Mixing
The hyper-derived functors $mathbb L_bullet F$ are just derived functors of $H_0F$?
$begingroup$
Problem
(Weibel's Introduction to Homological Algebra, Exercise 5.7.4,2) Let $mathbf{Ch}_{ge0}(mathcal A)$ be the subcategory of complexes $A$ with $A_p=0$ for $p<0$. Then the hyper-derived functors $mathbb L_iF$ retricted to $mathbf{Ch}_{ge0}$ are the left derived functors of the right exact functor $H_0F$.
Thoughts
First the $F$ in $H_0F$ should be the functor $mathcal B=mathbf{Ch}_{ge0}(mathcal A)tomathcal B$ induced by $F$, I think. We know that the projective objects in $mathcal B$ are just almost-acyclic ($H_n=0$ for $nneq0$) chain complexes of projective objects in $mathcal A$. However, it's quite hard for me to connect it with Cartan-Eilenberg resolutions. On the other hand, $mathbb L_iF(A)$, by definition, is $H_ioperatorname{Tot}(F(P))$ where $Pto A$ is a Cartan-Eilenberg resolution. In order to assimilate this with the expression of derived functor of $H_0F$, it seems to me that the projective resolution of $A$ should be chosen from $operatorname{Tot}(P)$, but I have no idea how to proceed.
Any idea? Thanks!
homological-algebra
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
$endgroup$
add a comment |
$begingroup$
Problem
(Weibel's Introduction to Homological Algebra, Exercise 5.7.4,2) Let $mathbf{Ch}_{ge0}(mathcal A)$ be the subcategory of complexes $A$ with $A_p=0$ for $p<0$. Then the hyper-derived functors $mathbb L_iF$ retricted to $mathbf{Ch}_{ge0}$ are the left derived functors of the right exact functor $H_0F$.
Thoughts
First the $F$ in $H_0F$ should be the functor $mathcal B=mathbf{Ch}_{ge0}(mathcal A)tomathcal B$ induced by $F$, I think. We know that the projective objects in $mathcal B$ are just almost-acyclic ($H_n=0$ for $nneq0$) chain complexes of projective objects in $mathcal A$. However, it's quite hard for me to connect it with Cartan-Eilenberg resolutions. On the other hand, $mathbb L_iF(A)$, by definition, is $H_ioperatorname{Tot}(F(P))$ where $Pto A$ is a Cartan-Eilenberg resolution. In order to assimilate this with the expression of derived functor of $H_0F$, it seems to me that the projective resolution of $A$ should be chosen from $operatorname{Tot}(P)$, but I have no idea how to proceed.
Any idea? Thanks!
homological-algebra
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
$endgroup$
add a comment |
$begingroup$
Problem
(Weibel's Introduction to Homological Algebra, Exercise 5.7.4,2) Let $mathbf{Ch}_{ge0}(mathcal A)$ be the subcategory of complexes $A$ with $A_p=0$ for $p<0$. Then the hyper-derived functors $mathbb L_iF$ retricted to $mathbf{Ch}_{ge0}$ are the left derived functors of the right exact functor $H_0F$.
Thoughts
First the $F$ in $H_0F$ should be the functor $mathcal B=mathbf{Ch}_{ge0}(mathcal A)tomathcal B$ induced by $F$, I think. We know that the projective objects in $mathcal B$ are just almost-acyclic ($H_n=0$ for $nneq0$) chain complexes of projective objects in $mathcal A$. However, it's quite hard for me to connect it with Cartan-Eilenberg resolutions. On the other hand, $mathbb L_iF(A)$, by definition, is $H_ioperatorname{Tot}(F(P))$ where $Pto A$ is a Cartan-Eilenberg resolution. In order to assimilate this with the expression of derived functor of $H_0F$, it seems to me that the projective resolution of $A$ should be chosen from $operatorname{Tot}(P)$, but I have no idea how to proceed.
