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I am self-reading a book, and was stuck in proving this property that the pullback of a monomorphism is a monomorphism. I am still not very familiar with how to use the universal mapping property (which I guess I need in this) to prove things, though I can understand some examples of using UMP.

Searching around, I only found this Math SE answer about what it means for pullbacks to preserve monomorphisms (which is the same as what the book says). Based on that answer:

Given a pullback square, $$\require{AMScd} \begin{CD} X' @> p_1 >> X \\ @V p_2 VV @VV f V \\ Y' @>> g > Y \end{CD}$$ if $f: X \to Y$ is a monomorphism, then $p_2: X' \to Y'$ is also a monomorphism.

How can I prove this property?

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1 Answers1

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Suppose $f$ is a monomorphism, and suppose we have some object $Z$ with two maps $\alpha, \beta: Z \to X'$ such that $p_2 \alpha = p_2 \beta$. We must show that $\alpha = \beta$. By hypothesis, it follows that $g p_2 \alpha = g p_2 \beta$, so $f p_1 \alpha = f p_1 \beta$ since the diagram commutes. But $f$ is a monomorphism, so $p_1 \alpha = p_1 \beta$. Now $\alpha$ and $\beta$ agree after applying either $p_1$ or $p_2$, so by the universal property of the fibered product, $\alpha = \beta$ as desired.

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