Take a commutative diagram in some category:

such that the right square is a pullback.

Theorem: The left square is a pullback if and only if the outer square is a pullback.

Proof: First, assume the outer square is a pullback. Let's label things:

So we have that $(Q, q_1, q_2)$ is a pullback over $B \xrightarrow{f} Z \xleftarrow{g} A$, and $(P, q_1 \circ p_1, p_2)$ is a pullback over $C \xrightarrow{f \circ h} Z \xleftarrow{g} A$. Take an arbitrary cone $(R, j, k)$ over $C \xrightarrow{h} B \xleftarrow{q_2} Q$. We need to produce a unique map (dotted) so that

commutes. Notice that $(R, q_1 \circ k, j)$ is a cone over $C \xrightarrow{f \circ h} Z \xleftarrow{g} A$. So we have a unique $\varphi_1 : R \rightarrow P$ making:

commute. We claim that $\varphi_1$ is exactly the unique map that makes (2) commute. The left triangle would commute by (3), so we need only show the top triangle of (2) -- that is $p_1 \circ \varphi_1 = k$.

To get more information, consider that $(R, q_1 \circ k, h \circ j)$ is a cone over $B \xrightarrow{f} Z \xleftarrow{g} A$. So we have a unique $\varphi_2 : R \rightarrow Q$ making:

commute.

Note that $p_1 \circ \varphi_1$ is another map from $R \rightarrow Q$. We claim that if we replace $\varphi_2$ by this map in (4), the diagram still commutes. To see this, note that the top triangle becomes that of (3) written slightly differently. For the left triangle, we need to verify $q_2 \circ p_1 \circ \varphi_1 = h \circ j$. Note that $q_2 \circ p_1 = h \circ p_2$ by (1), and $p_2 \circ \varphi_1 = j$ by (3), so $q_2 \circ p_1 \circ \varphi_1 = h \circ p_2 \circ \varphi_1 = h \circ j$.

Similarly, we can replace $\varphi_2$ with $k : R \rightarrow Q$ in (4) and the diagram still commutes. The top triangle commutes trivially. The left triangle commutes because it is really the square:

which commutes precisely because $(R, j, k)$ is a cone over $C \xrightarrow{h} B \xleftarrow{q_2} Q$.

Finally then, since $\varphi_2$ is unique we have $p_1 \circ \varphi_1 = \varphi_2 = k$. So the top of (2) commutes. So, (2) commutes if $\varphi_1$ is the dotted map, and $\varphi_1$ is unique, so the left square is a pullback.

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Now, suppose that both the left and right square are pullbacks. With the same labeling, we are now assuming that $(Q, q_1, q_2)$ is a pullback over $B \xrightarrow{f} Z \xleftarrow{g} A$, and $(P, p_1, p_2)$ is a pullback over $C \xrightarrow{h} B \xleftarrow{q_2} Q$. We wish to show the outer square is a pullback, so let $(R, j, k)$ be an arbitrary cone over $C \xrightarrow{f \circ h} Z \xleftarrow{g} A$. We need to produce a unique map (dotted) so that:

commutes. Since $(R, k, h \circ j)$ is a cone over $B \xrightarrow{f} Z \xleftarrow{g} A$, we have a unique $\varphi_1 : R \rightarrow A$ making:

commute. Now, we have that $(R, \varphi_1, j)$ is a cone over $C \xrightarrow{h} B \xleftarrow{q_2} Q$. So we have a unique $\varphi_2$ making:

commute. But then the whole diagram:

commutes. Of course, (5) is just a subdiagram of this, so taking the (unique) map $\varphi_2$ makes (5) commute, and the outer rectangle is a pullback.