Proving injectivity in a commutative diagram

Question

I have been given the following exercise:
Consider the following commutative diagram of modules over a ring R with exact rows:\

Let $\alpha$ and $g$ be surjective. Fix $b′ \in B′$ and $c \in C$ satisfying $g′(b′) = \gamma(c)$. Show that there exists $b \in B$ such that $\beta(b) = b′$ and $g(b) = c$.

I started by using the fact that $g$ is surjective. Thus $\exists b \in B: g(b) = c$.
Then $(\gamma \circ g)(b) = \gamma(c) \implies (g' \circ \beta)(b) = g'(b')$
Thus $g'(\beta(b)) = g'(b')$.

Now I need to use the rest of what I'm given to prove that $\beta(b) = b'$. One possible solution is to prove that $g'$ is injective. However I don't know how to continue from here. Can anyone help me?

Answer 1

As $g$ is surjective, there is a $b\in B$ such that $g(b)=c$. Then $g'(\beta(b)-b')=0$, because $g'(\beta(b))=\gamma(g(b))=\gamma(c)=g'(b')$. But the lower row is exact, so $\beta(b)-b'\in \operatorname{im}f'$, i.e. $f'(a')=\beta(b)-b'$ for some $a'\in A'$. Since $\alpha$ is surjective, there is an $a\in A$ satisfying $\alpha(a)=a'$. Consider now $b-f(a)\in B$:

• $g(b-f(a))=g(b)=c$ as $g\circ f=0$;
• $\beta(b-f(a))=\beta(b)-\beta(f(a))=$ $\beta(b)-f'(\alpha(a))=\beta(b)-f'(a')=$ $\beta(b)-(\beta(b)-b')=b'$.