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A map $e:X\to Y$ has this property if every commutative square with $e$ as its left side and a monomorphism $m:A\to B$ as its right side has a diagonal filler $t:Y\to A$. The filler is unique because $m$ is monic. This definition does not itself stipulate that $e$ is an epimorphism; binary products imply that conclusion by testing against the categorical diagonal.

A monomorphism is a morphism $m:X\to Y$ such that $mu=mv$ implies $u=v$ for every pair $u,v:Z\to X$. Dually, an epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$. A regular epimorphism is a coequalizer of some parallel pair. An isomorphism has a two-sided inverse.

For a product in a category, write ${\Delta }_A:A\to A\times A$ for its categorical diagonal. Since ${\pi }_1{\Delta }_A=1_A$, equality ${\Delta }_{A}u={\Delta }_{A}v$ gives $u={\pi }_1{\Delta }_{A}u={\pi }_1{\Delta }_{A}v=v$. Thus the diagonal is a monomorphism, indeed a split monomorphism.

Let $g:C\to D$ have the left lifting property against monomorphisms, and suppose $\alpha ,\beta :D\to B$ satisfy $\alpha g=\beta g$. Use the categorical diagonal ${\Delta }_B:B\to B\times B$, which is a monomorphism by part (a). The square with top arrow $\alpha g$, bottom arrow $⟨\alpha ,\beta ⟩$, left arrow $g$ and right arrow ${\Delta }_B$ commutes, since both product components are $\alpha g$.

Its lift $t:D\to B$ satisfies ${\Delta }_{B}t=⟨\alpha ,\beta ⟩$. Applying the two product in a category projections gives $t=\alpha$ and $t=\beta$. Thus $g$ is an epimorphism. This binary-product criterion for lifting-only strong epimorphisms requires binary products, rather than any assumption about equalizers.