$C^n(G,A)$ with $G$ a profinite group and $A$ a discrete $G$-module as direct limit
Let $C^n(G,A)$ be the set of continuous functions $G^n rightarrow A$ with $G$ a profinite group and $A$ a discrete $G$-module (these are the functions that are locally constant). I want to prove that $C^n(G,A) = varinjlim C^n(G/U,A^U)$, where $U$ runs through all open normal subgroups of $G$ and $A^U$ is the submodule fixed by $U$.
I think that the direct system you have to use is the following: let $U subset V$ be 2 open normal subgroups of $G$. Then we have a canonical projection $p_{UV}:G/U rightarrow G/V$, which is clearly continuous since both sides have the discrete topology. We also have a canonical inclusion $i_{UV}:A^V rightarrow A^U$ which is again continuous because of the discrete topology on both sides. So we can make a function $rho_{VU}:C^n(G/V,A^V) rightarrow C^n(G/U,A^U)$, defined by $rho_{VU}(phi)=i_{UV}phi p_{UV}$.
Now, I was trying to make an isomorphism from $C^n(G,A)$ to $varinjlim C^n(G/U,A^U)$, but I don't see how to do this. I must admit, I'm not very good with limits in categories.
Any help would be appreciated.
homological-algebra group-cohomology
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
add a comment |
Let $C^n(G,A)$ be the set of continuous functions $G^n rightarrow A$ with $G$ a profinite group and $A$ a discrete $G$-module (these are the functions that are locally constant). I want to prove that $C^n(G,A) = varinjlim C^n(G/U,A^U)$, where $U$ runs through all open normal subgroups of $G$ and $A^U$ is the submodule fixed by $U$.
I think that the direct system you have to use is the following: let $U subset V$ be 2 open normal subgroups of $G$. Then we have a canonical projection $p_{UV}:G/U rightarrow G/V$, which is clearly continuous since both sides have the discrete topology. We also have a canonical inclusion $i_{UV}:A^V rightarrow A^U$ which is again continuous because of the discrete topology on both sides. So we can make a function $rho_{VU}:C^n(G/V,A^V) rightarrow C^n(G/U,A^U)$, defined by $rho_{VU}(phi)=i_{UV}phi p_{UV}$.
Now, I was trying to make an isomorphism from $C^n(G,A)$ to $varinjlim C^n(G/U,A^U)$, but I don't see how to do this. I must admit, I'm not very good with limits in categories.
Any help would be appreciated.
homological-algebra group-cohomology
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
add a comment |
Let $C^n(G,A)$ be the set of continuous functions $G^n rightarrow A$ with $G$ a profinite group and $A$ a discrete $G$-module (these are the functions that are locally constant). I want to prove that $C^n(G,A) = varinjlim C^n(G/U,A^U)$, where $U$ runs through all open normal subgroups of $G$ and $A^U$ is the submodule fixed by $U$.
I think that the direct system you have to use is the following: let $U subset V$ be 2 open normal subgroups of $G$. Then we have a canonical projection $p_{UV}:G/U rightarrow G/V$, which is clearly continuous since both sides have the discrete topology. We also have a canonical inclusion $i_{UV}:A^V rightarrow A^U$ which is again continuous because of the discrete topology on both sides. So we can make a function $rho_{VU}:C^n(G/V,A^V) rightarrow C^n(G/U,A^U)$, defined by $rho_{VU}(phi)=i_{UV}phi p_{UV}$.
Now, I was trying to make an isomorphism from $C^n(G,A)$ to $varinjlim C^n(G/U,A^U)$, but I don't see how to do this. I must admit, I'm not very good with limits in categories.
Any help would be appreciated.
homological-algebra group-cohomology
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
Let $C^n(G,A)$ be the set of continuous functions $G^n rightarrow A$ with $G$ a profinite group and $A$ a discrete $G$-module (these are the functions that are locally constant). I want to prove that $C^n(G,A) = varinjlim C^n(G/U,A^U)$, where $U$ runs through all open normal subgroups of $G$ and $A^U$ is the submodule fixed by $U$.
