**Puzzle 167.**

I haven't solved the puzzle, but here is what I got so far; maybe someone can help me out.

From the definition of the left Kan extension, we know there is an one-to-one correspondence between

\[

\textrm{ morphisms from } \textrm{Lan}_!(H) \textrm{ to } F

\]

in category \\(\mathbf{Set}^1\\), and

\[

\textrm{ morphisms from } H \textrm{ to } F \circ !

\]

in category \\(\mathbf{Set}^\mathcal{D}\\).

I was thinking that maybe we can use the cardinality of the two sets of morphisms to get an idea of what \\(\textrm{Lan}_!(H)\\) is.

- In \\(\mathbf{Set}^1\\), the objects are sets and the morphisms are functions.

If we denote the set picked by \\(\textrm{Lan}_!(H)\\) by \\(T\\) and the set picked by \\(F\\) by \\(S\\), then the number of morphisms is \\(|S|^{|T|}\\).

- In \\(\mathbf{Set}^\mathcal{D}\\), the functor \\(F \circ !\\) picks the same set as \\(F\\);

more precisely, \\((F \circ !)(\textrm{Germans}) = S\\), \\((F \circ !)(\textrm{Italians}) = S\\) and \\((F \circ !)(\textrm{Friend}) = 1\\) (as Christopher has already pointed out).

The morphisms are natural transformations and they have to satisfy the naturality condition: \\(\alpha_\textrm{Germans} = \alpha_\textrm{Italians} \circ H(\textrm{Friend})\\).

I think that the number of natural transformations is given by the number of functions \\(\alpha_\textrm{Italians}\\) from \\(H(\textrm{Italians})\\) to \\(S\\), since the component for Germans is completely specified by the component for Italians.

So it seems that the cardinality of the set picked by the left Kan extension equals the cardinality of the set \\(H(\textrm{Italians})\\).

I haven't solved the puzzle, but here is what I got so far; maybe someone can help me out.

From the definition of the left Kan extension, we know there is an one-to-one correspondence between

\[

\textrm{ morphisms from } \textrm{Lan}_!(H) \textrm{ to } F

\]

in category \\(\mathbf{Set}^1\\), and

\[

\textrm{ morphisms from } H \textrm{ to } F \circ !

\]

in category \\(\mathbf{Set}^\mathcal{D}\\).

I was thinking that maybe we can use the cardinality of the two sets of morphisms to get an idea of what \\(\textrm{Lan}_!(H)\\) is.

- In \\(\mathbf{Set}^1\\), the objects are sets and the morphisms are functions.

If we denote the set picked by \\(\textrm{Lan}_!(H)\\) by \\(T\\) and the set picked by \\(F\\) by \\(S\\), then the number of morphisms is \\(|S|^{|T|}\\).

- In \\(\mathbf{Set}^\mathcal{D}\\), the functor \\(F \circ !\\) picks the same set as \\(F\\);

more precisely, \\((F \circ !)(\textrm{Germans}) = S\\), \\((F \circ !)(\textrm{Italians}) = S\\) and \\((F \circ !)(\textrm{Friend}) = 1\\) (as Christopher has already pointed out).

The morphisms are natural transformations and they have to satisfy the naturality condition: \\(\alpha_\textrm{Germans} = \alpha_\textrm{Italians} \circ H(\textrm{Friend})\\).

I think that the number of natural transformations is given by the number of functions \\(\alpha_\textrm{Italians}\\) from \\(H(\textrm{Italians})\\) to \\(S\\), since the component for Germans is completely specified by the component for Italians.

So it seems that the cardinality of the set picked by the left Kan extension equals the cardinality of the set \\(H(\textrm{Italians})\\).