How to show right adjoint additive functor $F:$ $R$-mod $\longrightarrow$ $R$-mod is left exact? Specifically, show that $\ker g_* = Im(f_*)$?

Question

If $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ is an exact sequence of $R$-modules ($R$ commutative), then we have a complex $0 \to F(A) \xrightarrow{f_*} F(B) \xrightarrow{g_*} F(C)$ (I have shown that $F(0)=0$ from the adjunction, and also $g_* \circ f_* = (g\circ f)_*=0$).

Since right adjoints commute with limits, then $\ker f_* = F(\ker f) = F(0)=0$.

However, I am having trouble showing that $\ker g_* = Im(f_*)$.

We can see that $\ker g_* \cong F(\ker g)=F(Im(f)) \cong F(A) \cong Im(f_*)$.

But we need to show $\ker g_* = Im(f_*)$; not just isomorphic.

How can I proceed?

Answer 1

Claim : $0\to A\to B\to C$ is exact if and only if $A\to B\rightrightarrows C$ is an equalizer.

Since $F$ is a right adjoint, it preserves equalizers, so this one in particular. So $F(A)\to F(B)\rightrightarrows F(C)$ is an equalizer, and so $0\to F(A)\to F(B)\to F(C)$ is exact.

To prove the claim is reasonably easy : check that $\ker(g)$ is the equalizer, so the sequence is exact if and only if $A$ is the equalizer.