Let $A$ be a commutative ring with 1, and suppose $$0 \longrightarrow M \longrightarrow N \longrightarrow P \longrightarrow 0$$ be a short exact sequence of $A$-modules. Let $D$ be an $A$-module.
I understand that if $Tor^{A}_1(D,P)=0$, then corresponding exact sequence $$0 \longrightarrow D\otimes_AM \longrightarrow D\otimes_AN \longrightarrow D\otimes_AP \longrightarrow 0 $$is exact, since Tor is left derived functor.
But conversely, if we know given sequence of tensor product is exact, we can always say that $Tor^{A}_1(D,P)=0$?
The answer is yes(in Atiyah says,) but I can't understand why this holds. I think it can be happended so that $Tor^{A}_1(D,P)\neq0$ but image of $Tor^{A}_1(D,P) \longrightarrow D\otimes_AM$ is zero, so that the mapping $D\otimes_AM \longrightarrow D\otimes_AN$ is injective.
We could have $N=P$ and $M=0$. This would make $$ 0 \to D\otimes_AM \to D\otimes_AN \to D\otimes_AP \to 0 $$ exact no matter what $P$ is.
I think, instead, what Atiyah meant was that if the tensored sequence is exact for any $M,N$ such that $$0\to M\to N\to P\to 0$$ is exact, then we get $\operatorname{Tor}_1^A(D,P)=0$.