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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.

Given $A,B$, their product in a category is the pullback of the unique arrows $A\to 1\leftarrow B$, where $1$ is a terminal object. Having constructed this product in a category, define the equalizer of $f,g:A\rightrightarrows B$ as the pullbackwhere ${\delta }_B$ is the categorical diagonal. A map $x:X\to A$ factors through this pullback exactly when $fx=gx$: the accompanying arrow to $B$ must equal this common composite. The factorization is unique, proving the equalizer property.

Thus the hypotheses give a terminal object, binary products in a category and all equalizers. Part (i) now gives all finite limits.

Let $1$ be a terminal object. The pullback in a category of $X\to 1\leftarrow Y$ has precisely the universal property of the binary product in a category $X\times Y$: every pair of morphisms to $X,Y$ is compatible over $1$. Repeated binary products in a category, together with $1$ as the empty product, give all finite products in a category.

For $f,g:X\rightrightarrows Y$, form the pullback in a category of the categorical diagonal ${\Delta }_Y:Y\to Y\times Y$ along $⟨f,g⟩:X\to Y\times Y$. Its projection $e:E\to X$ satisfies $fe=ge$. If $h:T\to X$ satisfies $fh=gh$, the pair $h,fh$ gives a unique morphism into this pullback in a category. Conversely any such morphism must project to that pair. Hence $e$ is the equalizer of $f,g$. The preceding construction now gives all finite limits.

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.