Pullback of Commutative Triangle

Theorem
Let $\mathbf C$ be a metacategory.

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


 * $\begin{xy}\xymatrix@+1em@L+2px{

A' \ar[rr]^*{h_\alpha} \ar[dd]_*{\alpha'} & & A \ar[rd]^*{\gamma} \ar[dd]^(.4)*{\alpha}

\\ & B' \ar[ld]^*{\beta'} \ar[rr] |{\hole} ^(.3)*{h_\beta} & & B \ar[ld]^*{\beta}

\\ C' \ar[rr]_*{h} & & C }\end{xy}$

and that the two squares in it are pullback diagrams.

Then there is a unique morphism $\gamma': A' \to B'$ making the following commute:


 * $\begin{xy}\xymatrix@+1em@L+2px{

A' \ar[rr]^*{h_\alpha} \ar[dd]_*{\alpha'} \ar@{-->}[rd]^*{\gamma'} & & A \ar[rd]^*{\gamma} \ar[dd]^(.4)*{\alpha}

\\ & B' \ar[ld]^*{\beta'} \ar[rr] |{\hole} ^(.3)*{h_\beta} & & B \ar[ld]^*{\beta}

\\ C' \ar[rr]_*{h} & & C }\end{xy}$

Furthermore, $\gamma'$ makes the following a pullback:


 * $\begin{xy}\xymatrix@+.5em@L+2px{

A' \ar[r]^*{h_\alpha} \ar[d]_*{\gamma'} & A \ar[d]^*{\gamma}

\\ B' \ar[r]_*{h_\beta} & B }\end{xy}$

Proof
The first diagram above can be distorted into:


 * $\begin{xy}\xymatrix@+.5em@L+2px{

A' \ar@/^/[rrd]^*{\gamma \circ h_\alpha} \ar@/_/[ddr]_*{\alpha'}

\\ & B' \ar[r]^*{h_\beta} \ar[d]^*{\beta'} & B \ar[d]^*{\beta}

\\ & C' \ar[r]_*{h} & C }\end{xy}$

and since the square is a pullback, there is a unique $\gamma': A' \to B'$ as desired.

That:


 * $\begin{xy}\xymatrix@+.5em@L+2px{

A' \ar[r]^*{h_\alpha} \ar[d]_*{\gamma'} & A \ar[d]^*{\gamma}

\\ B' \ar[r]_*{h_\beta} & B }\end{xy}$

is a pullback follows immediately from the Pullback Lemma.