Theorem 1   Every monomorphism in the category of sets is an injection.

Proof. Assume $f \colon A \to B$ is a monomorphism. Then, by definition of monomorphism, given any two maps $g,h \colon C \to A$ , if $f \circ g = f \circ h$ , then $g = h$ . Suppose $x$ and $y$ are two elements of $A$ such that $f(x) = f(y)$ . Let $C$ be a set with one element, let $g$ be the map which sends this one element to $x$ and let $h$ be the map which sends this one element to $y$ . Because $f(x) = f(y)$ , we have $f \circ g = f \circ h$ . Since $f$ is a monomorphism, $g = h$ , so $x = y$ . This implies that $f$ is injective.

Theorem 2   Every injection is a split monomorphism.

Proof. Assume $f \colon A \to B$ is injection. If $A$ is empty, the result is trivial, so we assume that $A$ is not empty; let $z$ be an element of $A$ . Set $$ g = \{ (f(x),x) \mid x \in A \} \cup \{(x,z) \mid x \in B \land (\forall y \in A) x \neq f(y)\} $$ We claim that $g$ is a function from $B$ into $A$ . Suppose that $x$ is an element of $B$ . If $x \neq f(y)$ for any $y \in A$ , then we have exactly one element of $g$ with $x$ as the first element, namely $(x,z)$ . If $x = f(y)$ for some $y \in A$ , then we the pair $(x,y)$ with $x$ as first element; were there another pair with $x$ as first element, then we would have $(f(x_1),x_1) = (f(x_2),x_2)$ but, as $f$ is an injection, $f(x_1) = f(x_2)$ would imply $x_1 = x_2$ , so this would not be a distinct pair. Hence $g$ is a function. Furthermore, by construction $g \circ f (x) = x$ for all $x \in A$ , so $f$ is a split monomorphism.