derive TOR long exact sequence without using right exactness of tensor Let $R$ be a commutative ring and $M$ be an $R$ module.  The only proof that I know of the right exactness of $ -\otimes_R M$ uses the left exactness of $Hom_R(-,M)$.  I wondered if there was a way of deriving the long exact sequence of tensor without using the right exactness of $ -\otimes_R M$.  This would allow me to deduce the right exactness of tensor.
Let $0 \to M \to N \to K \to 0$ be an exact sequence of $R$ modules.  Let $\mathfrak M$ be a free resolution of $M$, $\mathfrak N$ a free resolution of $N$, $\mathfrak K$ a  free resolution of $K$.  We get as usual that there is a short exact sequence of chain complexes $0 \to \mathfrak M \to \mathfrak N \to \mathfrak K \to 0$ which stays exact after tensoring with an $R$ module $L$.  
What I would appreciate assistance in is showing that the cohomology of $\mathfrak M \otimes_R L$ at the first entry is $M \otimes_R L$.  This is where I see the argument would ordinarily need the right exactness of tensor.
 A: 
What I would appreciate assistance in is showing that...

Slightly more generally, suppose you have some additive functor $F$ and a projective resolution of an object $A$
$$\cdots \to P^{-2} \to P^{-1} \to P^0 \to A \to 0$$
We look at the complex
$$F (P^\bullet)\colon \quad \cdots \to F (P^{-2}) \to F (P^{-1}) \to F (P^0) \to 0$$
The $0$-th cohomology is given by
$$H^0 (F (P^\bullet)) = \operatorname{coker} (F (P^{-1}) \to F (P^0))$$
You want to conclude that this is naturally isomorphic to $F (A)$, but this natural isomorphism corresponds to preservation of cokernels, i.e. right exactness:
$$\operatorname{coker} (F (P^{-1}) \to F (P^0)) \stackrel{???}{\cong} F (\operatorname{coker} (P^{-1} \to P^0)).$$
The only difference from right exactness is that you consider only cokernels of morphisms between projective objects; or free objects, if you take free resolutions of $R$-modules, as in your case. But checking this is not any easier than checking right exactness.


The only proof that I know of the right exactness of $-\otimes_R M$
  uses the left exactness of $\operatorname{Hom}_R (-,M)$.



*

*This is not 100% correct: the proof uses that $-\otimes_R M$ is left adjoint to $\operatorname{Hom}_R (M,-)$, and then any left adjoint is automatically right exact and any right adjoint is left exact.

*Alternatively, you can check right exactness of $-\otimes_R M$ by hand: just take a short exact sequence
$$0 \to N' \to N \to N'' \to 0$$
and look at the maps in the induced sequence
$$N'\otimes_R M \to N\otimes_R M \to N''\otimes_R M \to 0$$
