[proofplan] For each listed category, fix two objects and compare its morphisms with all functions between the underlying sets. The set of all functions between two underlying sets is a set, and the structure-preserving maps form a subclass of that set cut out by explicit algebraic, order-theoretic, topological, metric, or analytic conditions. Since any subclass of a set determined by such a predicate is a set, each hom-class is a set. [/proofplan]

[step:Reduce local smallness to a subset of an underlying function set]Let $\mathcal{C}$ be one of the categories in the statement, and let $A,B \in \operatorname{Ob}(\mathcal{C})$. Write $|A|$ and $|B|$ for the underlying sets of $A$ and $B$. Define the underlying function set \begin{align*} \operatorname{Set}(|A|,|B|) = \{f : |A| \to |B| \mid f \text{ is a function}\}. \end{align*} Because $|A|$ and $|B|$ are sets, $\operatorname{Set}(|A|,|B|)$ is a set. Every morphism $A \to B$ in any of the listed concrete categories is, by definition, a function $|A| \to |B|$ satisfying the relevant structure-preservation condition. Therefore \begin{align*} \operatorname{Hom}_{\mathcal{C}}(A,B) \subseteq \operatorname{Set}(|A|,|B|). \end{align*} Thus it remains only to identify, in each category, the defining predicate that cuts out the morphisms from the full function set.[/step]

[guided]The definition of local smallness asks for a set-theoretic statement: for each pair of objects, the morphisms between them must form a set. Since the categories in the theorem are concrete, each object $A$ has an underlying set $|A|$, and each morphism $A \to B$ has an underlying function from $|A|$ to $|B|$. So we first form the ambient set \begin{align*} \operatorname{Set}(|A|,|B|) = \{f : |A| \to |B| \mid f \text{ is a function}\}. \end{align*} This is a set because both $|A|$ and $|B|$ are sets. The key point is that the hom-class in each concrete category is not larger than this function set; it consists exactly of those functions satisfying the extra compatibility condition appropriate to the structures on $A$ and $B$. Therefore, once we show \begin{align*} \operatorname{Hom}_{\mathcal{C}}(A,B) \subseteq \operatorname{Set}(|A|,|B|), \end{align*} the hom-class is a subclass of a set. The remaining steps verify this inclusion explicitly for each category.[/guided]

[step:Check the algebraic examples by writing the homomorphism conditions]For $\mathbf{Set}$, if $A$ and $B$ are sets, then \begin{align*} \operatorname{Hom}_{\mathbf{Set}}(A,B) = \operatorname{Set}(A,B), \end{align*} so the hom-class is a set. For $\mathbf{Grp}$, let $G$ and $H$ be groups with multiplications denoted by juxtaposition and identity elements $e_G$ and $e_H$. Then $\operatorname{Hom}_{\mathbf{Grp}}(G,H)$ is the subclass of $\operatorname{Set}(|G|,|H|)$ consisting of all functions $f: |G| \to |H|$ such that $f(e_G)=e_H$ and, for all $g_1,g_2 \in |G|$, \begin{align*} f(g_1g_2)=f(g_1)f(g_2). \end{align*} Hence $\operatorname{Hom}_{\mathbf{Grp}}(G,H)$ is a subset of a set. For $\mathbf{Ring}$, let $S$ and $T$ be unital rings with units $1_S$ and $1_T$. Then $\operatorname{Hom}_{\mathbf{Ring}}(S,T)$ is the subclass of $\operatorname{Set}(|S|,|T|)$ consisting of all functions $f: |S| \to |T|$ such that $f(1_S)=1_T$ and, for all $s_1,s_2 \in |S|$, \begin{align*} f(s_1+s_2)&=f(s_1)+f(s_2),\\ f(s_1s_2)&=f(s_1)f(s_2). \end{align*} Hence $\operatorname{Hom}_{\mathbf{Ring}}(S,T)$ is a subset of a set. For $R\text{-}\mathbf{Mod}$, let $M$ and $N$ be left $R$-modules. Then $\operatorname{Hom}_{R\text{-}\mathbf{Mod}}(M,N)$ is the subclass of $\operatorname{Set}(|M|,|N|)$ consisting of all functions $f: |M| \to |N|$ such that, for all $m_1,m_2 \in |M|$ and all $r \in R$, \begin{align*} f(m_1+m_2)&=f(m_1)+f(m_2),\\ f(rm_1)&=rf(m_1). \end{align*} Hence $\operatorname{Hom}_{R\text{-}\mathbf{Mod}}(M,N)$ is a subset of a set.[/step]

