Implication details

Proof: Let $m : A \to B$ be a strict monomorphism in a category with pushouts. In particular, the pushout $B \sqcup_A B$ exists (and actually, we only need this pushout) with coprojections $i_1,i_2 : B \rightrightarrows B \sqcup_A B$ satisfying $i_1 \circ m = i_2 \circ m$. To show that $m$ is the equalizer of $i_1,i_2$, let $t : T \to B$ be a morphism with $i_1 \circ t = i_2 \circ t$. If $f,g : B \rightrightarrows C$ is any parallel pair with $f \circ m = g \circ m$, it induces a morphism $(f;g) : B \sqcup_A B \to C$ with $(f;g) \circ i_1 = f$ and $(f;g) \circ i_2 = g$. By composing these equations with $t$, we get $$f \circ t = (f;g) \circ i_1 \circ t = (f;g) \circ i_2 \circ t = g \circ t.$$ Thus, $t$ equalizes every parallel pair that is equalized by $m$. Since $m$ is a strict monomorphism, $t$ factors through $m$.