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Let $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ be a split-exact short exact sequence of $R$-modules, where $R$ is any ring. Let $T$ be an additive functor from $R$-modules to abelian groups. Then is it true that we still get a split-exact short exact sequence $0\rightarrow TA'\rightarrow TA\rightarrow TA''\rightarrow 0$ of abelian groups? (I just don't understand the reason why the zeros at both ends are preserved by $T$)

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A split short exact sequence is preserved by any additive functor between any two abelian categories whatsoever, because being a split short exact sequence is an equational condition. –  Zhen Lin Feb 29 '12 at 21:14

2 Answers 2

Because for $R$-modules $A\rightarrow A''$ has a right inverse if and only if $A'\rightarrow A$ has a left inverse!

Sometimes split-exact is also defined to mean that (up to isomorphism) $A=A'\oplus A''$ and the maps are inclusion and projection.

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Say $$0\to A'\stackrel{f}{\longrightarrow} A \stackrel{g}{\longrightarrow} A''\to 0$$ is split (and exact); then there exist $p\colon A\to A'$ and $q\colon A''\to A$ such that $gq=\mathrm{Id}_{A''}$ and $pf=\mathrm{Id}_{A'}$. In particular, $T(p)\circ T(f) = T(\mathrm{Id}_{A'}) = \mathrm{Id}_{T(A')}$, so $T(f)$ has a left inverse; therefore, $T(f)$ must be one-to-one.

Similarly, $T(g)\circ T(q) = T(\mathrm{Id}_{A''}) = \mathrm{Id}_{T(A'')}$, so $T(g)$ has a right inverse; therefore, $T(g)$ is onto.

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