$0\to A'\stackrel{f}{\longrightarrow} A \stackrel{g}{\longrightarrow} A''\to 0$ 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.
Best Answer
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.