= Solution
Use the <kernel squares in an abelian category> argument. If $b$ is monic and $r:X\to A$, $s:X\to K'$ satisfy $ar=f's$, then $bgr=g'ar=g'f's=0$, so $gr=0$. The <kernel in a category> property gives a unique $t:X\to K$ with $ft=r$. The equation $f'kt=af t=ar=f's$ and monicity of $f'$ give $kt=s$. This proves the left square is a <pullback in a category>.
Now suppose the right square is a pullback, without imposing the earlier monicity hypothesis on $b$. The pair $(f',0):K'\to A'\times_{B'}B$ gives a unique $r:K'\to A$ with $ar=f'$ and $gr=0$. Factor $r=f\ell$ through the kernel. Then $f'k\ell=af\ell=ar=f'$, so $k\ell=1_{K'}$. Also $r k$ and $f$ have the same two pullback projections, hence $rk=f$. Thus $f\ell k=f$ and $\ell k=1_K$. Therefore $k$ is an <isomorphism>. These arguments use only the relevant kernels, zero arrows and pullback properties; the abelian hypothesis supplies them.
Back to article page
