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Let $F=\operatorname {Hom}(S^2(\,\_\,),\mathbb{Q}/\mathbb{Z}):\mathsf{Ab} \to \mathsf{Ab}$ be the functor that sends an abelian group to the group of symmetric bilinear forms on this group. As far as I understood, this is neither left nor right exact, but preserves epimorphisms.

However, let $0 \to K \overset{\alpha}{\longrightarrow} G \overset{\beta}{\longrightarrow} H \to 0$ be an exact sequence of abelian groups. This question (and its comments) suggest that (after dualization) there is an exact sequence induced by the Koszul complex of the symmetric algebra $R=S(G)$: $$ 0 \to F(H) \to F(G) \to \operatorname{Hom}(G\otimes K,\mathbb{Q}/\mathbb{Z}) \to \operatorname{Alt}^2(K,\mathbb{Q}/\mathbb{Z}) \to 0. $$ My question is, if this is indeed true in the setting of (possibly finite) abelian groups. Unfortunately I don't really understand how this comes from the Koszul complex, so I am not sure what to check here..

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  • $\begingroup$ I guess that you need the flatness of $K,G,H$. In this case, you first deduce the result in the case $G\simeq K\oplus H$ and both $K$ and $H$ are finite free, then animate to get the general result. $\endgroup$
    – Z. M
    Commented Sep 20, 2022 at 11:14
  • $\begingroup$ Hm, my groups are actually all finite.. is flatness really the most general setting? $\endgroup$
    – Bipolar Minds
    Commented Sep 20, 2022 at 12:02

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