On cohomology of flasque sheaves of abelian groups












0












$begingroup$


$underline {Background}$: Suppose we have a short exact sequence of sheaves of abelian groups on $X$ as follows,
$0tomathcal {F}to mathcal {I}to mathcal {G}to 0$ with $mathcal F$ a flasque sheaf and $mathcal I$ injective.Then we have the following two exact sequences separately



$1)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to 0$ ,



and the usual cohomology long exact sequence



$2)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to H^1(X,mathcal {F})to H^1(X,mathcal {I})to H^1(X,mathcal {G})to.......to H^i(X,mathcal {F})to H^i(X,mathcal {I})to H^i(X,mathcal {G})to H^{i+1}(X,mathcal {F})to H^{i+1}(X,mathcal {I})to H^{i+1}(X,mathcal {G})to........$



If the maps between $mathcal {F}(X)$,$mathcal {I}(X)$ and $mathcal {G}(X)$ are same(in fact it is enough to have maps between $mathcal {I}(X)$,$mathcal {G}(X)$ are same)in both the cases then I have shown that $H^1(X,mathcal F)=0$.



But I only know that in $2)$,i.e in long exact sequence $mathcal{I}(X)$ is same as zeroth right derived functor of global section functor of $mathcal I$(same goes in case of $mathcal G$.$mathcal F$)



$underline {Question}$:$a)$Are the maps between $mathcal {I}(X)$,$mathcal {G}(X)$ same in both the cases(i.e in $1)$ and in $2)$)?.If it is so then how to see it explicitly?



$b)$ If the answer to the first question is no then how to see $H^1(X,mathcal F)=0$ ?



I tried to look at the process of the establishment of long exact sequence which uses snake lemma ,but it's still not very explicit to me that the maps are same in both the cases.



Any help from anyone is welcome.






algebraic-geometry homology-cohomology homological-algebra sheaf-cohomology







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    $underline {Background}$: Suppose we have a short exact sequence of sheaves of abelian groups on $X$ as follows,
    $0tomathcal {F}to mathcal {I}to mathcal {G}to 0$ with $mathcal F$ a flasque sheaf and $mathcal I$ injective.Then we have the following two exact sequences separately



    $1)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to 0$ ,



    and the usual cohomology long exact sequence



    $2)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to H^1(X,mathcal {F})to H^1(X,mathcal {I})to H^1(X,mathcal {G})to.......to H^i(X,mathcal {F})to H^i(X,mathcal {I})to H^i(X,mathcal {G})to H^{i+1}(X,mathcal {F})to H^{i+1}(X,mathcal {I})to H^{i+1}(X,mathcal {G})to........$



    If the maps between $mathcal {F}(X)$,$mathcal {I}(X)$ and $mathcal {G}(X)$ are same(in fact it is enough to have maps between $mathcal {I}(X)$,$mathcal {G}(X)$ are same)in both the cases then I have shown that $H^1(X,mathcal F)=0$.



    But I only know that in $2)$,i.e in long exact sequence $mathcal{I}(X)$ is same as zeroth right derived functor of global section functor of $mathcal I$(same goes in case of $mathcal G$.$mathcal F$)



    $underline {Question}$:$a)$Are the maps between $mathcal {I}(X)$,$mathcal {G}(X)$ same in both the cases(i.e in $1)$ and in $2)$)?.If it is so then how to see it explicitly?



    $b)$ If the answer to the first question is no then how to see $H^1(X,mathcal F)=0$ ?



    I tried to look at the process of the establishment of long exact sequence which uses snake lemma ,but it's still not very explicit to me that the maps are same in both the cases.



