[proofplan] We prove the contrapositive by assuming that two objects $A$ and $B$ admit at least two distinct arrows $A \to B$. Since $\mathcal C$ is small, the collection $M$ of all morphisms is a set, so we may form the $M$-indexed product of copies of $B$. The product universal property identifies arrows $A \to \prod_{m \in M} B$ with $M$-indexed families of arrows $A \to B$. Since there are at least two such arrows, a diagonal cardinality argument shows that there are more such families than there are morphisms in $\mathcal C$, contradicting smallness of the morphism set. [/proofplan] [step:Choose two parallel arrows and form the product indexed by all morphisms] Assume, for contradiction, that $\mathcal C$ is not a preorder. Then there exist objects $A,B \in \operatorname{Ob}(\mathcal C)$ and two distinct morphisms \begin{align*} h_0,h_1: A \to B \end{align*} in $\mathcal C$. Because $\mathcal C$ is small, its morphism collection \begin{align*} M := \operatorname{Mor}(\mathcal C) \end{align*} is a set. Define the constant $M$-indexed family of objects \begin{align*} F: M &\to \operatorname{Ob}(\mathcal C) \\ m &\mapsto B . \end{align*} Since $\mathcal C$ admits all small products, this family has a product. Choose a product object $P \in \operatorname{Ob}(\mathcal C)$ together with projection morphisms \begin{align*} \pi_m: P \to B \end{align*} for each $m \in M$. [guided] We begin by negating the conclusion. Saying that $\mathcal C$ is not a preorder means exactly that some hom-set has at least two elements. Thus we choose objects $A,B \in \operatorname{Ob}(\mathcal C)$ and distinct arrows \begin{align*} h_0,h_1: A \to B . \end{align*} The smallness hypothesis is used here in a specific way: it makes the total morphism collection \begin{align*} M := \operatorname{Mor}(\mathcal C) \end{align*} a set. Since products are assumed to exist for every set-indexed family of objects, we may index a product by $M$ itself. Define the constant family \begin{align*} F: M &\to \operatorname{Ob}(\mathcal C) \\ m &\mapsto B . \end{align*} The product hypothesis gives an object $P \in \operatorname{Ob}(\mathcal C)$ and projections \begin{align*} \pi_m: P \to B \end{align*} for every $m \in M$. This object is the product of one copy of $B$ for each morphism of the category. [/guided] [/step] [step:Identify arrows into the product with families of arrows into $B$] Let $H := \mathcal C(A,B)$ be the hom-set from $A$ to $B$. Define the set of $M$-indexed $H$-valued functions by \begin{align*} H^M := \{u: M \to H\}. \end{align*} The product projections define a map \begin{align*} \Phi: \mathcal C(A,P) &\to H^M \\ f &\mapsto \bigl(m \mapsto \pi_m \circ f\bigr). \end{align*} By the universal property of the product $P$, for every function $u: M \to H$ there exists a unique morphism $\bar u: A \to P$ such that \begin{align*} \pi_m \circ \bar u = u(m) \end{align*} for every $m \in M$. Hence $\Phi$ is bijective, and therefore \begin{align*} |\mathcal C(A,P)| = |H^M|. \end{align*} [guided] The product $P$ is designed so that maps into $P$ are exactly compatible families of maps into its factors. Here every factor is $B$, so maps $A \to P$ should correspond to $M$-indexed families of maps $A \to B$. Let \begin{align*} H := \mathcal C(A,B) \end{align*} and define \begin{align*} H^M := \{u: M \to H\}. \end{align*} Thus an element of $H^M$ assigns to each morphism $m \in M$ a morphism $u(m): A \to B$. Using the projections of the product, define \begin{align*} \Phi: \mathcal C(A,P) &\to H^M \\ f &\mapsto \bigl(m \mapsto \pi_m \circ f\bigr). \end{align*} For each $f: A \to P$ and each $m \in M$, the composite $\pi_m \circ f$ is a morphism $A \to B$, so $\Phi(f)$ is indeed an element of $H^M$. Now apply the universal property of the product $P$. Given any function $u: M \to H$, the family $(u(m))_{m \in M}$ is an $M$-indexed family of morphisms $A \to B$. The product universal property gives a unique morphism $\bar u: A \to P$ satisfying \begin{align*} \pi_m \circ \bar u = u(m) \end{align*} for every $m \in M$. This proves that every element of $H^M$ is hit by $\Phi$, and the uniqueness clause proves that two morphisms $A \to P$ with the same image under $\Phi$ must be equal. Therefore $\Phi$ is bijective, so \begin{align*} |\mathcal C(A,P)| = |H^M|. \end{align*} [/guided] [/step] [step:Prove that $H^M$ is too large to inject into $M$] [claim:The set $H^M$ has cardinality strictly larger than $M$] There is no injective map $H^M \to M$. [/claim] [proof] Let $\mathcal P(M)$ denote the power set of $M$, namely the set of all subsets of $M$. Since $h_0,h_1 \in H$ are distinct, define \begin{align*} \iota: \mathcal P(M) &\to H^M \\ S &\mapsto u_S, \end{align*} where $u_S: M \to H$ is the function \begin{align*} u_S(m)= \begin{cases} h_1, & m \in S,\\ h_0, & m \notin S. \end{cases} \end{align*} For subsets $S,T \subset M$, let $S \triangle T := (S \setminus T) \cup (T \setminus S)$ denote their symmetric difference. If $S,T \subset M$ and $S \neq T$, choose $m \in S \triangle T$. Then $u_S(m) \neq u_T(m)$ because $h_0 \neq h_1$, so $\iota(S) \neq \iota(T)$. Hence $\iota$ is injective. It remains to show that no injection $\mathcal P(M) \to M$ exists. Suppose that \begin{align*} j: \mathcal P(M) \to M \end{align*} is injective. Define \begin{align*} q: M &\to \mathcal P(M) \end{align*} by setting $q(j(S)) := S$ for $S \subset M$ and setting $q(m) := \varnothing$ for $m \notin j(\mathcal P(M))$. This is well-defined because $j$ is injective, and it is surjective because $q(j(S))=S$ for every $S \subset M$. Now define the diagonal subset \begin{align*} D := \{m \in M : m \notin q(m)\}. \end{align*} Since $q$ is surjective, there exists $a \in M$ such that $q(a)=D$. Then \begin{align*} a \in D \iff a \notin q(a) \iff a \notin D, \end{align*} which is impossible. Hence no injection $\mathcal P(M) \to M$ exists. Since $\mathcal P(M)$ injects into $H^M$, an injection $H^M \to M$ would compose with $\iota$ to give an injection $\mathcal P(M) \to M$, which has just been ruled out. Therefore there is no injective map $H^M \to M$. [/proof] [/step] [step:Derive the cardinality contradiction and conclude that the category is thin] The set $\mathcal C(A,P)$ is a subset of $M=\operatorname{Mor}(\mathcal C)$, because its elements are morphisms of $\mathcal C$. Therefore the inclusion \begin{align*} \mathcal C(A,P) \hookrightarrow M \end{align*} is injective. The bijection $\Phi: \mathcal C(A,P) \to H^M$ from the previous step gives an injection \begin{align*} H^M \to M \end{align*} by composing $\Phi^{-1}: H^M \to \mathcal C(A,P)$ with the inclusion $\mathcal C(A,P) \hookrightarrow M$. This contradicts the claim that no injection $H^M \to M$ exists. Hence the assumption that some hom-set $\mathcal C(A,B)$ contains two distinct morphisms is false. Therefore every hom-set of $\mathcal C$ has at most one element, so $\mathcal C$ is a preorder. Finally, if $\mathcal C$ is small and complete, then in particular it has all small products. Applying the result just proved, $\mathcal C$ is a preorder. [/step]