[guided]We now verify the algebraic cases by writing each hom-class as a subset of a set of underlying functions. For $\mathbf{Set}$, the objects $A$ and $B$ are sets and the morphisms are exactly the functions from $A$ to $B$. Therefore \begin{align*} \operatorname{Hom}_{\mathbf{Set}}(A,B) = \operatorname{Set}(A,B), \end{align*} which is a set. For $\mathbf{Grp}$, let $G$ and $H$ be groups with multiplications denoted by juxtaposition and identity elements $e_G$ and $e_H$. A group morphism from $G$ to $H$ is precisely a function $f: |G| \to |H|$ preserving the group structure. Explicitly, this means $f(e_G)=e_H$ and, for all $g_1,g_2 \in |G|$, \begin{align*} f(g_1g_2)=f(g_1)f(g_2). \end{align*} Thus $\operatorname{Hom}_{\mathbf{Grp}}(G,H)$ is obtained from the set $\operatorname{Set}(|G|,|H|)$ by imposing a predicate on its elements, so it is a subset of that set. For $\mathbf{Ring}$, let $S$ and $T$ be unital rings with units $1_S$ and $1_T$. A ring morphism from $S$ to $T$ is precisely a function $f: |S| \to |T|$ preserving the additive operation, the multiplicative operation, and the unit. Explicitly, $f(1_S)=1_T$ and, for all $s_1,s_2 \in |S|$, \begin{align*} f(s_1+s_2)&=f(s_1)+f(s_2),\\ f(s_1s_2)&=f(s_1)f(s_2). \end{align*} Therefore $\operatorname{Hom}_{\mathbf{Ring}}(S,T)$ is a subclass of the set $\operatorname{Set}(|S|,|T|)$. For $R\text{-}\mathbf{Mod}$, let $M$ and $N$ be left $R$-modules. An $R$-module morphism from $M$ to $N$ is precisely a function $f: |M| \to |N|$ preserving addition and scalar multiplication. Explicitly, for all $m_1,m_2 \in |M|$ and all $r \in R$, \begin{align*} f(m_1+m_2)&=f(m_1)+f(m_2),\\ f(rm_1)&=rf(m_1). \end{align*} Hence $\operatorname{Hom}_{R\text{-}\mathbf{Mod}}(M,N)$ is a subclass of $\operatorname{Set}(|M|,|N|)$.[/guided]

[step:Check the topological order metric and Hilbert space examples]For $\mathbf{Top}$, let $X$ and $Y$ be topological spaces with topologies $\tau_X$ and $\tau_Y$. Then $\operatorname{Hom}_{\mathbf{Top}}(X,Y)$ is the subclass of $\operatorname{Set}(|X|,|Y|)$ consisting of all functions $f: |X| \to |Y|$ such that, for every $V \in \tau_Y$, the preimage $f^{-1}(V)$ belongs to $\tau_X$. Hence $\operatorname{Hom}_{\mathbf{Top}}(X,Y)$ is a subset of a set. For $\mathbf{Pos}$, let $P$ and $Q$ be partially ordered sets with orders $\le_P$ and $\le_Q$. Then $\operatorname{Hom}_{\mathbf{Pos}}(P,Q)$ is the subclass of $\operatorname{Set}(|P|,|Q|)$ consisting of all functions $f: |P| \to |Q|$ such that, for all $p_1,p_2 \in |P|$, \begin{align*} p_1 \le_P p_2 \implies f(p_1) \le_Q f(p_2). \end{align*} Hence $\operatorname{Hom}_{\mathbf{Pos}}(P,Q)$ is a subset of a set. For $\mathbf{Met}$, let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Then $\operatorname{Hom}_{\mathbf{Met}}(X,Y)$ is the subclass of $\operatorname{Set}(|X|,|Y|)$ consisting of all non-expansive maps $f: |X| \to |Y|$, meaning that for all $x_1,x_2 \in |X|$, \begin{align*} d_Y(f(x_1),f(x_2)) \le d_X(x_1,x_2). \end{align*} Hence $\operatorname{Hom}_{\mathbf{Met}}(X,Y)$ is a subset of a set. For $\mathbf{Hilb}$, let $H$ and $K$ be Hilbert spaces over the same scalar field $\mathbb{F} \in \{\mathbb{R},\mathbb{C}\}$. Then $\operatorname{Hom}_{\mathbf{Hilb}}(H,K)$ is the subclass of $\operatorname{Set}(|H|,|K|)$ consisting of all functions $T: |H| \to |K|$ such that $T$ is $\mathbb{F}$-linear and bounded. Explicitly, for all $u,v \in |H|$ and all $\lambda \in \mathbb{F}$, \begin{align*} T(u+v)&=T(u)+T(v),\\ T(\lambda u)&=\lambda T(u), \end{align*} and there exists a real number $C \ge 0$ such that, for all $u \in |H|$, \begin{align*} \|T(u)\|_K \le C \|u\|_H, \end{align*} where $\|\cdot\|_H: H \to [0,\infty)$ and $\|\cdot\|_K: K \to [0,\infty)$ denote the Hilbert-space norms on $H$ and $K$. Hence $\operatorname{Hom}_{\mathbf{Hilb}}(H,K)$ is a subset of a set.[/step]

