Given a short exact sequence of abelian groups, \begin{eqnarray} 0\rightarrow A\xrightarrow{h}B\xrightarrow{k}C\rightarrow0, \end{eqnarray} we tensor it with an arbitrary abelian group $G$. How to prove (in an elementary way starting from definitions) the following sequence is exact: \begin{eqnarray} A\otimes G\xrightarrow{h\otimes 1}B\otimes G\xrightarrow{k\otimes 1}C\otimes G\rightarrow0, \end{eqnarray} where the leftmost zero inclusion has been discarded.
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1See Prop. 2.18 in Atiyah-Macdonald for example. This can help too: Proving that the tensor product is right exact – wormram Jul 31 '22 at 02:23
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@Michael Many thanks and that's really helpful! – Smart Yao Jul 31 '22 at 02:29
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1No problem. Those resources prove it for $R$-modules where $R$ is a ring, but the case for abelian groups is more simple: just forget the $R$-multiplication. – wormram Jul 31 '22 at 02:30