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$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\Ext}{\operatorname{Ext}}$$\newcommand{\Tor}{\operatorname{Tor}}$ Questions:

1. While the contravariant $\Ext$ functor is the first derived functor of the contravariant $\Hom$ functor, is the covariant $\Ext$ functor the first derived functor of the covariant $\Hom$ functor?

2. Does the fact that $ - \otimes X \simeq X \otimes -$, both covariant functors, imply in turn that: $$\Tor(-,X) \simeq \Tor(X,-),\text{ both covariant?} $$

Yes/no answers will suffice, although if an answer is no, I would also appreciate a suggestion for a reference that discusses these issues further, or even your own explanation if you feel up to it.

Background Information: Please correct me if I am wrong -- my current understanding is:

1. The tensor product functor $ - \otimes B$ and the co-variant Hom functor $\Hom(B, -)$ are an adjoint pair (tensor the left adjoint, co-variant Hom functor the right adjoint).

2. Any functor which is a left adjoint is right exact, and any functor which is a right adjoint is left exact. From 1., it follows as a special case that the tensor product functor is right exact and the co-variant Hom functor is left exact.

3. Because the tensor product functor is right exact but not left exact, we can use this "lack of left exactness" to define/form its first derived functor, the Tor functor.

4. Because the contra-variant Hom functor is left exact but not right exact, we can use this "lack of right exactness" to define/form its first derived functor, "the" Ext functor.

To clarify, I know 1. and 2. from my commutative algebra course, and 3. and 4. from my homology course. Since the two courses don't intersect or interact, I have to clarify any connections myself.

Consulting this page, I was reminded that one could define at least two different Ext functors: $\operatorname{Ext}(-,X)$ and $\operatorname{Ext}(X,-)$. Only the first one is derived from the contra-variant Hom functor (I think), and that page says that it is contravariant, while the other is covariant.

Corresponding to this, we could define at least two different Tor functors, $\operatorname{Tor}(X,-)$ or $\operatorname{Tor}(-,X)$. However, if the two tensor product functors are isomorphic and both covariant, one would expect (naively) that these two are also isomorphic and both covariant.

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  1. $\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\Ext}{\operatorname{Ext}}$$\newcommand{\Tor}{\operatorname{Tor}}$Yes, it is. For any $i\geq 0$ you can compute $\mathrm{Ext}^i(A,B)$ both as the $i-th$ derived functor of $\Hom(A,-)$ and as the $i-th$ derived functor of $\Hom(-,B)$. Of course the first is a right derived functor (of a left exact functor) while the second is a left derived functor (of a right exact functor). This result is called balancing of the Ext functor. Notice that this implies in particular that $\Ext^i(A,B)=0$ whenever $A$ is projective or $B$ is injective.

  2. In general, if $F$ and $G$ are two functors, both covariant (or both contravariant) and they are isomorphic, then their derived functors are all equal, in symbols $R^iF \simeq R^iG$ for every $i\geq 0$. This follows directly from the construction of the derived functors. In particular, since $-\otimes X$ and $X\otimes -$ are isomorphic covariant functors, $\Tor^i(-,X)\simeq \Tor^i(X,-)$.

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  • $\begingroup$ Thank you for the clarification, and in particular the new terminology which I will be able to use look into these issues further. I appreciate it! $\endgroup$ – Chill2Macht Dec 20 '16 at 16:35

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