# 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.

## Proof

### Necessary Condition

Suppose that the left square is a pullback.

To verify the outer rectangle is 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.

$\Box$

### Sufficient Condition

Suppose that the outer rectangle is a pullback.

To verify the left square is a pullback, suppose we are given the following commutative diagram:

$\begin{xy}\xymatrix@+0.5em{ P \ar@/^/[rrd]^*+{p_1} \ar@/_1em/[ddr]_*+{p_2} \\ & & E \ar[d]^*+{h'} \\ & A \ar[r]_*+{f} & B }\end{xy}$

Evidently, we can amend this diagram as follows:

$\begin{xy}\xymatrix@+0.5em{ P \ar@/^/[rrd]_*{p_1} \ar@/^1em/[rrrd]^*+{g' \circ p_1} \ar@/_1em/[ddr]_*+{p_2} \\ & & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h} \\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

Because the outer rectangle is a pullback, we thus obtain a unique $u: P \to F$ making the following commute:

$\begin{xy}\xymatrix@+0.5em{ P \ar@/^1em/[rrrd]^*+{g' \circ p_1} \ar@/_1em/[ddr]_*+{p_2} \ar@{-->}[rd]_*+{u} \\ & F \ar[d]_*+{h''} \ar[r]^*+{f'} & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h} \\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

If we now can prove that $f' \circ u = p_1$, then we're done.

To this end, remark that they both complete the following commutative diagram:

$\begin{xy}\xymatrix@+0.5em{ P \ar@{-->}[rrd] \ar@/^1em/[rrrd]^*+{g' \circ p_1} \ar@/_1em/[ddr]_*+{p_2} \\ & & E \ar[d]^*+{h'} \ar[r]^*+{g'} & D \ar[d]^*+{h} \\ & A \ar[r]_*+{f} & B \ar[r]_*+{g} & C }\end{xy}$

Since the right square is a pullback, there is only one such morphism.

Therefore, we conclude $f' \circ u = p_1$.

Hence the left square is also a pullback.

$\blacksquare$