Prove $\beta$ is surjective using commutative group diagram

Question

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.

Answer 1

I'll assume that these groups are all additive Abelian groups. If they aren't then you will have to amend accordingly

You have $\phi'(b')=\phi'(\beta(b))$. Rather than split cases according to whether $b'$ is in the kernel, instead note that $\phi'(b'-\beta(b))\in\text{Ker}(\phi')$ so $b'-\beta(b)=\psi'(a')$ for some $a'\in A'$. Now pick up the diagram chase again.

(By the way I do find the $\ni$ notation very hard to read.)

Answer 2

Write $b'=\beta(b)\tilde{b}$ for some $\tilde{b}\in B'$ (namely $\tilde{b}=b'(\beta(b))^{-1}$). Note if $b'=\beta(b)$, then $\tilde{b}$ is just the identity. Now you know

$$\varphi'(b')=\varphi'(\beta(b)\tilde{b})=\varphi'(\beta(b))\varphi'(b) = \varphi'(b')\varphi'(\tilde{b}) \Longrightarrow 1=\varphi'(\tilde{b})$$

so $\tilde{b}\in \ker(\varphi')$. Now use the fact which you proved that $\beta$ maps to the entire kernel of $\varphi'$.