Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Thinking out loud about the third question: can we define a category whose objects are modules with chosen projective resolutions and its morphisms are chosen liftings, guaranteed by the comparison lemma? –  Bruno Stonek Feb 10 '12 at 20:40
    
Dear Bruno: You may want to take a look at page 107 of Cartan-Eilenberg. –  Pierre-Yves Gaillard Feb 11 '12 at 5:44
    
@Pierre-YvesGaillard: thank you for the reference, however, could you point out more precisely where is the relevant part? I read the page and I don't get it... (it may be that there is some terminology I do not know, e.g. balanced or satellite). –  Bruno Stonek Feb 12 '12 at 13:12
    
Dear Bruno: You're welcome. I suggest that you look at Section V.8 pp 94-97. (I think you can ignore the notion of satellites, but not the notion of balanced functor.) –  Pierre-Yves Gaillard Feb 12 '12 at 13:43
    
@Bruno: In a category of modules there is in fact a canonical projective resolution: take the free module generated by the underlying set of the original module, then the free module generated by the underlying set of the relations module, etc. –  Zhen Lin Apr 10 '12 at 22:50

1 Answer 1

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

share|improve this answer
    
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. –  Bruno Stonek Feb 12 '12 at 13:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.