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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.2012-02-29

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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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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.