[guided]We now verify the remaining concrete categories by the same set-theoretic method: start with the set of all underlying functions, then impose the relevant compatibility condition. For $\mathbf{Top}$, let $X$ and $Y$ be topological spaces with topologies $\tau_X$ and $\tau_Y$. A continuous map from $X$ to $Y$ is a function $f: |X| \to |Y|$ such that, for every [open set](/page/Open%20Set) $V \in \tau_Y$, the preimage $f^{-1}(V)$ belongs to $\tau_X$. Hence $\operatorname{Hom}_{\mathbf{Top}}(X,Y)$ is a subclass of $\operatorname{Set}(|X|,|Y|)$. For $\mathbf{Pos}$, let $P$ and $Q$ be partially ordered sets with orders $\le_P$ and $\le_Q$. An order-preserving map from $P$ to $Q$ is a function $f: |P| \to |Q|$ such that, for all $p_1,p_2 \in |P|$, \begin{align*} p_1 \le_P p_2 \implies f(p_1) \le_Q f(p_2). \end{align*} Thus $\operatorname{Hom}_{\mathbf{Pos}}(P,Q)$ is a subclass of $\operatorname{Set}(|P|,|Q|)$. For $\mathbf{Met}$, let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A morphism in $\mathbf{Met}$ is a non-expansive map, meaning a function $f: |X| \to |Y|$ such that, for all $x_1,x_2 \in |X|$, \begin{align*} d_Y(f(x_1),f(x_2)) \le d_X(x_1,x_2). \end{align*} Therefore $\operatorname{Hom}_{\mathbf{Met}}(X,Y)$ is a subclass of $\operatorname{Set}(|X|,|Y|)$. For $\mathbf{Hilb}$, let $H$ and $K$ be Hilbert spaces over the same scalar field $\mathbb{F} \in \{\mathbb{R},\mathbb{C}\}$. Let $\|\cdot\|_H: H \to [0,\infty)$ and $\|\cdot\|_K: K \to [0,\infty)$ denote the Hilbert-space norms on $H$ and $K$. A morphism from $H$ to $K$ is a bounded $\mathbb{F}$-[linear map](/page/Linear%20Map), meaning a function $T: |H| \to |K|$ such that, for all $u,v \in |H|$ and all $\lambda \in \mathbb{F}$, \begin{align*} T(u+v)&=T(u)+T(v),\\ T(\lambda u)&=\lambda T(u), \end{align*} and there exists a real number $C \ge 0$ such that, for all $u \in |H|$, \begin{align*} \|T(u)\|_K \le C\|u\|_H. \end{align*} Hence $\operatorname{Hom}_{\mathbf{Hilb}}(H,K)$ is a subclass of $\operatorname{Set}(|H|,|K|)$.[/guided]

[step:Conclude that every listed hom-class is a set] In each case, $\operatorname{Hom}_{\mathcal{C}}(A,B)$ is a subclass of the set $\operatorname{Set}(|A|,|B|)$ defined by a precise predicate on functions. By separation, or equivalently by the principle that every subclass of a set cut out by a predicate is a set, $\operatorname{Hom}_{\mathcal{C}}(A,B)$ is a set. Since $A$ and $B$ were arbitrary objects of $\mathcal{C}$, each category in the list is locally small. [/step]