For a group homomorphism $alpha : Grightarrow G'$, does the pullback $alpha^# : Mod(G')rightarrow Mod(G)$...
$begingroup$
Let $alpha : Grightarrow G'$ be a group homomorphism.
There is a natural functor $alpha^# : Mod(G')rightarrow Mod(G)$ sending a $G'$-module $M$ to the $G$-module given by the same underlying abelian group as $M$, with $G$-action defined via $alpha$.
This functor is exact, but it doesn't obviously preserve projectives or injectives (at least the hypotheses of the adjoint functor criterion does not apply here).
However, it is stated in Ken Brown's book "Cohomology of Groups" (II.6), that $alpha^#$ sends $G'$-projectives to $G$-modules which are acyclic for group homology.
Is this clear? I don't see why.
As requested, here is the full statement:
"Given a homomorphism $alpha : Grightarrow G'$ and projective resolutions $F$ and $F'$ of $mathbb{Z}$ over $mathbb{Z}G$ and $mathbb{Z}G'$ respectively, we can regard $F'$ as a complex of $G$-modules via $alpha$. Then, $F'$ is acyclic (although not projective, in general, over $mathbb{Z}G$), so the fundamental lemma I.7.4 gives us an augmentation-preserving $G$-chain map $tau : Frightarrow F'$, well defined up to homotopy. The condition that $tau$ be a $G$-map is expressed by the formula $tau(gx) = alpha(g)tau(x)$ for $gin G,xin F$. Clearly $tau$ induces a map $F_Grightarrow F'_{G'}$, well-defined up to homotopy, hence we obtain a well-defined map $alpha_* : H_*(G)rightarrow H_*(G')$"
group-theory commutative-algebra homology-cohomology homological-algebra
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
$endgroup$
add a comment |
$begingroup$
Let $alpha : Grightarrow G'$ be a group homomorphism.
There is a natural functor $alpha^# : Mod(G')rightarrow Mod(G)$ sending a $G'$-module $M$ to the $G$-module given by the same underlying abelian group as $M$, with $G$-action defined via $alpha$.
This functor is exact, but it doesn't obviously preserve projectives or injectives (at least the hypotheses of the adjoint functor criterion does not apply here).
However, it is stated in Ken Brown's book "Cohomology of Groups" (II.6), that $alpha^#$ sends $G'$-projectives to $G$-modules which are acyclic for group homology.
Is this clear? I don't see why.
As requested, here is the full statement:
"Given a homomorphism $alpha : Grightarrow G'$ and projective resolutions $F$ and $F'$ of $mathbb{Z}$ over $mathbb{Z}G$ and $mathbb{Z}G'$ respectively, we can regard $F'$ as a complex of $G$-modules via $alpha$. Then, $F'$ is acyclic (although not projective, in general, over $mathbb{Z}G$), so the fundamental lemma I.7.4 gives us an augmentation-preserving $G$-chain map $tau : Frightarrow F'$, well defined up to homotopy. The condition that $tau$ be a $G$-map is expressed by the formula $tau(gx) = alpha(g)tau(x)$ for $gin G,xin F$. Clearly $tau$ induces a map $F_Grightarrow F'_{G'}$, well-defined up to homotopy, hence we obtain a well-defined map $alpha_* : H_*(G)rightarrow H_*(G')$"
group-theory commutative-algebra homology-cohomology homological-algebra
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
$endgroup$
$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04
add a comment |
$begingroup$
Let $alpha : Grightarrow G'$ be a group homomorphism.
There is a natural functor $alpha^# : Mod(G')rightarrow Mod(G)$ sending a $G'$-module $M$ to the $G$-module given by the same underlying abelian group as $M$, with $G$-action defined via $alpha$.
This functor is exact, but it doesn't obviously preserve projectives or injectives (at least the hypotheses of the adjoint functor criterion does not apply here).
However, it is stated in Ken Brown's book "Cohomology of Groups" (II.6), that $alpha^#$ sends $G'$-projectives to $G$-modules which are acyclic for group homology.
Is this clear? I don't see why.
As requested, here is the full statement:
"Given a homomorphism $alpha : Grightarrow G'$ and projective resolutions $F$ and $F'$ of $mathbb{Z}$ over $mathbb{Z}G$ and $mathbb{Z}G'$ respectively, we can regard $F'$ as a complex of $G$-modules via $alpha$. Then, $F'$ is acyclic (although not projective, in general, over $mathbb{Z}G$), so the fundamental lemma I.7.4 gives us an augmentation-preserving $G$-chain map $tau : Frightarrow F'$, well defined up to homotopy. The condition that $tau$ be a $G$-map is expressed by the formula $tau(gx) = alpha(g)tau(x)$ for $gin G,xin F$. Clearly $tau$ induces a map $F_Grightarrow F'_{G'}$, well-defined up to homotopy, hence we obtain a well-defined map $alpha_* : H_*(G)rightarrow H_*(G')$"
group-theory commutative-algebra homology-cohomology homological-algebra
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
$endgroup$
Let $alpha : Grightarrow G'$ be a group homomorphism.
There is a natural functor $alpha^# : Mod(G')rightarrow Mod(G)$ sending a $G'$-module $M$ to the $G$-module given by the same underlying abelian group as $M$, with $G$-action defined via $alpha$.
This functor is exact, but it doesn't obviously preserve projectives or injectives (at least the hypotheses of the adjoint functor criterion does not apply here).
However, it is stated in Ken Brown's book "Cohomology of Groups" (II.6), that $alpha^#$ sends $G'$-projectives to $G$-modules which are acyclic for group homology.
Is this clear? I don't see why.
As requested, here is the full statement:
"Given a homomorphism $alpha : Grightarrow G'$ and projective resolutions $F$ and $F'$ of $mathbb{Z}$ over $mathbb{Z}G$ and $mathbb{Z}G'$ respectively, we can regard $F'$ as a complex of $G$-modules via $alpha$. Then, $F'$ is acyclic (although not projective, in general, over $mathbb{Z}G$), so the fundamental lemma I.7.4 gives us an augmentation-preserving $G$-chain map $tau : Frightarrow F'$, well defined up to homotopy. The condition that $tau$ be a $G$-map is expressed by the formula $tau(gx) = alpha(g)tau(x)$ for $gin G,xin F$. Clearly $tau$ induces a map $F_Grightarrow F'_{G'}$, well-defined up to homotopy, hence we obtain a well-defined map $alpha_* : H_*(G)rightarrow H_*(G')$"
group-theory commutative-algebra homology-cohomology homological-algebra
group-theory commutative-algebra homology-cohomology homological-algebra
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
edited Jan 6 at 0:51
stupid_question_bot
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
asked Jan 6 at 0:27
stupid_question_botstupid_question_bot
1,940415
1,940415
$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04
add a comment |
$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04
$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04
add a comment |
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$begingroup$
Do you make no assumption on $alpha$? For example, are these groups finite? Can you include the full statement in Brown's book?
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 0:39
$begingroup$
@PedroTamaroff I've added the paragraph in which he made this statement.
$endgroup$
– stupid_question_bot
Jan 6 at 0:52
$begingroup$
@PedroTamaroff Actually now it seems like if $G' = 1$ and $G$ nontrivial, then the statement is clearly false, so now I'm wondering what he meant...
$endgroup$
– stupid_question_bot
Jan 6 at 1:00
$begingroup$
Yes. If you assume that the map is a monomorphism with finite index, which is always true for inclusions between finite groups, then what you want is Shapiro's lemma.
$endgroup$
– Pedro Tamaroff♦
Jan 6 at 1:04