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Given a ring $R$, we can consider the following functors:

any $A\in Mod-R$ and choice of projective resolutions $P_\bullet(B)$ for every $B\in R-Mod$ defines a functor $Tor_n^R(A,-):R-Mod\to Ab$,

any $B\in R-Mod$ and choice of projective resolutions $Q_\bullet(A)$ for every $A\in Mod-R$ defines a functor $tor_n^R(-,B):Mod-R\to Ab$.

It is known that $Tor_n^R(A,B)\simeq tor_n^R(A,B)$ as abelian groups for every $A\in Mod-R$, $B\in R-Mod$.

Question 1: can we define a bifunctor $Mod-R\times R-Mod \to Ab$ that, such that $(A,B)\mapsto Tor_n^R(A,B)$?

This first question is related to the second comment darij grinberg made in this question at MO.

Question 2: If $R$ is a commutative ring, then given an $R$-module $A$, we can consider $Tor_n^R(A,-), tor_n^R(-,A): R-Mod\to Ab$. Are these functors naturally isomorphic?

I think the answer to this question is yes, provided that the choices $Q_\bullet$ and $P_\bullet$ are the same. But what if they are different? If they are not necessarily naturally isomorphic, is there any condition that guarantees this is the case?

Now for a bonus question. The Tor functor involves an arbitrary choice in its definition that makes me quite uneasy. Formally, it doesn't seem to be well defined, since if we write $Tor_n^R(A,-):R-Mod\to Ab$, we aren't really taking into account the choice of projective resolutions.

Question 3 (bonus): Is there a way to fix this? Has this been considered in some treatise on the Tor functor?

EDIT: I've striked down the third question since I think I now understand how it goes: $Tor_n^R(A,-)$ (or any derived functor for that matter) is defined as a functor such that this and that. You then prove that every pair of functors that satisfy those conditions are naturally isomorphic, so you can be at rest that even if your choice was a bit arbitrary, you're not losing much without considering the other choices of projective resolutions.

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    If $A$ is a $k$-algebra and $N$ is a left $A$-module, then the bar construction $B_*(A,A,N)$ gives an $A$-resolution of $N$ as with augmentation given by the action map. Moreover, if $A$ and $N$ are $k$-flat, this is a flat resolution. This gives a nice way to compute Tor under some hypotheses.2015-12-01

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Q1: Yes, take projective resolutions $P^*,Q^*$ of $A,B$ resp., build the tensor bicomplex $R^{*,*}$ and take the total homology of that bicomplex. This is how you can define Tor as a bifunctor.

Q2: Yes, and the choices of the resolutions don't matter. You can prove that the total homology of the bicomplex in Q1 is the same as the homology of $B \otimes P^*$ is the same as the homology of $A \otimes Q^*$.

Q3: The choice of projective resolution doesn't matter, since there is a chain homotopy between any two projective resolutions of the same object. This is proved using induction on the length of the resolution and the universal property of a projective object. This guarantees that any projective resolution produces the same Tor's, and more generally that the left derived functors of a right exact functor are always well defined.

Edit: spelling

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    For the first two questions: thank you for your answer; however, I don't understand them yet... I'll reread them when I know what a bicomplex, or the "total homology" is. Sorry for my ignorance! As for the third question, I've added a pertinent edit to the question, would you mind reading it? Thank you.2012-02-12