---
tag: 60A5CC7S3A
kind: statement
role: lemma
name: Pullback pasting lemma
msc: [18A30]
uses: [A2DE66937D, WV898P4X16, Q05K04QJ71]
about: [Q05K04QJ71]
sources:
  - url: https://en.wikipedia.org/wiki/Pullback_(category_theory)
---

## Statement

Let $\mathcal{C}$ be a [[category|A2DE66937D]], and consider a [[commutative diagram|WV898P4X16]] of two squares side by side:
$$\begin{array}{ccccc} A & \xrightarrow{a} & B & \xrightarrow{b} & C \\ \downarrow{\scriptstyle u} & & \downarrow{\scriptstyle v} & & \downarrow{\scriptstyle w} \\ D & \xrightarrow{d} & E & \xrightarrow{e} & F \end{array}$$
so that $v \circ a = d \circ u$ and $w \circ b = e \circ v$.
Assume that the right square is a [[pullback|Q05K04QJ71]].
The outer rectangle has top side $b \circ a$, bottom side $e \circ d$ and vertical sides $u$ and $w$.
Then the left square is a pullback if and only if the outer rectangle is a pullback.

## Remarks

Both directions are checked with the universal property.
A cone over the outer rectangle gives, through the right pullback, a cone over the left square, and back again.
The uniqueness parts match because the right square is a pullback.
In the category of sets, a pullback along an inclusion is a preimage.
The lemma then says that the preimage under $e \circ d$ is the preimage under $d$ of the preimage under $e$.

The hypothesis on the right square is needed.
Let $A$, $C$, $D$, $E$ and $F$ be one-point sets, let $B$ have two points, and let $a$ pick one of them.
The outer rectangle is then a pullback, but the left square is not, because the pullback of $B \to E \leftarrow D$ has two points while $A$ has one.

## Explanations

### L1 human

Pullbacks can be stacked.
If you glue a pullback square to a second square, the big rectangle is a pullback exactly when the second square is.
It means that pulling back in two steps is the same as pulling back once along the composite map.

### L2 human

The lemma is the main tool for computing with pullbacks in diagrams.
It shows, for example, that pullback along $g \circ f$ is the same as pullback along $g$ followed by pullback along $f$, up to isomorphism.
In algebraic geometry it gives the transitivity of base change, $(X \times_S S') \times_{S'} S'' \cong X \times_S S''$.
The dual statement holds for pushouts: if the left square is a pushout, then the right square is a pushout if and only if the outer rectangle is.