Dan Oneata wrote:

>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.

What you did above I think is correct.

However, note that for all sets, there's a function from the empty set into any other set, therefore we must have,

\\[
\begin{matrix}
\mathbf{Cat}[ & Disc(\varnothing) & \rightarrow & \mathcal{C} & ] \\\\
& & & & \\\\
& \Uparrow & & \Downarrow & \\\\
& & & & \\\\
\mathbf{Set}[ & \varnothing & \rightarrow & Ob(\mathcal{C}) & ]
\end{matrix}
\\]

which must hold for every category. However, the discrete category on an empty set is the empty category, which does have a functor into every other category.

This gives an interesting special case,

\\[
\begin{matrix}
\mathbf{Cat}[ & Disc(\varnothing) & \rightarrow & Disc(\varnothing) & ] \\\\
& & & & \\\\
& \Uparrow & & \Downarrow & \\\\
& & & & \\\\
\mathbf{Set}[ & \varnothing & \rightarrow & Ob(Disc(\varnothing)) & ]
\end{matrix}
\\]

Likewise, for any one object, one morphism category there exists a unique functor into from any other category, therefore we must have,

\\[
\begin{matrix}
\mathbf{Cat}[ & Disc(S) & \rightarrow & \mathbf{1} & ] \\\\
& & & & \\\\
& \Uparrow & & \Downarrow & \\\\
& & & & \\\\
\mathbf{Set}[ & S & \rightarrow & Ob(\mathbf{1}) & ]
\end{matrix}
\\]

which must hold for every set. However, any object set on a one object category is just a one element set, for which every other set maps uniquely into.

This gives an interesting special case,

\\[
\begin{matrix}
\mathbf{Cat}[ & Disc(Ob(\mathbf{1})) & \rightarrow & \mathbf{1} & ] \\\\
& & & & \\\\
& \Uparrow & & \Downarrow & \\\\
& & & & \\\\
\mathbf{Set}[ & Ob(\mathbf{1}) & \rightarrow & Ob(\mathbf{1}) & ]
\end{matrix}
\\]