Mixing
Are properties of Tor, Ext analogous to those of tensor product and Hom in the following two ways?
$begingroup$
$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.
reference-request commutative-algebra category-theory homological-algebra
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
$endgroup$
add a comment |
$begingroup$
$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.
reference-request commutative-algebra category-theory homological-algebra
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
$endgroup$
add a comment |
$begingroup$
$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.
reference-request commutative-algebra category-theory homological-algebra
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
$endgroup$
$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.
reference-request commutative-algebra category-theory homological-algebra
reference-request commutative-algebra category-theory homological-algebra
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
edited Dec 20 '16 at 12:34
Chill2Macht
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
asked Dec 19 '16 at 17:50
Chill2MachtChill2Macht
11.6k91868
11.6k91868
add a comment |
add a comment |
1 Answer
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$begingroup$
$newcommand{Hom}{operatorname{Hom}}$$newcommand{Ext}{operatorname{Ext}}$$newcommand{Tor}{operatorname{Tor}}$Yes, it is. For any $igeq 0$ 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.
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 $igeq 0$. This follows directly from the construction of the derived functors. In particular, since $-otimes X$ and $Xotimes -$ are isomorphic covariant functors, $Tor^i(-,X)simeq Tor^i(X,-)$.
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
$newcommand{Hom}{operatorname{Hom}}$$newcommand{Ext}{operatorname{Ext}}$$newcommand{Tor}{operatorname{Tor}}$Yes, it is. For any $igeq 0$ 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.
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 $igeq 0$. This follows directly from the construction of the derived functors. In particular, since $-otimes X$ and $Xotimes -$ are isomorphic covariant functors, $Tor^i(-,X)simeq Tor^i(X,-)$.
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$newcommand{Hom}{operatorname{Hom}}$$newcommand{Ext}{operatorname{Ext}}$$newcommand{Tor}{operatorname{Tor}}$Yes, it is. For any $igeq 0$ 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.
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 $igeq 0$. This follows directly from the construction of the derived functors. In particular, since $-otimes X$ and $Xotimes -$ are isomorphic covariant functors, $Tor^i(-,X)simeq Tor^i(X,-)$.
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$newcommand{Hom}{operatorname{Hom}}$$newcommand{Ext}{operatorname{Ext}}$$newcommand{Tor}{operatorname{Tor}}$Yes, it is. For any $igeq 0$ 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.
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 $igeq 0$. This follows directly from the construction of the derived functors. In particular, since $-otimes X$ and $Xotimes -$ are isomorphic covariant functors, $Tor^i(-,X)simeq Tor^i(X,-)$.
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
$endgroup$
$newcommand{Hom}{operatorname{Hom}}$$newcommand{Ext}{operatorname{Ext}}$$newcommand{Tor}{operatorname{Tor}}$Yes, it is. For any $igeq 0$ 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.
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 $igeq 0$. This follows directly from the construction of the derived functors. In particular, since $-otimes X$ and $Xotimes -$ are isomorphic covariant functors, $Tor^i(-,X)simeq Tor^i(X,-)$.
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
edited Jan 29 at 13:29
Martin Sleziak
44.9k10122277
44.9k10122277
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
answered Dec 20 '16 at 15:46
Alessandro17Alessandro17
37816
37816
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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