I begun reading the “Applied Category Theory“ by Fong and Spivak, and I got stuck on this excercise (1.74):
Let $P$ and $Q$ be preorders, and $f \colon P \rightarrow Q$ be a monotone map. Then we can define a monotone map $f^* \colon \mathsf{U}(Q) \rightarrow \mathsf{U}(P)$ sending an upper set $U \subseteq Q$ to the upper set $f^{-1}(U)\subseteq P$. We call this the pullback along $f$. Viewing upper sets as monotone maps to $\mathbb{B}$, the proposition can be understood in terms of composition. Indeed, show that $f^*$ is defined by taking $u \colon Q \rightarrow \mathbb{B}$ to $(f;u) \colon P \rightarrow \mathbb{B}$.
The solution states:
Given an upper set $U\subseteq Q$, let $u \colon Q \rightarrow \mathbb{B}$ the corresponding monotone map, which sends $q \mapsto \mathtt{true}$ iff $q\in U$. Then $(f;u) \colon P \rightarrow B$ sends $p \mapsto \mathtt{true}$ iff $f(p) \in U$; it corresponds to the upper set $\{p\in P \mid f(p)\in U \}$ which is exactly $f^{-1}(U)$.
I don’t understand how $(f;u)$ can be $f^*$ if the former’s domain is $P$ while the latter’s is $\mathsf{U}(Q)$.
You’re misreading what the text is trying to say. The map $f^*$ does not correspond to $(f;u)$, but to the mapping $u \mapsto (f;u)$.
Let’s be a bit more explicit than the text:
For any two preorders $P$ and $Q$ let us denote the set of monotone maps from $P$ to $Q$ by $\mathrm{Mon}(P, Q)$. According to Proposition 1.73 (and its proof), we have for every preorder $P$ a one-to-one correspondence between monotone maps from $P$ to $$ and upper sets in $P$, given by $$ Φ_P \colon \mathrm{Mon}(P, ) \longrightarrow \mathsf{U}(P) \,, \quad u \longmapsto u^{-1}(\mathtt{true}) \,. $$
Let now $P$ and $Q$ be two preorders and let $f \colon P \to Q$ be a monotone map. We then have the induced pull-back map $$ f^* \colon \mathsf{U}(Q) \longrightarrow \mathsf{U}(P) \,, \quad U \longmapsto f^{-1}(U) \,, $$ as explained in the question. We can now consider the composite $$ \mathrm{Mon}(P, ) \xrightarrow{\enspace Φ_P \enspace} \mathsf{U}(P) \xrightarrow{\enspace f^* \enspace} \mathsf{U}(Q) \xrightarrow{\enspace Φ_Q^{-1} \enspace} \mathrm{Mon}(Q, ) \,, $$ which is a map from $\mathrm{Mon}(P, )$ to $\mathrm{Mon}(Q, )$. The text claims that this map is given by $u \mapsto (f;u)$.
Graphically, the overall situation may be depicted as follows: $$ \require{AMScd} \begin{CD} \mathrm{Mon}(P, ) @>{\enspace u \mapsto (f;u) \enspace}>> \mathrm{Mon}(Q, ) \\ @V{Φ_P}VV @VV{Φ_Q}V \\ \mathsf{U}(P) @>>{f^*}> \mathsf{U}(Q) \end{CD} $$