# Additive functors preserve direct summands

My question is simple - I just can't seem to prove it for the life of me. Let $$0\rightarrow L \rightarrow M \rightarrow N\rightarrow 0$$ be a split short exact sequence of $R$-modules. Let also $F:R-Mod \rightarrow R-Mod$ be an additive covariant endofunctor. Now, I know that $$0\rightarrow FL \rightarrow FM \rightarrow FN\rightarrow 0$$is also a split short exact sequence since $F$ is additive.

My question is: Suppose that $M$ and $FM$ are isomorphic as $R$-modules. How can I show that $L$ and $FL$, as well as $N$ and $FN$, are also isomorphic?

• are you saying additive functor is exact? – user312648 Sep 16 '17 at 18:30

I doubt this. Let $F(A)=A\oplus A$ (with obvious effect on homomorphisms) and $M$ the direct sum of infinitely many copies of $R$, $L=R$ viewed as first componentn, i.e., a direct summand. Then $M\cong F(M)$, but $R=L\not\cong FL=R^2$