I'm trying to prove in a locally small category $\mathsf{C}$ that given $f:a\to b$ and $g:c\to d$ that the set of commutative squares
$$
Sq(f,g) \;\; =\;\; \left \{
\begin{array}{ccc}
a & \rightarrow & c \\
\downarrow^f & \quad & \downarrow^g \\
b& \rightarrow &d \\
\end{array}
\right \}
$$
can be realized as a pullback in the category $\mathsf{Set}$. The one result that I'm trying to use here is the fact that limits of $\mathsf{Set}$-valued functors can be realized as equalizers of the form
$$
\lim F \;\; \longrightarrow\;\; \prod_{j \in Ob(J)} Fj \;\; \rightrightarrows \;\; \prod_{f \in Mor(J)} F(cod\; f)
$$
Trying to interpret this equalizer statement in the context of the statement, my best guess was to interpret
$$
\prod_{j \in Ob(J)} Fj \;\; =\;\; a\times c \hspace{2pc} \text{and} \hspace{2pc} \prod_{f \in Mor(J)} F(cod\; f) \;\; =\;\; b\times d.
$$
Then the the two non-identity morphisms should be maps that send $a\to b$ and $c\to d$, but then the proof statement seems vacuously true. I'm honestly lost and looking for guidance. This is Riehl exercise 3.2.iii.
Best Answer
The trick to translating a set into a universal property is to note that "the set of all $P$" can be pieced apart:
Just to illustrate, for $X$ and $Y$ sets, $X\times Y$ is the "set of all pairs $(x,y)$ for $x\in X$ and $y\in Y$", the pairs part being the diagram $\{X\quad Y\}$, then "set of all" is taking the limit of this diagram.
In this case, we're trying to take the set of all commutative squares with $f$ and $g$ parallel. The interpretation of the result you state about a limit being an equaliser of products tells you that an arbitrary limit can be constructed (in $\mathbf{Set}$) by collecting all the ingredients (i.e., form those products) and then subject them to constraints (i.e., form the equaliser). For a pullback of $A\xrightarrow pB\xleftarrow qC$, the equaliser diagram is $$ A\times_BC \to A\times B\times C \rightrightarrows B\times B $$ where the maps $A\times B\times C\to B\times B$ are given by $A\times B\times C\xrightarrow\pi A\times B\xrightarrow{p\times B} B\times B$ (where $\pi$ is a projection) and similarly for $q$. You might notice this is a bit redundant, and we can really take the pullback to be the equaliser $$ A\times_BC \to A\times C\rightrightarrows B $$ of the maps $A\times C\to A\to B$ and $A\times C\to C\to B$. In any case, this allows us to interpret the pullback $A\times_BC$ as "the set of all pairs $(a,c)$ for which $p(a)=q(c)$".
Now, what is a commutative square involving $f$ and $g$? Well, it's a pair of morphisms $u:a\to c$ and $v:b\to d$ such that $\require{AMScd}$ \begin{CD} a @>u>> c \\ @VfVV @VVgV \\ b @>>v> d \end{CD} commutes; that is, $g\circ u=v\circ f$. So, $\operatorname{Sq}(f,g)$ is "the set of all pairs $(u,v)$ such that $g\circ u=v\circ f$," which is exactly what a pullback describes!
To finish the job, we just need to specify the sets. $u$ comes from $\def\Hom{\operatorname{Hom}}\Hom(a,c)$ and $v$ from $\Hom(b,d)$, while the two composites lie in $\Hom(a,d)$, so we get the pullback square \begin{CD} \operatorname{Sq}(f,g) @>>> \Hom(a,c) \\ @VVV @VVg\circ(-)V \\ \Hom(b,d) @>>(-)\circ f> \Hom(a,d) \end{CD}