TITLE: Additive functor over a short split exact sequence.
QUESTION [8 upvotes]: $0\to A'\stackrel{f}{\longrightarrow} A \stackrel{g}{\longrightarrow} A''\to 0$ 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.

REPLY [9 votes]: Suppose we have morphisms $f : A' \to A$, $g : A \to A''$, $r : A \to A'$, $s : A'' \to A$ in an abelian category. We have a split exact sequence if and only if these equations hold:
\begin{align}
r \circ f & = \textrm{id}_{A'} &
g \circ f & = 0 \\
r \circ s & = 0 &
g \circ s & = \textrm{id}_{A''} \\
\end{align}
$$f \circ r + s \circ g = \textrm{id}_A$$
These are precisely the same equations required to make $A$ into the direct sum $A' \oplus A''$, and in particular
$$0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0$$
is an exact sequence. It is clear that $f$ is monic and $g$ is epic, and $g \circ f = 0$ implies $\operatorname{im} f \subseteq \ker g$; we only need to check that $\operatorname{im} f \supseteq \ker g$ now. So suppose $g \circ x = 0$ for some $x : X \to A$. Then, $s \circ g \circ x = 0$ as well, so 
$$x = (f \circ r + s \circ g) \circ x = f \circ r \circ x$$
and therefore $\ker g \subseteq \operatorname{im} f$ indeed.