I was always quite confused by natural transformations, but it seems they're just a selection of morphisms in the target category? How interesting!

**Puzzle 126.** Define \$$\alpha_{\textrm{People}}\$$ to be the inclusion map between the sets \$$\\{\textrm{Alice}, \textrm{Bob}, \textrm{Stan}, \textrm{Tyler}\\}\$$ and \$$\\{\textrm{Alice}, \textrm{Bob}, \textrm{Stan}, \textrm{Tyler}, \textrm{Mei-Chu}\\}\$$. As a restriction of the identity function, all squares commute.

Practically speaking, the inclusion map allows us to extend a database in a consistent way. As long as all relations in the source database are preserved by the target database, we can add new rows.

**Puzzle 127.** Let \$$\alpha_{\textrm{People}}\$$ be the identity map on everyone except Stan, who instead gets mapped to Bob. It is notable that the database instance \$$H\$$ replaces Stan with Bob as Tyler's friend, which ensures that all induced squares commute.

This seems to be a kind of reduction or compaction, where equivalent rows are coalesced and existing relationships involving the coalesced rows are updated. It reminds me a lot of [minimization of deterministic finite-state automata](https://en.wikipedia.org/wiki/DFA_minimization).

**Puzzle 128.** We can leave everything the same (identity) or project Tyler onto Alice. We cannot project Alice onto Tyler or swap Tyler and Alice, since \$$\mathrm{FriendOf}(\alpha(\textrm{Bob})) = \textrm{Alice} \neq \textrm{Tyler} = \alpha(\mathrm{FriendOf}(\textrm{Bob}))\$$.