I think that the direct system you have to use is the following: let $U subset V$ be 2 open normal subgroups of $G$. Then we have a canonical projection $p_{UV}:G/U rightarrow G/V$, which is clearly continuous since both sides have the discrete topology. We also have a canonical inclusion $i_{UV}:A^V rightarrow A^U$ which is again continuous because of the discrete topology on both sides. So we can make a function $rho_{VU}:C^n(G/V,A^V) rightarrow C^n(G/U,A^U)$, defined by $rho_{VU}(phi)=i_{UV}phi p_{UV}$.
Now, I was trying to make an isomorphism from $C^n(G,A)$ to $varinjlim C^n(G/U,A^U)$, but I don't see how to do this. I must admit, I'm not very good with limits in categories.
Any help would be appreciated.
homological-algebra group-cohomology
homological-algebra group-cohomology
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
edited Apr 2 '12 at 11:44
user38268
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
asked Apr 2 '12 at 10:12
KevinDLKevinDL
831521
831521
add a comment |
add a comment |
1 Answer
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This looks like it might be some work.
First of all we don't only have maps $C^n(G/V,A^V) to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) to varinjlim C^n(G/V,A^V)$.
Putting it all together you have a diagram like this:
where $Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $Psi$ is both injective and surjective.
Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).
edited 2 days ago
Glorfindel
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
add a comment |
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1 Answer
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This looks like it might be some work.
First of all we don't only have maps $C^n(G/V,A^V) to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) to varinjlim C^n(G/V,A^V)$.
Putting it all together you have a diagram like this:
where $Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $Psi$ is both injective and surjective.
Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).
edited 2 days ago
Glorfindel
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
add a comment |
This looks like it might be some work.
First of all we don't only have maps $C^n(G/V,A^V) to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) to varinjlim C^n(G/V,A^V)$.
Putting it all together you have a diagram like this:
where $Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $Psi$ is both injective and surjective.
Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).
edited 2 days ago
Glorfindel
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
add a comment |
This looks like it might be some work.
First of all we don't only have maps $C^n(G/V,A^V) to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) to varinjlim C^n(G/V,A^V)$.
Putting it all together you have a diagram like this:
where $Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $Psi$ is both injective and surjective.
Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).
edited 2 days ago
Glorfindel
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
This looks like it might be some work.
First of all we don't only have maps $C^n(G/V,A^V) to C^n(G/U,A^U)$ you also have maps $C^n(G/V,A^V) to C^n(G,A)$ and cannonical maps $C^n(G/V,A^V) to varinjlim C^n(G/V,A^V)$.
Putting it all together you have a diagram like this:
where $Psi$ is induced by the universal property, and everything commutes. Your task now is to show that $Psi$ is an isomorphism. Unfortunately, I don't see this as an easy task! It seems you need to go through the whole verification that $Psi$ is both injective and surjective.
Since it looks like a bit of work, I offer up an alternative. If you have institutional access to SpringerLink you should be (at least I can) able to view the book 'Profinite Groups' by Ribes and Zalesskii. Theorem 5.1.4(a) is a proof of a very similar statement - in essence it appears that following their steps should work here (with suitable modifications).
edited 2 days ago
Glorfindel
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
edited 2 days ago
Glorfindel
3,41981830
edited 2 days ago
Glorfindel
3,41981830
edited 2 days ago
Glorfindel
3,41981830
3,41981830
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
answered Apr 2 '12 at 11:27
Juan SJuan S
6,5592159
6,5592159
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
add a comment |
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Thanks, after going through it again I got the same idea, but you're right, I don't see how to prove that $Psi$ is injective or surjective.
– KevinDL
Apr 2 '12 at 11:40
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
Do you have access to Ribes and Zalesskii? I think it could be nutted out from that, but it looks like it might take me a while and I don't have enough time!
– Juan S
Apr 2 '12 at 11:53
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
No, I can't seem to find it. I'm wondering if this is the best way, since it's only an exercise in 'Introduction to homological algebra' by Weibel, I don't think this needs a long proof.
– KevinDL
Apr 2 '12 at 11:57
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
That's interesting. What is the reference in Weibel? Although it looks like it might be a bit of work - it's probably just 'following your nose' - i.e. messy, but maybe not that hard
– Juan S
Apr 2 '12 at 12:01
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
Exercise 6.11.10, which is necessary for the proof of Theorem 6.11.13.
– KevinDL
Apr 2 '12 at 12:06
add a comment |
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