>**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}. \$