Chain complexes question between free $K$-modules and almost zero chain.












1












$begingroup$


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:




  1. The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?

  2. 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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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:




    1. The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?

    2. 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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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:




      1. The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?

      2. 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







      share|cite|improve this question









      $endgroup$




      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:




      1. The chain complex $Z$ will be $Zleftarrow 0 leftarrow 0leftarrow dots$ with $Z=mathbb{Z}$?

      2. 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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 10:04









      MonsieurGaloisMonsieurGalois

      3,4581334




      3,4581334






















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          1. 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)


          2. $epsilon$ 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







          share|cite|improve this answer









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            $begingroup$


            1. 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)


            2. $epsilon$ 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







            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$


              1. 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)


              2. $epsilon$ 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







              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$


                1. 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)


                2. $epsilon$ 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







                share|cite|improve this answer









                $endgroup$




                1. 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)


                2. $epsilon$ 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








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                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 10:20









                MaxMax

                15.9k11144




                15.9k11144






























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