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Misunderstanding in definition of homology of groups
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I am following Brown's 'Cohomology of groups' and the homology is defined as follows:
Let $cdots rightarrow F_{n}rightarrow F_{n-1}rightarrow cdots rightarrow F_{1}rightarrow F_{0}rightarrow mathbb{Z}rightarrow 0$ be a projective resolution of $mathbb{Z}$ over $mathbb{Z}G$.
Apply the functor $mathbb{Z} otimes_{mathbb{Z}G} -$, so we have
$cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n-1}rightarrow cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{1}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{0}rightarrow mathbb{Z}otimes_{mathbb{Z}G}mathbb{Z}rightarrow 0$.
Then, $H_{i}G=H_{i}(F_{G})$.
I am not understanding something very basic: If projective resolutions are exact by definition and $mathbb{Z} otimes_{mathbb{Z}G} -$ is a right-exact functor, why is not the homology 0?
Sorry for the stupid question and thanks in advance!
homology-cohomology group-cohomology
asked Jan 15 at 21:30
KarenKaren
946
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add a comment |
$begingroup$
I am following Brown's 'Cohomology of groups' and the homology is defined as follows:
Let $cdots rightarrow F_{n}rightarrow F_{n-1}rightarrow cdots rightarrow F_{1}rightarrow F_{0}rightarrow mathbb{Z}rightarrow 0$ be a projective resolution of $mathbb{Z}$ over $mathbb{Z}G$.
Apply the functor $mathbb{Z} otimes_{mathbb{Z}G} -$, so we have
$cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n-1}rightarrow cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{1}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{0}rightarrow mathbb{Z}otimes_{mathbb{Z}G}mathbb{Z}rightarrow 0$.
Then, $H_{i}G=H_{i}(F_{G})$.
I am not understanding something very basic: If projective resolutions are exact by definition and $mathbb{Z} otimes_{mathbb{Z}G} -$ is a right-exact functor, why is not the homology 0?
Sorry for the stupid question and thanks in advance!
homology-cohomology group-cohomology
asked Jan 15 at 21:30
KarenKaren
946
$endgroup$
2
$begingroup$
Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37
add a comment |
$begingroup$
I am following Brown's 'Cohomology of groups' and the homology is defined as follows:
Let $cdots rightarrow F_{n}rightarrow F_{n-1}rightarrow cdots rightarrow F_{1}rightarrow F_{0}rightarrow mathbb{Z}rightarrow 0$ be a projective resolution of $mathbb{Z}$ over $mathbb{Z}G$.
Apply the functor $mathbb{Z} otimes_{mathbb{Z}G} -$, so we have
$cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n-1}rightarrow cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{1}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{0}rightarrow mathbb{Z}otimes_{mathbb{Z}G}mathbb{Z}rightarrow 0$.
Then, $H_{i}G=H_{i}(F_{G})$.
I am not understanding something very basic: If projective resolutions are exact by definition and $mathbb{Z} otimes_{mathbb{Z}G} -$ is a right-exact functor, why is not the homology 0?
Sorry for the stupid question and thanks in advance!
homology-cohomology group-cohomology
asked Jan 15 at 21:30
KarenKaren
946
$endgroup$
I am following Brown's 'Cohomology of groups' and the homology is defined as follows:
Let $cdots rightarrow F_{n}rightarrow F_{n-1}rightarrow cdots rightarrow F_{1}rightarrow F_{0}rightarrow mathbb{Z}rightarrow 0$ be a projective resolution of $mathbb{Z}$ over $mathbb{Z}G$.
Apply the functor $mathbb{Z} otimes_{mathbb{Z}G} -$, so we have
$cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{n-1}rightarrow cdots rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{1}rightarrow mathbb{Z}otimes_{mathbb{Z}G}F_{0}rightarrow mathbb{Z}otimes_{mathbb{Z}G}mathbb{Z}rightarrow 0$.
Then, $H_{i}G=H_{i}(F_{G})$.
I am not understanding something very basic: If projective resolutions are exact by definition and $mathbb{Z} otimes_{mathbb{Z}G} -$ is a right-exact functor, why is not the homology 0?
Sorry for the stupid question and thanks in advance!
homology-cohomology group-cohomology
homology-cohomology group-cohomology
asked Jan 15 at 21:30
KarenKaren
946
asked Jan 15 at 21:30
KarenKaren
946
asked Jan 15 at 21:30
KarenKaren
946
asked Jan 15 at 21:30
KarenKaren
946
asked Jan 15 at 21:30
KarenKaren
946
946
2
$begingroup$
Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37
add a comment |
2
$begingroup$
Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37
2
2
$begingroup$
Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37
$begingroup$
Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37
add a comment |
1 Answer
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A right exact functor does not preserve arbitrary exact sequences, but only those of the form $Ato Bto Cto 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).
answered Jan 15 at 21:39
PedroPedro
2,9291720
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$begingroup$
A right exact functor does not preserve arbitrary exact sequences, but only those of the form $Ato Bto Cto 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).
answered Jan 15 at 21:39
PedroPedro
2,9291720
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add a comment |
$begingroup$
A right exact functor does not preserve arbitrary exact sequences, but only those of the form $Ato Bto Cto 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).
answered Jan 15 at 21:39
PedroPedro
2,9291720
$endgroup$
add a comment |
$begingroup$
A right exact functor does not preserve arbitrary exact sequences, but only those of the form $Ato Bto Cto 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).
answered Jan 15 at 21:39
PedroPedro
2,9291720
$endgroup$
A right exact functor does not preserve arbitrary exact sequences, but only those of the form $Ato Bto Cto 0$.
More precisely, a right exact functor is by definition a functor that preserves finite colimits. We can build all finite colimits from cokernels and finite coproducts, so an additive functor (i.e. a functor that preserves finite products and coprpducts) is right exact if and only if it preserves cokernels, hence if and only if it preserves a sequence as the one above (because such a sequence is exact if and only if the map on the right is the cokernel of the map on the left).
answered Jan 15 at 21:39
PedroPedro
2,9291720
answered Jan 15 at 21:39
PedroPedro
2,9291720
answered Jan 15 at 21:39
PedroPedro
2,9291720
answered Jan 15 at 21:39
PedroPedro
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2,9291720
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Since the functor is only right-exact, the new complex you get will not necessarily be exact.
$endgroup$
– Wojowu
Jan 15 at 21:37