Definition of Tor functor in an arbitrary abelian category

Question

Let $\mathcal{C}$ be an abelian category. We define $\text{Ext}_\mathcal{C}^n(A,B)$ for objects $A$ and $B$ via $n$-extensions of $A$ by $B$ as described here, with $\text{Ext}_{\mathcal{C}}^0(A,B):=\text{Hom}_{\mathcal{C}}(A,B).$ This coincides with the definition of $\text{Ext}$ in terms of resolutions if $\mathcal{C}$ has enough projectives or enough injectives.

Is there a corresponding definition for $\text{Tor}_\mathcal{C}^n(A,B)$? $\text{Ext}$ starts with the abelian group of morphisms $A\to B$, whereas Tor starts with the tensor product $A\otimes B$. Yet there is no tensor product in an arbitrary abelian category, so I don't see how Tor can work in this abstract setting. What am I missing?

score 5 · Accepted Answer · answered Jan 08 '23 at 19:36

5

You don't miss anything. It is not possible to define $\mathrm{Tor}^n$ in an arbitrary abelian category. You need an abelian $\otimes$-category with enough projectives.

answered Jan 08 '23 at 19:36

Martin Brandenburg

181,922

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@downvoter: please explain what I can improve. – Martin Brandenburg Jan 08 '23 at 19:43
It is really strange, I get downvotes on a regular basis without any reason, often directly after posting my answer. – Martin Brandenburg Jan 08 '23 at 23:53
I would say that it is not necessary to assume the existence of enough projectives, but certainly one should have $\otimes$ at least. – Zhen Lin Jan 09 '23 at 03:47
How do you define (and construct) Tor when there are not enough projectives? I mean without leaving the abelian category. – Martin Brandenburg Jan 09 '23 at 18:21
The same way as for any other derived functor: define it by representability of some other functor. See here. – Zhen Lin Jan 09 '23 at 22:40
Thanks! I will check it out. But this is what I meant with "construct". I wanted to have a criterion which enables us to construct Tor, not just characterize it. – Martin Brandenburg Jan 10 '23 at 07:25

Definition of Tor functor in an arbitrary abelian category

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