>**Puzzle 124.** Show there is exactly one functor \\(G : \mathcal{D} \to \mathcal{C}\\). What does this functor do, in concrete terms? Hint: on objects it forgets the difference between Germans and Italians... but what does it do to morphisms?

As you stated, \\(G\\) maps \\(\text{Italians}\\) and \\(\text{Germans}\\) to the same place, since in \\(\mathcal{C}\\) there is only one place to go, \\(\text{People}\\), we must have,

\\[
G(\text{Italians}) = G(\text{Germans}) = \text{People},
\\]

and likewise, for the same reasoning,
\\[
G(\text{FriendOf}\_\mathcal{D}) = G(\text{FriendOf}'\_\mathcal{D})
= \text{FriendOf}\_\mathcal{C},
\\]

that follows the functorial constraints,
\\[
G(\text{FriendOf}\_\mathcal{D} \circ \text{FriendOf}'\_\mathcal{D}) \\\\
= G(\text{FriendOf}\_\mathcal{D}) \circ G(\text{FriendOf}'\_\mathcal{D}) \\\\
= \text{FriendOf}\_\mathcal{C} \circ \text{FriendOf}\_\mathcal{C}
\\]

and,
\\[
G(\text{FriendOf}'\_\mathcal{D} \circ \text{FriendOf}\_\mathcal{D}) \\\\
= G(\text{FriendOf}'\_\mathcal{D}) \circ G(\text{FriendOf}\_\mathcal{D}) \\\\
= \text{FriendOf}\_\mathcal{C} \circ \text{FriendOf}\_\mathcal{C}.
\\]