Unclear use of the right exactness of $\otimes_R$, isomorphism,projective exact sequence

Question

Suppose that for $R-\mathbb{Mod}$ $M$ and $N$ we have a projective resolution

$...\to P_1\to P_0\to M \to 0$

How do I use the right exactness of $\otimes_R$ to show that for the deleted (last term at $M$) complex $...\to P_1\otimes_RN\to P_0\otimes_RN\to 0$

its 0-th homology group (i.e. $\mathbb{Tor}^0_R(M,N)$) is isomorphic to $M\otimes_R N$?

Se 28th page here.

Answer 1

The zeroth homology group is the cokernel of $$P_1\otimes_R N\to P_0\otimes_R N.$$ By right extactness $$P_1\otimes_R N\to P_0\otimes_R N\to M\otimes_R N\to0$$ is exact, so that cokernel is $M\otimes_R N$.

Answer 2

Let $f:P_1\to P_0$ be the homomorphism in the projective resolution, so $\operatorname{coker} f\cong M$, where $\operatorname{coker} f$ is defined to be $P_0/\operatorname{im} f$ (codomain modulo image).

The $0$th homology group of the resolution tensored by $N$ is $\operatorname{coker} (f\otimes_R N)$, where $f\otimes_R N$ is the map $f$ through the functor $({-})\otimes_R N$. Right exactness means $\operatorname{coker}(f\otimes_R N)\cong\operatorname{coker}(f)\otimes_R N$, which is $M\otimes_R N$.

To elaborate on this reasoning: let $F$ be a right-exact functor in the category of modules. Given $f:A\to B$, we have an associated exact sequence $$A\xrightarrow{f} B\xrightarrow{q} \operatorname{coker} f\to 0$$ where $q:B\to B/\operatorname{im}f$ is the quotient map. Applying $F$, we get another exact sequence$$F(A)\xrightarrow{F(f)} F(B)\to F(\operatorname{coker}f)\to 0$$ and exactness means $\operatorname{coker} F(f)\cong F(\operatorname{coker}f)$.