Any idea? Thanks!
homological-algebra
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
$endgroup$
Problem
(Weibel's Introduction to Homological Algebra, Exercise 5.7.4,2) Let $mathbf{Ch}_{ge0}(mathcal A)$ be the subcategory of complexes $A$ with $A_p=0$ for $p<0$. Then the hyper-derived functors $mathbb L_iF$ retricted to $mathbf{Ch}_{ge0}$ are the left derived functors of the right exact functor $H_0F$.
Thoughts
First the $F$ in $H_0F$ should be the functor $mathcal B=mathbf{Ch}_{ge0}(mathcal A)tomathcal B$ induced by $F$, I think. We know that the projective objects in $mathcal B$ are just almost-acyclic ($H_n=0$ for $nneq0$) chain complexes of projective objects in $mathcal A$. However, it's quite hard for me to connect it with Cartan-Eilenberg resolutions. On the other hand, $mathbb L_iF(A)$, by definition, is $H_ioperatorname{Tot}(F(P))$ where $Pto A$ is a Cartan-Eilenberg resolution. In order to assimilate this with the expression of derived functor of $H_0F$, it seems to me that the projective resolution of $A$ should be chosen from $operatorname{Tot}(P)$, but I have no idea how to proceed.
Any idea? Thanks!
homological-algebra
homological-algebra
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
edited Dec 2 '17 at 21:57
Frank Science
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
asked Mar 7 '15 at 16:38
Frank ScienceFrank Science
4,75511356
4,75511356
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's solved now.
Suppose that $Pto A$ is a projective resolution in $mathbf{Ch}_{ge0}(mathcal A)$, and $P_{bullet,bullet}$ is the associated double complex (with a sign convention). Take $Q_i=operatorname{coker}d_{i,1}^h$, then $Q_bullet$ is a chain complex of projective objects in $mathcal A$. The first and second spectral sequence associated with $P_{bullet,bullet}$ converges to $H_bullet(operatorname{Tot}(P_{bullet,bullet}))$, therefore $Q_bullet$ is quasi-isomorphic to $A_bullet$. Now it follows that $mathbb L_iF(A)=H_i(FQ_bullet)=H_i(H_0FP)=L_i(H_0F)(A)$.
A lemma is used here: Suppose $Q_bullet$ is a chain complex of projective objects quasi-isomorphic to $A_bullet$, then $mathbb L_iFA=H_i(FQ_bullet)$. A general reference for this is Iversen's Cohomology of Sheaves, Chapter 1, Theorem 6.2.
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
$endgroup$
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
add a comment |
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$begingroup$
It's solved now.
Suppose that $Pto A$ is a projective resolution in $mathbf{Ch}_{ge0}(mathcal A)$, and $P_{bullet,bullet}$ is the associated double complex (with a sign convention). Take $Q_i=operatorname{coker}d_{i,1}^h$, then $Q_bullet$ is a chain complex of projective objects in $mathcal A$. The first and second spectral sequence associated with $P_{bullet,bullet}$ converges to $H_bullet(operatorname{Tot}(P_{bullet,bullet}))$, therefore $Q_bullet$ is quasi-isomorphic to $A_bullet$. Now it follows that $mathbb L_iF(A)=H_i(FQ_bullet)=H_i(H_0FP)=L_i(H_0F)(A)$.
A lemma is used here: Suppose $Q_bullet$ is a chain complex of projective objects quasi-isomorphic to $A_bullet$, then $mathbb L_iFA=H_i(FQ_bullet)$. A general reference for this is Iversen's Cohomology of Sheaves, Chapter 1, Theorem 6.2.
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
$endgroup$
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
add a comment |
$begingroup$
It's solved now.
Suppose that $Pto A$ is a projective resolution in $mathbf{Ch}_{ge0}(mathcal A)$, and $P_{bullet,bullet}$ is the associated double complex (with a sign convention). Take $Q_i=operatorname{coker}d_{i,1}^h$, then $Q_bullet$ is a chain complex of projective objects in $mathcal A$. The first and second spectral sequence associated with $P_{bullet,bullet}$ converges to $H_bullet(operatorname{Tot}(P_{bullet,bullet}))$, therefore $Q_bullet$ is quasi-isomorphic to $A_bullet$. Now it follows that $mathbb L_iF(A)=H_i(FQ_bullet)=H_i(H_0FP)=L_i(H_0F)(A)$.
