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[parent] monomorphisms are pullback stable (Theorem)

In this entry we show that the pullback of a monomorphism is also a monomorphism. Let $m:T\to Y$ be a monomorphism, let $f:X\to Y$ be an arbitrary morphism, and suppose that $S=T\times_Y X$ is a pullback (refer to the diagram below) with $\hat{m}:S\to X$ and $\hat{f}:S\to T$ the pullback projections. Let $a,b:A\to S$ be morphisms such that

\begin{displaymath}\hat{m} a = \hat{m} b =d.\end{displaymath}

Our goal is to show that, necessarily, $a=b$.

By the commutativity of the pullback square, we have

\begin{displaymath}m \hat{f} a = m \hat{f} b.\end{displaymath}

Since $m$ is a mono, this implies that
\begin{displaymath}\hat{f} a = \hat{f} b;\end{displaymath}

we name this morphism $c$. By construction,
\begin{displaymath}fd = mc .\end{displaymath}

Hence, by the universality of the pullback there exists a unique morphism from $A$ to $S$ that makes everything commute. Hence, both $a$ and $b$ are equal to that morphism, and hence are equal to each other.
Figure: pullback of a monomorphism


\includegraphics[width=5cm]{pullbackofmonomorphism-fig}

Bibliography

1
J. Adámek, H. Herrlich and G.E. Strecker, ''Abstract and Concrete Categories'', http://katmat.math.uni-bremen.de/acc/acc.pdf



"monomorphisms are pullback stable" is owned by rmilson.
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Cross-references: universality, implies, pullback square, commutativity, projections, morphism, monomorphism, pullback
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This is version 10 of monomorphisms are pullback stable, born on 2006-02-18, modified 2006-06-06.
Object id is 7633, canonical name is PullbackOfAMonomorphismIsAMonomorphism.
Accessed 838 times total.

Classification:
AMS MSC: 18A20 (Category theory; homological algebra :: General theory of categories and functors :: Epimorphisms, monomorphisms, special classes of morphisms, null morphisms)
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