[proofplan] We prove both directions directly from the universal property of pullbacks. For the forward direction, a cone over the outer rectangle first factors uniquely through the right pullback, and that factor then factors uniquely through the left pullback. For the converse, a cone over the left square gives a cone over the outer rectangle; the resulting mediating morphism is then shown to have the correct map into $B$ by uniqueness in the right pullback. [/proofplan] [step:Factor an arbitrary cone over the outer rectangle through the two smaller pullbacks] Assume that the left square and the right square are pullbacks. Let $N$ be an object of $\mathcal C$, and let \begin{align*} x: N \to C, \qquad y: N \to D \end{align*} be morphisms satisfying the compatibility condition for the outer rectangle: \begin{align*} c \circ x = u \circ t \circ y. \end{align*} The morphisms $x: N \to C$ and $t \circ y: N \to E$ are compatible with the right square, since \begin{align*} c \circ x = u \circ (t \circ y). \end{align*} Because the right square is a pullback, there exists a unique morphism \begin{align*} z: N \to B \end{align*} such that \begin{align*} s \circ z = x, \qquad b \circ z = t \circ y. \end{align*} Now the morphisms $z: N \to B$ and $y: N \to D$ are compatible with the left square, since \begin{align*} b \circ z = t \circ y. \end{align*} Because the left square is a pullback, there exists a unique morphism \begin{align*} w: N \to A \end{align*} such that \begin{align*} r \circ w = z, \qquad a \circ w = y. \end{align*} Therefore \begin{align*} (s \circ r) \circ w = s \circ (r \circ w) = s \circ z = x, \qquad a \circ w = y. \end{align*} Thus $w$ is a mediating morphism from $N$ to $A$ for the cone $(x,y)$ over the outer rectangle. [/step] [step:Use uniqueness twice to prove the outer rectangle is a pullback] We prove that the mediating morphism $w: N \to A$ constructed above is unique. Let \begin{align*} w': N \to A \end{align*} be another morphism satisfying \begin{align*} (s \circ r) \circ w' = x, \qquad a \circ w' = y. \end{align*} Then the morphism $r \circ w': N \to B$ satisfies \begin{align*} s \circ (r \circ w') = x \end{align*} and, using commutativity of the left square, \begin{align*} b \circ (r \circ w') = (b \circ r) \circ w' = (t \circ a) \circ w' = t \circ y. \end{align*} Hence $r \circ w'$ and $z$ are both morphisms $N \to B$ mediating the same compatible pair $(x,t \circ y)$ for the right pullback. By uniqueness in the right pullback, \begin{align*} r \circ w' = z. \end{align*} Now $w'$ and $w$ are both morphisms $N \to A$ satisfying \begin{align*} r \circ w' = z = r \circ w, \qquad a \circ w' = y = a \circ w. \end{align*} By uniqueness in the left pullback, \begin{align*} w' = w. \end{align*} Therefore the outer rectangle satisfies the universal property of the pullback. [/step] [step:Construct the mediating morphism for the left square from the outer pullback] Conversely, assume that the right square and the outer rectangle are pullbacks. Let $N$ be an object of $\mathcal C$, and let \begin{align*} z: N \to B, \qquad y: N \to D \end{align*} be morphisms satisfying the compatibility condition for the left square: \begin{align*} b \circ z = t \circ y. \end{align*} Then the morphisms $s \circ z: N \to C$ and $y: N \to D$ are compatible with the outer rectangle, because \begin{align*} c \circ (s \circ z) = (c \circ s) \circ z = (u \circ b) \circ z = u \circ (b \circ z) = u \circ (t \circ y) = (u \circ t) \circ y. \end{align*} Since the outer rectangle is a pullback, there exists a unique morphism \begin{align*} w: N \to A \end{align*} such that \begin{align*} (s \circ r) \circ w = s \circ z, \qquad a \circ w = y. \end{align*} [/step] [step:Recover the required map into $B$ by uniqueness in the right pullback] It remains to prove that $r \circ w = z$. The morphisms $r \circ w: N \to B$ and $z: N \to B$ have the same image under $s$, since \begin{align*} s \circ (r \circ w) = (s \circ r) \circ w = s \circ z. \end{align*} They also have the same image under $b$, since \begin{align*} b \circ (r \circ w) = (b \circ r) \circ w = (t \circ a) \circ w = t \circ (a \circ w) = t \circ y = b \circ z. \end{align*} Thus $r \circ w$ and $z$ are both mediating morphisms for the same compatible pair $(s \circ z, b \circ z)$ over the right pullback. By uniqueness in the right pullback, \begin{align*} r \circ w = z. \end{align*} Hence $w$ satisfies \begin{align*} r \circ w = z, \qquad a \circ w = y, \end{align*} so $w$ is a mediating morphism for the original cone $(z,y)$ over the left square. [/step] [step:Use uniqueness in the outer pullback to prove the left square is a pullback] We prove uniqueness of the mediating morphism for the left square. Let \begin{align*} w': N \to A \end{align*} be any morphism satisfying \begin{align*} r \circ w' = z, \qquad a \circ w' = y. \end{align*} Then \begin{align*} (s \circ r) \circ w' = s \circ z, \qquad a \circ w' = y. \end{align*} Thus $w'$ and $w$ are both mediating morphisms for the same compatible pair $(s \circ z,y)$ over the outer pullback. By uniqueness in the outer pullback, \begin{align*} w' = w. \end{align*} Therefore the left square satisfies the universal property of the pullback. This proves the converse and completes the proof. [/step]