Clarification Regarding the Tor Functor involved in a Finite Exact Sequence Let $\cdots\rightarrow F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$ be a free resolution of the $A$-module $M$. Let $N$ be an $A$-module. I saw in some notes that we have an exact sequence $0 \rightarrow \operatorname{Tor}(M,N) \rightarrow F_1 \otimes N \rightarrow F_0 \otimes N \rightarrow M \otimes N \rightarrow 0$. Why is that true?
Edited:
Let me explain the source of my confusion. In my study, $\operatorname{Tor}_n(M,N)$ was defined as the homology of dimension $n$ of the double complex $K_{p,q}=F_p \otimes Q_q$, where $F_{\cdot}, Q_{\cdot}$ are projective resolutions of $M,N$ respectively. It was shown that $\operatorname{Tor}_n(M,N) = \operatorname{Tor}_n(F_{\cdot} \otimes N)=H_n(F_{\cdot} \otimes N) = \frac{Ker(F_n \otimes N \rightarrow F_{n-1} \otimes N)}{Im(F_{n+1} \otimes N \rightarrow F_{n} \otimes N)}$. Now if the sequence 
$0 \rightarrow \operatorname{Tor}(M,N) \rightarrow F_1 \otimes N \rightarrow F_0 \otimes N \rightarrow M \otimes N \rightarrow 0$ is exact, then $\operatorname{Tor}_1(M,N) = \operatorname{Ker}(F_1 \otimes N \rightarrow F_{0} \otimes N)$, which means that $\operatorname{Im}(F_{2} \otimes N \rightarrow F_{1} \otimes N)=0$, which does not make sense. What am I missing?
PS: The reference that i am using is Matsumura's Commutative Ring Theory Appendix B. That's as far as i have gone so far with homological algebra.
 A: The tensor functor is right-exact. We have an exact sequence
$F_1\to F_0\to M\to 0$,
implying that the complex
$F_2\otimes N\to F_1\otimes N\to F_0\otimes N\to M\otimes N\to 0$,
is exact except possibly at $F_1\otimes N$. Now how can you calculate $\mathrm{Tor}_1(M,N)$? You should be able to figure it out from here.
A: You are right to be confused because the statement isn't true as written, for  the reason you say.  Example: take $A=k[x]/x^2$ and $M=k$ and $N=A$.  Then $\operatorname{Tor}_A(M,N)=0$ because $N$ is free therefore flat.  A free resolution for $M=k$ begins
$$ \cdots A \stackrel{x}\to A \stackrel{x}\to A \to k \to 0 $$
Tensoring with $N=A$ gives
$$ \cdots \to A \otimes_A A \to A \otimes_A A \to k \otimes_A A \to 0$$
Your sequence is 
$$0 \to 0 \to A \to A \to k \to 0 $$
(because the Tor group vanished and $A\otimes _A A \cong A$ and $A \otimes_A k\cong k$) which can't possibly be exact just by considering dimensions.
The correct statement is that 
$$ 0 \to \operatorname{Tor}_1^A(M,N) \to \ker d_0 \otimes_A N \to F_0 \otimes_A N \to M\otimes_A N \to 0 $$ is exact, where $d_0$ is the map $F_0 \to N$ in the free resolution. This is the long exact sequence for associated to $0 \to \ker d_0 \to F_0 \to N \to 0$; the 0 on the left is $\operatorname{Tor}_1^A(M,F_0)$ which vanishes by flatness of $F_0$.
