$0\to A'\stackrel{f}{\longrightarrow} A \stackrel{g}{\longrightarrow} A''\to 0$ is a short split exact sequence, where $A'$, $A$, $A''$ are $R$-modules, and $T$ is an additive functor from $R$-$\mathsf{mod}$ to $\mathsf{Ab}$.
Then we have the sequence $0\to TA'\stackrel{T(f)}{\longrightarrow} TA \stackrel{T(g)}{\longrightarrow} TA''\to 0$ is a split exact sequence again.
I know that the second sequence is split. That $T(f)$ is injective and $T(g)$ is surjective is also clear for me too.
My question is why $\ker T(g)=\mathrm{im}T(f)$.
1. Some comments I found from the Internet told that an additive functor preserves binary direct sum.Does it help here? I know very few about module theory.I don't know whether this is the reason why I get stuck here.
2.I have checked from this math.SE post. But the answer there seems to be not suitable for my question.
Thanks in advance.