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

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Category theory can be fun, but I hesitate to call it addictive... as for split exact sequences – just knowing that you have a sequence of morphisms that satisfies the defining equations means the sequence is exact. No ifs or buts. –  Zhen Lin Dec 24 '12 at 4:55
    
@ZhenLin:Yes,but i am stuck with why $kerT(g)=imT(f)$. If I can solve this equation,I will get clear with all. And I will know the second sequence is exact then. –  Andylang Dec 24 '12 at 5:02

1 Answer 1

up vote 5 down vote accepted

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.

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Perhaps not "precisely" in the literal sense; it seems that $r\circ s=0$ or $g\circ f=0$ (but not both) can be omitted. –  ashpool Sep 7 at 22:27

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