$R$ ring, $R-MOD$ is the category of $R-$modules anf $F:R-Mod\to R-Mod$ a functor such that the induced map$$Hom(M,N)\to Hom(F(M),F(N))$$ is a homomorphism of abelian groups. $F$ satisfies $kerF(f)\cong F(ker(f))$ for all $R-MOD$-morphisms $f:M\to M''$.
My question: If $$0\to M'\to M\to M''$$ is an exact sequence, why is $$0\to F(M')\to F(M)\to F(M'')$$ exact?
I know how to prove that $F(M')\to F(M)$ is injective I think: If we have an exact sequence $0\to M'\to M\to M''$ with $g:M'\to M$, it is $F(ker(g))=ker(F(g))=ker(F(M')\to F(M))$ and $F(ker(g))=0$. So we obtain the exactness in $F(M')$.
How to prove the exactness in $F(M)$?
Edit: For $f:M\to M''$ it is $f\circ g=0$. Therefore and because $F$ is a homomorphism of groups, which implies $F(0)=0$, it is $F(f)\circ F(g)=0$. We obtain $Im(F(g))\subseteq Ker(F(g))$. But how to prove the other inclusion?
Is $Im(F(g))=F(Im(g))$? If yes, I know how to prove it.
Edit 2: Ok, everything is clear now: $g $ is injective and therefore $g:M'\to Im(g)$ is an isomorphism. It follows that $F(g):F(M')\to F(Im(g))$ is an isomorphism, we get $ImF(g)=F(Im(g))$