A lemma is used here: Suppose $Q_bullet$ is a chain complex of projective objects quasi-isomorphic to $A_bullet$, then $mathbb L_iFA=H_i(FQ_bullet)$. A general reference for this is Iversen's Cohomology of Sheaves, Chapter 1, Theorem 6.2.
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
$endgroup$
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
add a comment |
$begingroup$
It's solved now.
Suppose that $Pto A$ is a projective resolution in $mathbf{Ch}_{ge0}(mathcal A)$, and $P_{bullet,bullet}$ is the associated double complex (with a sign convention). Take $Q_i=operatorname{coker}d_{i,1}^h$, then $Q_bullet$ is a chain complex of projective objects in $mathcal A$. The first and second spectral sequence associated with $P_{bullet,bullet}$ converges to $H_bullet(operatorname{Tot}(P_{bullet,bullet}))$, therefore $Q_bullet$ is quasi-isomorphic to $A_bullet$. Now it follows that $mathbb L_iF(A)=H_i(FQ_bullet)=H_i(H_0FP)=L_i(H_0F)(A)$.
A lemma is used here: Suppose $Q_bullet$ is a chain complex of projective objects quasi-isomorphic to $A_bullet$, then $mathbb L_iFA=H_i(FQ_bullet)$. A general reference for this is Iversen's Cohomology of Sheaves, Chapter 1, Theorem 6.2.
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
$endgroup$
It's solved now.
Suppose that $Pto A$ is a projective resolution in $mathbf{Ch}_{ge0}(mathcal A)$, and $P_{bullet,bullet}$ is the associated double complex (with a sign convention). Take $Q_i=operatorname{coker}d_{i,1}^h$, then $Q_bullet$ is a chain complex of projective objects in $mathcal A$. The first and second spectral sequence associated with $P_{bullet,bullet}$ converges to $H_bullet(operatorname{Tot}(P_{bullet,bullet}))$, therefore $Q_bullet$ is quasi-isomorphic to $A_bullet$. Now it follows that $mathbb L_iF(A)=H_i(FQ_bullet)=H_i(H_0FP)=L_i(H_0F)(A)$.
A lemma is used here: Suppose $Q_bullet$ is a chain complex of projective objects quasi-isomorphic to $A_bullet$, then $mathbb L_iFA=H_i(FQ_bullet)$. A general reference for this is Iversen's Cohomology of Sheaves, Chapter 1, Theorem 6.2.
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
edited Jan 28 at 19:36
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
answered Mar 13 '15 at 10:47
Frank ScienceFrank Science
4,75511356
4,75511356
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
add a comment |
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
This lemma was very helpful! I am wondering if Prof. Weibel had a different proof in his mind avoiding this lemma. It seems that if one starts with Cartan-Eilenberg double complex P.,. for A. and applies F to it, then the II spectral sequence would imply the result, if one can show (H_n F) (P.,k)=0 for n>0. I guess that should follow from the projectivity of P.,. ...
$endgroup$
– Bananeen
Sep 27 '17 at 14:16
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
Why $P_bullet$ is quasi-isomorphic to $A_bullet$? We know isomorphisms between homologies, but what is the chain map which induces these isomorphisms?
$endgroup$
– Rafael Holanda
Dec 1 '17 at 20:24
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
@RafaelHolanda I don't have Weibel at hand. I think that what I obtained is that $P_bullet$ and $A_bullet$ are both quasi-isomorphic to $operatorname{Tot}(P_{bullet,bullet})$, therefore they are quasi-isomorphic. To be more precise, if there is a chain map $fcolon C_bulletto P_bullet$, where $P_bullet$ is made up of projective objects, which induces isomorphisms on homologies, then there is also a chain map $P_bulletto C_bullet$ which induces isomorphisms, by considering the isomorphism $[P_bullet,f]colon[P_bullet,C_bullet]to[P_bullet,P_bullet]$.
$endgroup$
– Frank Science
Dec 2 '17 at 21:43
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
$begingroup$
How is the last equality obtained?
$endgroup$
– davik
Jan 28 at 18:10
add a comment |
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