Pullback pasting lemma

Remarks

Both directions are checked with the universal property. A cone over the outer rectangle gives, through the right pullback, a cone over the left square, and back again. The uniqueness parts match because the right square is a pullback. In the category of sets, a pullback along an inclusion is a preimage. The lemma then says that the preimage under $e \circ d$ is the preimage under $d$ of the preimage under $e$.

The hypothesis on the right square is needed. Let $A$, $C$, $D$, $E$ and $F$ be one-point sets, let $B$ have two points, and let $a$ pick one of them. The outer rectangle is then a pullback, but the left square is not, because the pullback of $B \to E \leftarrow D$ has two points while $A$ has one.

Going deeper

For a student

The lemma is the main tool for computing with pullbacks in diagrams. It shows, for example, that pullback along $g \circ f$ is the same as pullback along $g$ followed by pullback along $f$, up to isomorphism. In algebraic geometry it gives the transitivity of base change, $(X \times_S S') \times_{S'} S'' \cong X \times_S S''$. The dual statement holds for pushouts: if the left square is a pushout, then the right square is a pushout if and only if the outer rectangle is.