Pullback Lemma

Theorem
Let $\mathbf C$ be a metacategory.

Suppose that the following is a commutative diagram in $\mathbf C$:


 * $\begin{xy}\xymatrix@+0.5em{

F \ar[d]_*+{h''} \ar[r]^*+{f'} & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h}

\\ A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

Suppose furthermore that the right square is a pullback.

Then the left square is a pullback iff the outer rectangle is.

Necessary Condition
Suppose that the left square is a pullback.

To verify the outer rectangle to be a pullback, suppose we are given the following commutative diagram:


 * $\begin{xy}\xymatrix@+0.5em{

P \ar@/^1em/[rrrd]^*+{p_1} \ar@/_1em/[ddr]_*+{p_2}

\\ & & & D \ar[d]^*+{h}

\\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

The right square being a pullback implies there is a unique $v: P \to E$ making the following commute:


 * $\begin{xy}\xymatrix@+0.5em{

P \ar@/^1em/[rrrd]^*+{p_1} \ar@/_1em/[ddr]_*+{p_2} \ar@{-->}@/^/[rrd]_*+{v}

\\ & & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h}

\\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

Since the left square is also a pullback, we subsequently find a unique $u: P \to F$ making the following commute:


 * $\begin{xy}\xymatrix@+0.5em{

P \ar@/^1em/[rrrd]^*+{p_1} \ar@/_1em/[ddr]_*+{p_2} \ar@/^/[rrd]_*+{v} \ar@{-->}[rd]_*+{u}

\\ & F \ar[d]_*+{h''} \ar[r]^*+{f'} & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h}

\\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

In conclusion, the outer rectangle is a pullback.