[proofplan] We prove both implications directly from the categorical definition of a monomorphism. If $f$ is injective, then equality after composing with $f$ forces pointwise equality of any two maps into $X$. Conversely, if $f$ is monic, we test the monomorphism condition against maps from a singleton set to show that two elements of $X$ with the same image under $f$ must be equal. [/proofplan] [step:Use injectivity to prove left cancellation against arbitrary maps] Assume that $f: X \to Y$ is injective. To prove that $f$ is a monomorphism in $\mathbf{Set}$, let $Z$ be a set and let \begin{align*} g: Z &\to X, \\ h: Z &\to X \end{align*} be functions such that $f \circ g = f \circ h$ as functions $Z \to Y$. For each $z \in Z$, equality of the two composite functions gives \begin{align*} f(g(z)) = (f \circ g)(z) = (f \circ h)(z) = f(h(z)). \end{align*} Since $f$ is injective, this implies $g(z) = h(z)$. Therefore $g$ and $h$ agree at every element of $Z$, so $g = h$ as functions $Z \to X$. Hence $f$ is left-cancellative with respect to all maps into $X$, and so $f$ is a monomorphism in $\mathbf{Set}$. [/step] [step:Test monicity on a singleton to prove injectivity] Assume that $f: X \to Y$ is a monomorphism in $\mathbf{Set}$. Let $x_0, x_1 \in X$ satisfy \begin{align*} f(x_0) = f(x_1). \end{align*} We prove $x_0 = x_1$. Let $1 := \{*\}$ be the singleton set. Define functions \begin{align*} g: 1 &\to X, & * &\mapsto x_0, \\ h: 1 &\to X, & * &\mapsto x_1. \end{align*} Then the composites $f \circ g: 1 \to Y$ and $f \circ h: 1 \to Y$ satisfy \begin{align*} (f \circ g)(*) = f(x_0) = f(x_1) = (f \circ h)(*). \end{align*} Since $1$ has exactly one element, this proves $f \circ g = f \circ h$ as functions $1 \to Y$. Because $f$ is a monomorphism, the equality $f \circ g = f \circ h$ implies $g = h$. Evaluating both sides at $*$ gives \begin{align*} x_0 = g(*) = h(*) = x_1. \end{align*} Thus any two elements of $X$ with the same image under $f$ are equal, so $f$ is injective. [/step] [step:Combine the two implications] The first step proves that every injective function $f: X \to Y$ is a monomorphism in $\mathbf{Set}$. The second step proves that every monomorphism $f: X \to Y$ in $\mathbf{Set}$ is injective. Therefore $f$ is a monomorphism in $\mathbf{Set}$ if and only if $f$ is injective. [/step]