Mixing
Chain complexes question between free $K$-modules and almost zero chain.
$begingroup$
I have this sentence from the article Resolutions for extensions of groups by C.T.C. Wall:
Let $Z(K)$ denote the group ring of the group $K$ over the ring $Z$ of the integers. Let $otimes_K$ denote tensor product over the ring $Z(K)$. Let $Z$ also denote the chain complex in which the $0$-th chain group is $Z$ and all the other are zero. Let $B$ be a chain complex of free $K$-modules
$$ 0leftarrow B_0 leftarrow B_1 leftarrow B_r leftarrow dots $$
and $epsilon colon Brightarrow Z$ a $K$-map of chain complexes ($K$ acts trivially over $Z$) inducing an isomorphism of homology. Then we call $B$ a free resolution for $K$ and $epsilon$ its augmentation.
My question is this:
- The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?
- If the chain complex $Z$ is like in 1, then how is $epsilon$ and why it induces an isomorphism in homology?
abstract-algebra group-cohomology
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
$endgroup$
add a comment |
$begingroup$
I have this sentence from the article Resolutions for extensions of groups by C.T.C. Wall:
Let $Z(K)$ denote the group ring of the group $K$ over the ring $Z$ of the integers. Let $otimes_K$ denote tensor product over the ring $Z(K)$. Let $Z$ also denote the chain complex in which the $0$-th chain group is $Z$ and all the other are zero. Let $B$ be a chain complex of free $K$-modules
$$ 0leftarrow B_0 leftarrow B_1 leftarrow B_r leftarrow dots $$
and $epsilon colon Brightarrow Z$ a $K$-map of chain complexes ($K$ acts trivially over $Z$) inducing an isomorphism of homology. Then we call $B$ a free resolution for $K$ and $epsilon$ its augmentation.
My question is this:
- The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?
- If the chain complex $Z$ is like in 1, then how is $epsilon$ and why it induces an isomorphism in homology?
abstract-algebra group-cohomology
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
$endgroup$
add a comment |
$begingroup$
I have this sentence from the article Resolutions for extensions of groups by C.T.C. Wall:
Let $Z(K)$ denote the group ring of the group $K$ over the ring $Z$ of the integers. Let $otimes_K$ denote tensor product over the ring $Z(K)$. Let $Z$ also denote the chain complex in which the $0$-th chain group is $Z$ and all the other are zero. Let $B$ be a chain complex of free $K$-modules
$$ 0leftarrow B_0 leftarrow B_1 leftarrow B_r leftarrow dots $$
and $epsilon colon Brightarrow Z$ a $K$-map of chain complexes ($K$ acts trivially over $Z$) inducing an isomorphism of homology. Then we call $B$ a free resolution for $K$ and $epsilon$ its augmentation.
My question is this:
- The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?
- If the chain complex $Z$ is like in 1, then how is $epsilon$ and why it induces an isomorphism in homology?
abstract-algebra group-cohomology
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
$endgroup$
I have this sentence from the article Resolutions for extensions of groups by C.T.C. Wall:
Let $Z(K)$ denote the group ring of the group $K$ over the ring $Z$ of the integers. Let $otimes_K$ denote tensor product over the ring $Z(K)$. Let $Z$ also denote the chain complex in which the $0$-th chain group is $Z$ and all the other are zero. Let $B$ be a chain complex of free $K$-modules
$$ 0leftarrow B_0 leftarrow B_1 leftarrow B_r leftarrow dots $$
and $epsilon colon Brightarrow Z$ a $K$-map of chain complexes ($K$ acts trivially over $Z$) inducing an isomorphism of homology. Then we call $B$ a free resolution for $K$ and $epsilon$ its augmentation.
My question is this:
- The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?
- If the chain complex $Z$ is like in 1, then how is $epsilon$ and why it induces an isomorphism in homology?
abstract-algebra group-cohomology
abstract-algebra group-cohomology
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
asked Jan 30 at 10:04
MonsieurGaloisMonsieurGalois
3,4581334
3,4581334
add a comment |
add a comment |
1 Answer
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$begingroup$
It's actually $0leftarrow Z leftarrow 0leftarrow 0 leftarrow dots$ You want $epsilon$ to induce a map $B_0to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0to Z$ which isn't an isomorphism)
$epsilon$ is given, there's no statement that there's always such an $epsilon$ (for starters, for such an $epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $epsilon$ exists, and then we call $B$ a free resolution
answered Jan 30 at 10:20
MaxMax
15.9k11144
$endgroup$
add a comment |
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1 Answer
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$begingroup$
It's actually $0leftarrow Z leftarrow 0leftarrow 0 leftarrow dots$ You want $epsilon$ to induce a map $B_0to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0to Z$ which isn't an isomorphism)
$epsilon$ is given, there's no statement that there's always such an $epsilon$ (for starters, for such an $epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $epsilon$ exists, and then we call $B$ a free resolution
answered Jan 30 at 10:20
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
It's actually $0leftarrow Z leftarrow 0leftarrow 0 leftarrow dots$ You want $epsilon$ to induce a map $B_0to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0to Z$ which isn't an isomorphism)
$epsilon$ is given, there's no statement that there's always such an $epsilon$ (for starters, for such an $epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $epsilon$ exists, and then we call $B$ a free resolution
answered Jan 30 at 10:20
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
It's actually $0leftarrow Z leftarrow 0leftarrow 0 leftarrow dots$ You want $epsilon$ to induce a map $B_0to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0to Z$ which isn't an isomorphism)
$epsilon$ is given, there's no statement that there's always such an $epsilon$ (for starters, for such an $epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $epsilon$ exists, and then we call $B$ a free resolution
answered Jan 30 at 10:20
MaxMax
15.9k11144
$endgroup$
It's actually $0leftarrow Z leftarrow 0leftarrow 0 leftarrow dots$ You want $epsilon$ to induce a map $B_0to Z$ (otherwise it couldn't induce an ismorphism in homology, because you would have $0to Z$ which isn't an isomorphism)
$epsilon$ is given, there's no statement that there's always such an $epsilon$ (for starters, for such an $epsilon$ to exist, the complex $B$ must be exact in positive degrees) : we're assuming that such an $epsilon$ exists, and then we call $B$ a free resolution
answered Jan 30 at 10:20
MaxMax
15.9k11144
answered Jan 30 at 10:20
MaxMax
15.9k11144
answered Jan 30 at 10:20
MaxMax
15.9k11144
answered Jan 30 at 10:20
MaxMax
15.9k11144
15.9k11144
add a comment |
add a comment |
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