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I am trying to show the following:

Let $f:M\to N$ and $g:N\to M$ be module homomorphisms such that $g\circ f=Id_M$. Prove that $N=Im(f)\oplus\ker(g)$.

I know that $M/\ker(f)\cong Im(f)$, 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!

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$\def\im{\operatorname{im}}$*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 $n - (f\circ g)(n)$?
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    Okay, the first part certainly follows since $f(m)=n=f(0)$ and the additive identity must map to the additive identity. I see my problem on the second part, thanks for pointing it out!2012-12-07