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# properties of pullback

This entry lists and proves some basic properties of categorical pullbacks. Fix a category $\mathcal{C}$.

###### Proposition 1.

If $P$ is a pullback of $f:A\to C$ and $g:B\to C$, then $P$ is unique (up to isomorphism). This is the justification that $P$ may be written as $A\times_{C}B$.

###### Proof.

Let

$\xymatrix@+=1.5cm{P\ar[r]^{-}{p_{A}}\ar[d]_{{p_{B}}}&A\ar[d]^{f}\\ B\ar[r]_{g}&C}$ |

be the corresponding pullback diagram. First note that if $h:P\to P$ is a morphism such that $p_{A}=p_{A}\circ h$ and $p_{B}=p_{B}\circ h$, then $h=1_{P}$. This follows from the universal property of pullbacks (for more detail, see the proof of the uniqueness of product here).

Now if $Q$ is another pullback of $f:A\to C$ and $g:B\to C$ with pullback diagram

$\xymatrix@+=1.5cm{Q\ar[r]^{-}{q_{A}}\ar[d]_{{q_{B}}}&A\ar[d]^{f}\\ B\ar[r]_{g}&T}$ |

then there are unique morphisms $x:P\to Q$ and $y:Q\to P$ such that $p_{A}=q_{A}\circ x$, $p_{B}=q_{B}\circ x$ and $q_{A}=p_{A}\circ y$, $q_{B}=p_{B}\circ y$. So $p_{A}=p_{A}\circ(y\circ x)$ and $p_{B}=p_{B}\circ(y\circ x)$. Therefore, $y\circ x=1_{P}$. Similarly $x\circ y=1_{Q}$. As a result $P$ is isomorphic to $Q$. ∎

###### Proposition 2.

Let $I$ be the disjoint union of non-empty sets $J,K$. Let $I^{{\prime}}=\{x_{i}:C_{i}\to C\mid i\in I\}$ be a set of morphisms in $\mathcal{C}$. Let $X,Y$, and $Z$ be the generalized pullbacks of $I^{{\prime}}$, $J^{{\prime}}=\{x_{i}\mid i\in J\}$, and $K^{{\prime}}=\{x_{i}\mid i\in K\}$ respectively. Then $X\cong Y\times_{C}Z$.

Note: we are not asserting the existence of $X,Y$ and $Z$. We are merely saying that if they exist, we have an isomorphism.

###### Sketch of Proof..

This proof is analogous to the proof of a similar property regarding arbitrary products (see here), so we will skip the diagrams and be brief here. First, note that there are unique morphisms $y:X\to Y$ and $z:X\to Z$, which results in a unique morphism $f:X\to Y\times_{C}Z$. On the other hand, the morphisms $Y\to C_{j}$ and $Z\to C_{k}$ give us a well-defined collection of morphisms $Y\times_{C}Z\to C_{i}$ for all $i\in I$, which results in a unique morphism $g:Y\times_{C}Z\to X$. There are a number of commutative diagrams relating $f$ and $g$. In the end, one proves that $g\circ f=1_{C}$ and $f\circ g=1_{{Y\times_{C}Z}}$. ∎

###### Corollary 1.

(commutativity of pullbacks) $X\times_{C}Y\cong Y\times_{C}X$, provided that one of them (and hence the other) exists.

###### Corollary 2.

(associativity of pullbacks) $X\times_{C}(Y\times_{C}Z)\cong X\times_{C}Y\times_{C}Z\cong(X\times_{C}Y)% \times_{C}Z$, provided that they exist.

###### Corollary 3.

If $\mathcal{C}$ has pullbacks, then it has finite generalized pullbacks.

###### Proposition 3.

Let $\{x_{i}:C_{i}\to C\mid i\in I\}$ be a collection of morphisms indexed by a set $I$. Let $\alpha,\beta:J\to I$ be two surjections. Suppose $D,E$ are the generalized pullbacks of $\{x_{{\alpha(j)}}\mid j\in J\}$ and $\{x_{{\beta(k)}}\mid k\in J\}$ respectively. Then $D\cong E$.

###### Sketch of Proof..

For every $i\in I$, there are $j,k\in J$, such that $\alpha(j)=\beta(k)$. So

$E\to C_{{\beta(k)}}=E\to C_{{\alpha(j)}},$ |

and its composition with $x_{{\beta(k)}}$ is the same as the composition with $x_{{\alpha(j)}}$. By the universality of generalized pullbacks, we get a unique morphism $f:E\to D$ with

$E\stackrel{f}{\longrightarrow}D\longrightarrow C_{{\beta(k)}}=E\to C_{{\beta(k% )}}.$ |

Dually, we have a unique morphism $g:D\to E$ with

$D\stackrel{g}{\longrightarrow}E\longrightarrow C_{{\alpha(j)}}=D\to C_{{\alpha% (j)}}.$ |

As a result, $f\circ g=1_{D}$ and $g\circ f=1_{E}$. ∎

## Mathematics Subject Classification

18A30*no label found*

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