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Let $$0\longrightarrow L^{(i)}\longrightarrow M^{(i)}\longrightarrow N^{(i)}\longrightarrow 0$$ be a short exact sequence of abelian groups for every index $i$. Clearly if I take finite direct products, then $$0\longrightarrow \prod_iL^{(i)}\longrightarrow\prod_i M^{(i)}\longrightarrow \prod_iN^{(i)}\longrightarrow 0$$ is a short exact sequence. But what about infinite direct product? Is the exactness preserved?

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up vote 3 down vote accepted

In a general (abelian) category, the product of epimorphisms (if it exists!) may not be an epimorphism – that is why we sometimes assume (AB4*) as an extra axiom. As it turns out, the category of $R$-modules (for any ring $R$) always satisfies the (AB4*) axiom.

Proposition. If the (AB4*) axiom is satisfied in an abelian category, then products of short exact sequences are short exact sequences.

Proof. Consider short exact sequences $$0 \longrightarrow L^{(i)} \longrightarrow M^{(i)} \longrightarrow N^{(i)} \longrightarrow 0$$ By general abstract nonsense, one can show that the kernel of a product is the product of the kernels, so we have a left exact sequence $$0 \longrightarrow \prod_i L^{(i)} \longrightarrow \prod_i M^{(i)} \longrightarrow \prod_i N^{(i)}$$ but by the (AB4*) assumption, the last homomorphism is an epimorphism, so we in fact have a short exact sequence.

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+1, and a gold star for "general abstract nonsense". –  Cameron Buie Jun 19 '12 at 18:50

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