Let ${\displaystyle \alpha ,\,\beta }$ be such that ${\displaystyle \alpha \,e=\beta \,e}$, one needs to show that ${\displaystyle \alpha =\beta }$. Since the equalizer of ${\displaystyle (\alpha ,\beta )}$ exists, ${\displaystyle e}$ factorizes as ${\displaystyle e=q\,e'}$ with ${\displaystyle q}$ monic. But then ${\displaystyle f=(m\,q)\,e'}$ is a factorization of ${\displaystyle f}$ with ${\displaystyle (m\,q)}$ monomorphism. Hence by the universal property of the image there exists a unique arrow ${\displaystyle v:I\to Eq_{\alpha ,\beta }}$ such that ${\displaystyle m=m\,q\,v}$ and since ${\displaystyle m}$ is monic ${\displaystyle {\text{id}}_{I}=q\,v}$. Furthermore, one has ${\displaystyle m\,q=(mqv)\,q}$ and by the monomorphism property of ${\displaystyle mq}$ one obtains ${\displaystyle {\text{id}}_{Eq_{\alpha ,\beta }}=v\,q}$.

This means that ${\displaystyle I\equiv Eq_{\alpha ,\beta }}$ and thus that ${\displaystyle {\text{id}}_{I}=q\,v}$ equalizes ${\displaystyle (\alpha ,\beta )}$, whence ${\displaystyle \alpha =\beta }$.

First definition implies the second: Assume that (1) holds with ${\displaystyle m}$ regular monomorphism.

• Equalization: one needs to show that ${\displaystyle i_{1}\,m=i_{2}\,m}$ . As the cokernel pair of ${\displaystyle f,\ i_{1}\,f=i_{2}\,f}$ and by previous proposition, since ${\displaystyle C}$ has all equalizers, the arrow ${\displaystyle e}$ in the factorization ${\displaystyle f=m\,e}$ is an epimorphism, hence ${\displaystyle i_{1}\,f=i_{2}\,f\ \Rightarrow \ i_{1}\,m=i_{2}\,m}$.
• Universality: in a category with all colimits (or at least all pushouts) ${\displaystyle m}$ itself admits a cokernel pair ${\displaystyle (Y\sqcup _{I}Y,c_{1},c_{2})}$

Moreover, as a regular monomorphism, ${\displaystyle (I,m)}$ is the equalizer of a pair of morphisms ${\displaystyle b_{1},b_{2}:Y\longrightarrow B}$ but we claim here that it is also the equalizer of ${\displaystyle c_{1},c_{2}:Y\longrightarrow Y\sqcup _{I}Y}$.

Indeed, by construction ${\displaystyle b_{1}\,m=b_{2}\,m}$ thus the "cokernel pair" diagram for ${\displaystyle m}$ yields a unique morphism ${\displaystyle u':Y\sqcup _{I}Y\longrightarrow B}$ such that ${\displaystyle b_{1}=u'\,c_{1},\ b_{2}=u'\,c_{2}}$. Now, a map ${\displaystyle m':I'\longrightarrow Y}$ which equalizes ${\displaystyle (c_{1},c_{2})}$ also satisfies ${\displaystyle b_{1}\,m'=u'\,c_{1}\,m'=u'\,c_{2}\,m'=b_{2}\,m'}$, hence by the equalizer diagram for ${\displaystyle (b_{1},b_{2})}$, there exists a unique map ${\displaystyle h':I'\to I}$ such that ${\displaystyle m'=m\,h'}$.

Finally, use the cokernel pair diagram (of ${\displaystyle f}$) with ${\displaystyle j_{1}:=c_{1},\ j_{2}:=c_{2},\ Z:=Y\sqcup _{I}Y}$ : there exists a unique ${\displaystyle u:Y\sqcup _{X}Y\longrightarrow Y\sqcup _{I}Y}$ such that ${\displaystyle c_{1}=u\,i_{1},\ c_{2}=u\,i_{2}}$. Therefore, any map ${\displaystyle g}$ which equalizes ${\displaystyle (i_{1},i_{2})}$ also equalizes ${\displaystyle (c_{1},c_{2})}$ and thus uniquely factorizes as ${\displaystyle g=m\,h'}$. This exactly means that ${\displaystyle (I,m)}$ is the equalizer of ${\displaystyle (i_{1},i_{2})}$.

Second definition implies the first:

• Factorization: taking ${\displaystyle m':=f}$ in the equalizer diagram (${\displaystyle m'}$ corresponds to ${\displaystyle g}$), one obtains the factorization ${\displaystyle f=m\,h}$.
• Universality: let ${\displaystyle f=m'\,e'}$ be a factorization with ${\displaystyle m'}$ regular monomorphism, i.e. the equalizer of some pair ${\displaystyle (d_{1},d_{2})}$.

Then ${\displaystyle d_{1}\,m'=d_{2}\,m'\ \Rightarrow \ d_{1}\,f=d_{1}\,m'\,e=d_{2}\,m'\,e=d_{2}\,f}$ so that by the "cokernel pair" diagram (of ${\displaystyle f}$), with ${\displaystyle j_{1}:=d_{1},\ j_{2}:=d_{2},\ Z:=D}$, there exists a unique ${\displaystyle u'':Y\sqcup _{X}Y\longrightarrow D}$ such that ${\displaystyle d_{1}=u''\,i_{1},\ d_{2}=u''\,i_{2}}$.

Now, from ${\displaystyle i_{1}\,m=i_{2}\,m}$ (m from the equalizer of (i₁, i₂) diagram), one obtains ${\displaystyle d_{1}\,m=u''\,i_{1}\,m=u''\,i_{2}\,m=d_{2}\,m}$, hence by the universality in the (equalizer of (d₁, d₂) diagram, with f replaced by m), there exists a unique ${\displaystyle v:Im\longrightarrow I'}$ such that ${\displaystyle m=m'\,v}$.