I'm working on a problem from Dummit and Foote's Abstract Algebra, 3rd Edition, reproduced below.
Suppose that this is a commutative diagram of groups and that the rows are exact.
Prove that if $\phi, \alpha,$ and $\gamma$ are surjective, then $\beta$ is surjective.
Here's what I've got so far:
If $b'\in B'$ then the surjectivity of $\gamma$ implies $\exists$ $ c\in C$ such that $\gamma(c)=\varphi'(b')$.
The surjectivity of $\varphi$ implies $\exists$ $ b\in B$ such that $\varphi(b)=c$.
Thus, $\varphi'(b')=\gamma(\varphi(b))$ and since the diagram commutes $\gamma(\varphi(b))=\varphi'(\beta(b))$.
This gives us that $\varphi'(b')=\varphi'(\beta(b))$
But we don't have that $\varphi'$ is injective, so we don't know that $b'=\beta(b)$, which would give us surjectivity of $\beta$.
Secondly, if $b'\in Ker(\varphi')$ then since the rows are exact we know that for some $a'\in A'$ it is true that $\psi'(a)=b'$.
Also, since $\alpha$ is surjective we have that $\exists$ $a\in A$ such that $ \alpha(a)=a'$.
Thus, $\psi'(\alpha(a))=b'$, and since the diagram commute we also have that $\beta(\psi(a))=\psi'(\alpha(a))=b'$.
Thus, we at least know that $\beta$ maps to the entire kernel of $\varphi'$.
But what about when $b'\notin Ker(\varphi')$?
Hope you guys can help me, because I really don't know what I'm missing.