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Suppose we have the following morphism of short exact sequences in $R$-Mod:

$$\begin{matrix}0\to&L&\stackrel{f'} \to& M'&\stackrel{g'}\to &N' & \to 0\\ &\;||&&\downarrow\rlap{\scriptsize\alpha'}&&\;\downarrow\rlap{\scriptsize\alpha}\\ 0\to &L&\stackrel{f}\to& M&\stackrel{g}\to& N& \to 0 \end{matrix} $$

I have to show that the right square is a pullback.

What I have done: Suppose we have $K \in R$-Mod and $ \phi : K \to N' $, $\lambda : K \to M $ that verifiy $$\alpha \circ \phi = g \circ \lambda$$

For $r \in K $ exists $m' \in M' $ with $g'(m') = \phi(k) $ because $g'$ is surjective.

So I defined $z : K \to M' $ with $$z(k) = m'$$

But now I'm stuck, how to proceed ?

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You've made an arbitrary choice to pick $m'$: you could add any element of $\operatorname{im}(f')$. If you make the wrong choices, $z$ won't be a homomorphism and $\alpha'\circ z$ won't be equal to $\lambda$. Try to show that there's a unique choice for which $\alpha'\circ z=\lambda$, and that this is a homomorphism. – Jeremy Rickard Jun 18 '14 at 10:49
up vote 2 down vote accepted

If $(m,n') \in M \times_N N'$, choose a lift $m' \in M'$ of $n'$ and consider the image $\tilde{m} \in M$. Then $\tilde{m},m$ have the same image in $N$, hence there is a unique $l \in L$ such that $m=\tilde{m}+f(l)$. Then $m' + f'(l)$ is a the unique element of $M'$ which maps to $m$ and to $n'$.

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