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Assume we want to define $\mathrm{Tor}_n (M,N)$ where $M,N$ are $R$-modules and $R$ is a commutative unital ring.

We take a projective resolution of $M$:

$ \dots \to P_1 \to P_0 \to M \to 0$

Now does it matter whether we apply $-\otimes N$ or $N \otimes -$ to this? It shouldn't because we have $N \otimes P \cong P \otimes N$. Right? Thanks.

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    @BenjaLim That tag was added in response to the comment.2012-07-24

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I guess the point here is that $N \otimes_R -$ and $- \otimes_R N$ are naturally isomorphic functors. Therefore, you get an isomorphism of chain complexes $N \otimes_R P^{\bullet} \cong P^{\bullet} \otimes_R N$, which implies that the two complexes have isomorphic homology. So, it doesn't matter if you apply $N \otimes_R -$ or $- \otimes_R N$.

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    As you use chain complexes (as opposed to cochain complexes), it should indeed be homology instead of cohomology. I'll edit my answer accordingly.2012-07-24