# Commuting two pullbacks

I have stumbled upon some interesting exercise whilst reading the "Category Theory for Scientists" book. Below is the universal property of fiber products: By using the universal property, I can try placing some set A, connect it to the X, X', Y, Y' through some functions and thus have two unique arrows spawning from it to the W and W'. However, nothing that could be used to create a direct connection between the W and W' comes out from that.

Any suggestions how this could be solved?

Thanks!

## 2 Answers

Consider the morphisms $W \to Y \to Y'$ and $W \to X \to X'$. Evidently they induce the same morphism $W \to Z'$. Hence, the universal property of $W'$ tells us that that they induce a morphism $W \to W'$ such that everything commutes (we don't need even need that $W$ is a pullback).

• So basically the "W" pullback in the exercise can be looked at as the "A" set in the universal property definition, and the W -> Y -> Y' and W -> X -> X' morphisms as the "f" and "g" functions? :) – spacemonkey Feb 7 '14 at 0:11
• Yes. Remember that names of variables can be freely interchanged. – Martin Brandenburg Feb 7 '14 at 0:46

Hint: Since the square on the face behind is a pullback, look for a pair of maps $W\to Y',\ W\to X'$ to form a commuting square with $Y'\to Z'$ and $X'\to Z'$.