 Solution

Keep the notation of the preceding part. For $u:T\to D$ and $w:T\to B$ satisfying $fu=gw$, we have $efu=egw=dcw$. The outside pullback in a category gives a unique $z:T\to A$ with $az=u$ and $cbz=cw$. Moreover $gbz=faz=fu=gw$. The morphisms $bz,w:T\to B$ thus have the same composites with $g$ and $c$. Uniqueness in the right pullback in a category gives $bz=w$. Any factorization through the left square also factors through the outside rectangle, so it is unique there. The left square is a pullback. Together these two arguments prove the pullback pasting lemma.