= 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. \b[The left square is a pullback]. Together these two arguments prove the <pullback pasting lemma>.
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