) \circ \alpha_{S,\mathcal{C}} = \alpha_{T,\mathcal{C}} \circ \mathbf{Cat}(\mathrm{Disc}(f), \mathcal{C}).
\]

For any functor \\(G : \mathrm{Disc}(S) \to \mathcal{C}\\), we have:

\[
\begin{align}
& (\mathbf{Set}(f, \mathrm{Ob}(\mathcal{C})) \circ \alpha_{S,\mathcal{C}})(G) \\\\
& = \text{{ composition }} \\\\
& \mathbf{Set}(f, \mathrm{Ob}(\mathcal{C}))(\alpha_{S,\mathcal{C}})(G)) \\\\
& = \text{{ definition of the hom functor }} \\\\
& f \circ \alpha_{S,\mathcal{C}}(G) \\\\
& = \left\\{ \; \alpha \text{ behaves like Ob on functors } \right\\} \\\\
& f \circ \mathrm{Ob}(G) \\\\
& = \left\\{ \mathrm{Ob} \circ \mathrm{Disc} = 1 \right\\} \\\\
& \mathrm{Ob}(\mathrm{Disc}(f)) \circ \mathrm{Ob}(G) \\\\
& = \text{{ Ob preserves composition }} \\\\
& \mathrm{Ob}(\mathrm{Disc}(f) \circ G) \\\\
& = \left\\{ \; \alpha \text{ behaves like Ob on functors } \right\\} \\\\
& \alpha_{T,\mathcal{C}}(\mathrm{Disc}(f) \circ G) \\\\
& = \text{{ definition of the hom functor }} \\\\
& \alpha_{T,\mathcal{C}}(\mathbf{Cat}(\mathrm{Disc}(f), \mathcal{C})(G)) \\\\
& = \text{{ composition }} \\\\
& (\alpha_{T,\mathcal{C}} \circ \mathbf{Cat}(\mathrm{Disc}(f), \mathcal{C}))(G).
\end{align}
\]