Short five lemma by pullback without members

Form the pullback of $p^{\mathrm{\prime }}$ along $h$ shown above. Because $p^{\mathrm{\prime }}g=hp$, there is a unique comparison map $u:B\to P$ with $\alpha u=p$ and $\beta u=g$. Since $p^{\mathrm{\prime }}$ is the cokernel of $i^{\mathrm{\prime }}$ by [L1], it is epic, so [L3] makes the square also cocartesian, and [L2] identifies $\ker (\beta )$ with $\ker (h)$.

Assume $f$ and $h$ are monic. Then $\ker (h)=0$, so step 1.1 makes $\beta$ monic. If $ut=0$, then $pt=\alpha ut=0.$ Because $i=\ker (p)$ by [L1], there is $s$ with $t=is$. Now $0=\beta ut=gis=i^{\mathrm{\prime }}fs\mathrm{.}$ The map $i^{\mathrm{\prime }}$ is monic by [L1], and $f$ is monic by hypothesis, so $s=0$ and hence $t=0$. Thus $u$ is monic, and therefore $g=\beta u$ is monic.

Still with the pullback square of step 1.1, let $j:A^{\mathrm{\prime }}\to P$ be a kernel of $\alpha$; by [L2] this exists and agrees with the induced map from $\ker (p^{\mathrm{\prime }})=\mathrm{im}(i^{\mathrm{\prime }})$ to $P$. Assume now that $f$ and $h$ are epic. To show that $u$ is epic, let $t:P\to T$ satisfy $tu=0$. Then $tjf=tui=0.$ Since $f$ is epic, $tj=0$. Because $j=\ker (\alpha )$, there is $r:C\to T$ with $r\alpha =t$. But then $0=tu=r\alpha u=rp,$ and $p$ is epic by [L1], so $r=0$ and hence $t=0$. Therefore $u$ is epic.

To show that $\beta$ is epic under the same hypotheses, let $s:B^{\mathrm{\prime }}\to T$ satisfy $s\beta =0$. Since the square of step 1.1 is cocartesian by [L3], the compatible pair of maps $0:C\to T$ and $s:B^{\mathrm{\prime }}\to T$ induces a unique $v:C^{\mathrm{\prime }}\to T$ with $vh=0$ and $v{p}^{\mathrm{\prime }}=s$. Because $h$ is epic, $v=0$, so $s=0$. Thus $\beta$ is epic, and therefore $g=\beta u$ is epic.

If $f$ and $h$ are isomorphisms, steps 2.1 and 3.1 show that $g$ is both monic and epic. Therefore [L4] makes $g$ an isomorphism.

This gives the short five lemma again, now by a pullback-and-pushout argument and without any use of members.