Module Homomorphisms and a Direct Sum

Question

I am trying to show the following:

Let $f:M\to N$ and $g:N\to M$ be module homomorphisms such that $g\circ f={\rm Id}_M$. Prove that $N={\rm im}(f)\oplus\ker(g)$.

I know that $M/\ker(f)\cong{\rm im}(f)$, but I'm not sure if this is even helpful. I also think that $g$ must be surjective, so $N/\ker(g)\cong M$, and that is a sum of what I've determined.

Any suggestions would be helpful!

Answer 1

$\def\im{\operatorname{im}}$*Hint*: I don't see why you use quotients. The direct way should work: To show $N = \im f \oplus \ker g$ you have to show two things:

• $\ker g \cap \im f = 0$ (if $g(n) = 0$ and $f(m) = n$, we have $m = \mathrm{id}_M(m) = \cdots$.
• $N = \ker g + \im f$. For $n \in N$, consider $(f \circ g)(n) \in \im f$. What can you say about $n - (f\circ g)(n)$?