Epimorphism is surjection in category of Abelian groups

Question

I was trying to prove that epimorphisms are surjections in the category of abelian groups.

I do not know how I should start.

Assume $f:A \rightarrow B$ is an epimorphism, where $A,B$ are abelian groups. Then $\beta' \circ f=\beta''\circ f \implies \beta'=\beta''$.

Now I want to show that this implies for each $b \in B$, there is an $a \in A$ such that $f(a)=b$. I am stuck and do not know how to proceed.

Should I look at specific functions $\beta',\beta''$ in order to come up with a solution? Does this require a completely different approach?

Any hints are appreciated. I was looking at help on this link. Do I take $β′$ to be quotient map from $B$ to $B/im(f)$? Then I take $β′′$ to be the map $β′′:B→{0}$. Then since $β′∘f=β′′∘f$ implies $β′=β′′$ does this imply that $f$ is surjective?

Answer 1

Your idea is nearly correct! Looking at specific choices of $\beta'$ and $\beta''$ is the right idea. However, $\beta'$ and $\beta''$ must have the same codomain in order to have the equality $\beta'\circ f = \beta''\circ f$ make sense.

The choice of $\beta'$ you had initially (the projection $B\to B/\operatorname{im}(f)$) is good, but the choice of $\beta''$ needs to be modified slightly (in your initial post, it was the map $B\to\{0\}$). Do you see how to choose $\beta''$ so that $\beta''$ is also a map $\beta'' : B\to B/\operatorname{im}(f),$ but will still guarantee that $f\circ\beta''(a) = f\circ\beta'(a)$ for all $a\in A$?

Hint: Think about what the composition $\beta'\circ f$ is!

Try to figure this out for yourself before looking at the spoilers below! The first box describes an appropriate choice of $\beta'',$ and the second outlines why this shows that $f$ must be a surjection.

Let $\beta''$ be the map which sends each $b\in B$ to $0 + \operatorname{im}(f)\in B/\operatorname{im}(f)$ -- i.e., $\beta''$ is the trivial (zero) morphism from $B$ to $B/\operatorname{im}(f).$

Since $f$ is an epimorphism, we must have $\beta' = \beta'',$ so that $\beta'(b) = \beta''(b) = 0 + \operatorname{im}(f)$ for all $b\in B.$ But $\beta'$ is the canonical projection $B\to B/\operatorname{im}(f),$ so that $\beta'(b) = b + \operatorname{im}(f).$ This implies that $b + \operatorname{im}(f) = \operatorname{im}(f)\in B/\operatorname{im}(f)$ for all $b\in B.$ Unwinding the definition of equality in a quotient group, this means $b\in\operatorname{im}(f)$ for all $b\in B,$ which is precisely what was to be shown.