Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
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 at 10:49

1 Answer 1

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 $n'$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.