Isbell's condition for concretizability

Proof. Let $U : \C \to \Set$ be a faithful functor. A span $$A \xleftarrow{a} X \xrightarrow{b} B$$ commutes with a cospan $$A \xrightarrow{a'} Y \xleftarrow{b'} B$$ if and only if $a' \circ a = b' \circ b$. Since $U$ is faithful, this is equivalent to $$U(a') \circ U(a) = U(b') \circ U(b).$$ Define $P_{a,b} \subseteq U(A) \times U(B)$ to be the set of pairs $$(U(a)(x),U(b)(x))$$ for $x \in U(X)$. Then the span commutes with the cospan if and only if $$U(a')(u) = U(b')(v)$$ for every $(u,v) \in P_{a,b}$. Thus, whether a span commutes with a given cospan depends only on the subset $P_{a,b}$.

Call a subset of $U(A) \times U(B)$ realizable if it is of the form $P_{a,b}$ for some span $(a,b)$. Since $U(A) \times U(B)$ is a set, there is a set of realizable subsets. As we have just seen, the map that sends a realizable subset $P_{a,b}$ to the equivalence class of $(a,b)$ is well-defined, and it is clearly surjective. Hence, the collection of equivalence classes of spans over $(A,B)$ is essentially small. $\square$