A question about the definition of the Tor functor

Question

On pg 278 of Bredon's "Topology and Geometry" says the following"

Let $0\to A'\to A\to A''\to 0$ be a short exact sequence of abelian groups, and let $M$ be another abelian group. Then the following long exact sequence is induced: $0\to Tor(M,A')\to Tor(M,A)\to Tor(M,A'')\to A'\otimes M\to A\otimes M\to A''\otimes M\to 0$

How do we know that the long exact sequence ends at $Tor(M,A')$? That $Tor(M,A')\to Tor(M,A)$ is injective?

Indeed! In general, the left-derived functors of a right-exact functor like $-\otimes M$ do not become zero ... — paul garrett, Sep 24 '20 at 21:26
Every abelian group has a free (so in particular flat) resolution of length $2$. — Qiaochu Yuan, Sep 24 '20 at 22:38

score 0 · Answer 1 · answered Sep 24 '20 at 22:55

0

Abelian groups are the same as $\mathbf Z$-modules, and all P.I.D.s have (weak) global dimension $1$, which means every module over a P.I.D. has a flat resolution of length $\le 1$.

answered Sep 24 '20 at 22:55

Bernard

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A question about the definition of the Tor functor

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