I'm stuck with a proposition which seems simple but I can't get through. I have $C,D$ abelian categories, $R:C\to D$ & $L:D \to C$ additive adjoint functors. I want to prove that $L$ is exact iff it preserves monomorphisms.
So I know $L$ is right exact since it's a left adjoint, it is additive so it preserves finites products. We only need to show that preserving monomorphism implies that $L$ preserves kernels.
That is I need to show that for any $f \in Hom_D(A,B)$ $L kerf=ker Lf$. Preserving monomorphisms implies that $L kerf \hookrightarrow ker Lf$ but I can't get more. I also tried to prove that $L kerf$ satisfies the universal property of $ker Lf$ with units and counits but I didn't succeed. Thanks for any hint
Given $0\to A\to B\to C\to 0$ short exact, from right exactness of $L$ we know that $LA\to LB\to LC\to 0$ is exact. That is, for exactness at $LB,$ the image of $LA\to LB$ is the kernel of $LB\to LC.$ So the only possible failure of $0\to LA\to LB\to LC\to 0$ to be exact would be for $LA\to LB$ to be non-monomorphic, that is, for $0\to LA\to LB$ to be non-exact. This is what happens, for instance, with a functor like $(-)\otimes \mathbb Z/2\mathbb Z,$ which turns the exact sequence $0\to \mathbb Z\stackrel{2}{\to} \mathbb Z\to \mathbb Z/2\mathbb Z\to 0$ into the inexact sequence $0\to \mathbb Z/2\mathbb Z\stackrel{0}{\to}\mathbb Z/2\mathbb Z\stackrel{1}{\to}\mathbb Z/2\mathbb Z\to 0$: the first map is still surjective on the (trivial) kernel of the second. But now we can apply the assumption that $L$ preserves monomorphisms to rule out this possibility.