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CoRes $circ$ Res $ = [G:H]Id$ on the cohomology groups of a profinite group

Interpretations of the first cohomology groupOther differentials for group cohomology other than the standard one.Question on $operatorname{res}^G_U circ operatorname{cor}^U_G = N_{G/U}$Inductive definition of group cohomology?Group cohomology via resolutionsWhether a functor is exact?Cohomology groups of unramified field extensionsGalois group is a profinite groupGroup Cohomology is a universal $delta$-FunctorHow to prove $C^0(G,A)$ isomorphic to $A$

1

$begingroup$

Let $G$ be a profinite group, $H leq G$ open. It is known that thus $H$ is closed and has finite index in $G$.

Any $G-$module is an $H-$module and one can construct the restriction map as the extension of $H^0(G,A) = A^G to A^H = H^0(H,A)$, using the universality of the cohomological functor.

The trace map works the other way: we have a natural map CoRes$:H^0(H,-) Rightarrow H^0(G, -)$ given by $a mapsto sum ag_i$ where $g_i$ are representative of the right cosets of $H$ in $G$.

Clearly it follows that in dimension $0$, CoRes $circ$ Res $ = [G:H]Id$.

In the book by Ribes he states that the reason this implies equality in all dimensions of cohomology is because of the following proposition:

If $eta_0:H^0(G,-) Rightarrow H^0(G,-)$ is an isomorphism, then its
extension to the cohomology complex is also an isomorphism.

However,

1. I don't see why it follows from the above, does it?




2. Doesn't it simply follow by the universality of the cohomological functor? The extensions of these natural transformations must equal. Am I missing something?

homology-cohomology group-cohomology profinite-groups

share|cite|improve this question

edited Mar 14 at 21:11

Mariah

asked Mar 14 at 20:42

MariahMariah

2,0431718

$endgroup$

add a comment | 

1

$begingroup$

Let $G$ be a profinite group, $H leq G$ open. It is known that thus $H$ is closed and has finite index in $G$.

Any $G-$module is an $H-$module and one can construct the restriction map as the extension of $H^0(G,A) = A^G to A^H = H^0(H,A)$, using the universality of the cohomological functor.

The trace map works the other way: we have a natural map CoRes$:H^0(H,-) Rightarrow H^0(G, -)$ given by $a mapsto sum ag_i$ where $g_i$ are representative of the right cosets of $H$ in $G$.

Clearly it follows that in dimension $0$, CoRes $circ$ Res $ = [G:H]Id$.

In the book by Ribes he states that the reason this implies equality in all dimensions of cohomology is because of the following proposition:

If $eta_0:H^0(G,-) Rightarrow H^0(G,-)$ is an isomorphism, then its
extension to the cohomology complex is also an isomorphism.

However,

1. I don't see why it follows from the above, does it?




2. Doesn't it simply follow by the universality of the cohomological functor? The extensions of these natural transformations must equal. Am I missing something?

homology-cohomology group-cohomology profinite-groups

share|cite|improve this question

edited Mar 14 at 21:11

Mariah

asked Mar 14 at 20:42

MariahMariah

2,0431718

$endgroup$

add a comment | 

$begingroup$

Let $G$ be a profinite group, $H leq G$ open. It is known that thus $H$ is closed and has finite index in $G$.

Any $G-$module is an $H-$module and one can construct the restriction map as the extension of $H^0(G,A) = A^G to A^H = H^0(H,A)$, using the universality of the cohomological functor.

The trace map works the other way: we have a natural map CoRes$:H^0(H,-) Rightarrow H^0(G, -)$ given by $a mapsto sum ag_i$ where $g_i$ are representative of the right cosets of $H$ in $G$.

Clearly it follows that in dimension $0$, CoRes $circ$ Res $ = [G:H]Id$.

In the book by Ribes he states that the reason this implies equality in all dimensions of cohomology is because of the following proposition:

If $eta_0:H^0(G,-) Rightarrow H^0(G,-)$ is an isomorphism, then its
extension to the cohomology complex is also an isomorphism.

However,

1. I don't see why it follows from the above, does it?




2. Doesn't it simply follow by the universality of the cohomological functor? The extensions of these natural transformations must equal. Am I missing something?

homology-cohomology group-cohomology profinite-groups

share|cite|improve this question

edited Mar 14 at 21:11

Mariah

asked Mar 14 at 20:42

MariahMariah

2,0431718

$endgroup$

share|cite|improve this question

edited Mar 14 at 21:11

Mariah

asked Mar 14 at 20:42

MariahMariah

2,0431718

share|cite|improve this question

edited Mar 14 at 21:11

Mariah

asked Mar 14 at 20:42

MariahMariah

2,0431718

asked Mar 14 at 20:42

MariahMariah

2,0431718

asked Mar 14 at 20:42

MariahMariah

2,0431718