Theorem: A map $f : X \rightarrow Y$ is monic if and only if the square:

is a pullback.

Proof: Assume the square is a pullback. Let $g, g' : A \rightarrow X$ be maps with $f \circ g = f \circ g'$. Then $(A, g, g')$ is a cone over $X \xrightarrow{f} Y \xleftarrow{f} X$. Thus, we have a unique $\varphi : A \rightarrow X$ with $1 \circ \varphi = g$ and $1 \circ \varphi = g'$, so $\varphi = g = g'$.

Assume $f$ is monic. Let $(A, g, g')$ be an arbitrary cone over $X \xrightarrow{f} Y \xleftarrow{f} X$. We need a unique $\varphi$ such that:

commutes. Note that since $(A, g, g')$ is a cone, we have $f \circ g = f \circ g'$ and since $f$ is monic, $g = g'$. Thus, taking $\varphi = g = g'$ makes the diagram commute. Clearly then, this is the unique such map, because if any other $\varphi = h$ made the diagram commute, we would have $h = g = g'$.