Exact sequence of double complexes induces exact sequence on total complexes
$begingroup$
This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)
Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 to A to B to C to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) to Tot(B)$ is a quasi-isomorphism.
The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form
$ldots to H_{i+1}C(=0) to H_i(A) to H_i(B) to H_i(C)(=0) $
so the induced morphism on homology is an isomorphism.
The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 to A_{ij} to B_{ij} to C_{ij} to 0$ is exact for all i,j and everything commutes.
Let $alpha:A to B$ be a morphism of double complexes. The induced morphism between the total complexes, $alpha^*: Tot(A) to Tot(B)$, is then defined, I'd assume, as $alpha^*=d_B^h alpha + d_B^v alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.
"EDIT:" After some thought, I guess a good first step would be to show that $beta^* circ alpha^* = 0$, which shouldn't be too difficult. (where $alpha^*,beta^*$ denotes the induced morphisms of total complexes)
Edit2: I've clarified my notation a bit.
homological-algebra
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
|
show 4 more comments
$begingroup$
This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)
Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 to A to B to C to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) to Tot(B)$ is a quasi-isomorphism.
The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form
$ldots to H_{i+1}C(=0) to H_i(A) to H_i(B) to H_i(C)(=0) $
so the induced morphism on homology is an isomorphism.
The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 to A_{ij} to B_{ij} to C_{ij} to 0$ is exact for all i,j and everything commutes.
Let $alpha:A to B$ be a morphism of double complexes. The induced morphism between the total complexes, $alpha^*: Tot(A) to Tot(B)$, is then defined, I'd assume, as $alpha^*=d_B^h alpha + d_B^v alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.
"EDIT:" After some thought, I guess a good first step would be to show that $beta^* circ alpha^* = 0$, which shouldn't be too difficult. (where $alpha^*,beta^*$ denotes the induced morphisms of total complexes)
Edit2: I've clarified my notation a bit.
homological-algebra
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
1
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
1
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58
|
show 4 more comments
$begingroup$
This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)
Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 to A to B to C to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) to Tot(B)$ is a quasi-isomorphism.
The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form
$ldots to H_{i+1}C(=0) to H_i(A) to H_i(B) to H_i(C)(=0) $
so the induced morphism on homology is an isomorphism.
The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 to A_{ij} to B_{ij} to C_{ij} to 0$ is exact for all i,j and everything commutes.
Let $alpha:A to B$ be a morphism of double complexes. The induced morphism between the total complexes, $alpha^*: Tot(A) to Tot(B)$, is then defined, I'd assume, as $alpha^*=d_B^h alpha + d_B^v alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.
"EDIT:" After some thought, I guess a good first step would be to show that $beta^* circ alpha^* = 0$, which shouldn't be too difficult. (where $alpha^*,beta^*$ denotes the induced morphisms of total complexes)
Edit2: I've clarified my notation a bit.
homological-algebra
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
$endgroup$
This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)
Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 to A to B to C to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) to Tot(B)$ is a quasi-isomorphism.
The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form
$ldots to H_{i+1}C(=0) to H_i(A) to H_i(B) to H_i(C)(=0) $
so the induced morphism on homology is an isomorphism.
The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 to A_{ij} to B_{ij} to C_{ij} to 0$ is exact for all i,j and everything commutes.
Let $alpha:A to B$ be a morphism of double complexes. The induced morphism between the total complexes, $alpha^*: Tot(A) to Tot(B)$, is then defined, I'd assume, as $alpha^*=d_B^h alpha + d_B^v alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.
"EDIT:" After some thought, I guess a good first step would be to show that $beta^* circ alpha^* = 0$, which shouldn't be too difficult. (where $alpha^*,beta^*$ denotes the induced morphisms of total complexes)
Edit2: I've clarified my notation a bit.
homological-algebra
homological-algebra
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
edited Feb 2 '11 at 4:30
Fredrik Meyer
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
asked Feb 2 '11 at 2:59
Fredrik MeyerFredrik Meyer
15.3k24165
15.3k24165
$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
1
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
1
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58
|
show 4 more comments
$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
1
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
1
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58
$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
1
1
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
1
1
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58
|
show 4 more comments
1 Answer
1
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oldest
votes
$begingroup$
Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $mathrm{Tot}(f)$ for a map $f:Cto D$ of double complexes works. Rather, an element in $mathrm{Tot}(C)$ of the form $(ldots,c_{0,n},c_{1,n-1},ldots)$ gets sent to $(ldots,fc_{0,n},fc_{1,n-1},ldots)$.
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
$endgroup$
add a comment |
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$begingroup$
Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $mathrm{Tot}(f)$ for a map $f:Cto D$ of double complexes works. Rather, an element in $mathrm{Tot}(C)$ of the form $(ldots,c_{0,n},c_{1,n-1},ldots)$ gets sent to $(ldots,fc_{0,n},fc_{1,n-1},ldots)$.
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
$endgroup$
add a comment |
$begingroup$
Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $mathrm{Tot}(f)$ for a map $f:Cto D$ of double complexes works. Rather, an element in $mathrm{Tot}(C)$ of the form $(ldots,c_{0,n},c_{1,n-1},ldots)$ gets sent to $(ldots,fc_{0,n},fc_{1,n-1},ldots)$.
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
$endgroup$
add a comment |
$begingroup$
Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $mathrm{Tot}(f)$ for a map $f:Cto D$ of double complexes works. Rather, an element in $mathrm{Tot}(C)$ of the form $(ldots,c_{0,n},c_{1,n-1},ldots)$ gets sent to $(ldots,fc_{0,n},fc_{1,n-1},ldots)$.
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
$endgroup$
Note that exactness depends only on the underlying objects, and not on the differentials. It follows that if each sequence at coordinate is exact, then the sequence of total complexes is too, because the direct sum of exact sequences is exact. On the other hand, the morphism you have defined is not how the map $mathrm{Tot}(f)$ for a map $f:Cto D$ of double complexes works. Rather, an element in $mathrm{Tot}(C)$ of the form $(ldots,c_{0,n},c_{1,n-1},ldots)$ gets sent to $(ldots,fc_{0,n},fc_{1,n-1},ldots)$.
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
answered Dec 28 '18 at 15:28
Pedro Tamaroff♦Pedro Tamaroff
97.4k10153297
97.4k10153297
add a comment |
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$begingroup$
What are alpha and beta? The morphism of total complexes is the only possible one!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:02
1
$begingroup$
By the way, you know what a morphism of complexes is, and a double complex a complex of complexes, so you know what a morphism of double complexes is!
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:14
$begingroup$
As I said, I'd assumed that the definition of a morphism of total complexes was "the only" possible one. My main problem is to show that exactness is carried over to total complexes.
$endgroup$
– Fredrik Meyer
Feb 2 '11 at 3:18
$begingroup$
No, that is not what I meant. Your fifth paragraph starts with what you assume is the map between total complexes, giving a formula for alpha involving differentials and, afaict, alpha itself (!) well, that is not the map you want.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 3:54
1
$begingroup$
I really don't know why you think $alpha^*$ is defined like that. The correct definition does not involve differentials at all---what you wrote is not even a map of degree zero! Look at what you have and consider what you want: there is exactly one way to do it.
$endgroup$
– Mariano Suárez-Álvarez
Feb 2 '11 at 4:58