    Any help from anyone is welcome.






    algebraic-geometry homology-cohomology homological-algebra sheaf-cohomology







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $underline {Background}$: Suppose we have a short exact sequence of sheaves of abelian groups on $X$ as follows,
      $0tomathcal {F}to mathcal {I}to mathcal {G}to 0$ with $mathcal F$ a flasque sheaf and $mathcal I$ injective.Then we have the following two exact sequences separately



      $1)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to 0$ ,



      and the usual cohomology long exact sequence



      $2)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to H^1(X,mathcal {F})to H^1(X,mathcal {I})to H^1(X,mathcal {G})to.......to H^i(X,mathcal {F})to H^i(X,mathcal {I})to H^i(X,mathcal {G})to H^{i+1}(X,mathcal {F})to H^{i+1}(X,mathcal {I})to H^{i+1}(X,mathcal {G})to........$



      If the maps between $mathcal {F}(X)$,$mathcal {I}(X)$ and $mathcal {G}(X)$ are same(in fact it is enough to have maps between $mathcal {I}(X)$,$mathcal {G}(X)$ are same)in both the cases then I have shown that $H^1(X,mathcal F)=0$.



      But I only know that in $2)$,i.e in long exact sequence $mathcal{I}(X)$ is same as zeroth right derived functor of global section functor of $mathcal I$(same goes in case of $mathcal G$.$mathcal F$)



      $underline {Question}$:$a)$Are the maps between $mathcal {I}(X)$,$mathcal {G}(X)$ same in both the cases(i.e in $1)$ and in $2)$)?.If it is so then how to see it explicitly?



      $b)$ If the answer to the first question is no then how to see $H^1(X,mathcal F)=0$ ?



      I tried to look at the process of the establishment of long exact sequence which uses snake lemma ,but it's still not very explicit to me that the maps are same in both the cases.



      Any help from anyone is welcome.






      algebraic-geometry homology-cohomology homological-algebra sheaf-cohomology







      share|cite|improve this question











      $endgroup$




      $underline {Background}$: Suppose we have a short exact sequence of sheaves of abelian groups on $X$ as follows,
      $0tomathcal {F}to mathcal {I}to mathcal {G}to 0$ with $mathcal F$ a flasque sheaf and $mathcal I$ injective.Then we have the following two exact sequences separately



      $1)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to 0$ ,



      and the usual cohomology long exact sequence



      $2)$ $0tomathcal {F}(X)to mathcal {I}(X)to mathcal {G}(X)to H^1(X,mathcal {F})to H^1(X,mathcal {I})to H^1(X,mathcal {G})to.......to H^i(X,mathcal {F})to H^i(X,mathcal {I})to H^i(X,mathcal {G})to H^{i+1}(X,mathcal {F})to H^{i+1}(X,mathcal {I})to H^{i+1}(X,mathcal {G})to........$



      If the maps between $mathcal {F}(X)$,$mathcal {I}(X)$ and $mathcal {G}(X)$ are same(in fact it is enough to have maps between $mathcal {I}(X)$,$mathcal {G}(X)$ are same)in both the cases then I have shown that $H^1(X,mathcal F)=0$.



      But I only know that in $2)$,i.e in long exact sequence $mathcal{I}(X)$ is same as zeroth right derived functor of global section functor of $mathcal I$(same goes in case of $mathcal G$.$mathcal F$)



      $underline {Question}$:$a)$Are the maps between $mathcal {I}(X)$,$mathcal {G}(X)$ same in both the cases(i.e in $1)$ and in $2)$)?.If it is so then how to see it explicitly?



      $b)$ If the answer to the first question is no then how to see $H^1(X,mathcal F)=0$ ?



      I tried to look at the process of the establishment of long exact sequence which uses snake lemma ,but it's still not very explicit to me that the maps are same in both the cases.



      Any help from anyone is welcome.






      algebraic-geometry homology-cohomology homological-algebra sheaf-cohomology




      algebraic-geometry homology-cohomology homological-algebra sheaf-cohomology






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 3 '18 at 13:56







      HARRY

















      asked Dec 3 '18 at 13:50









      HARRYHARRY

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          $begingroup$

          They are literally the same, since your long exact sequence actually comes from injective resolutions for your sheaves, in particular this long exact sequence coincides everywhere, with the left exact sequence $$0 to mathcal{F}(X) to mathcal{I}(X)to mathcal{G}(X).$$
          And now you can finish your argument, since your short exact sequence makes said left exact sequence into a sequence itself.
          About the precise mechanics about injective resolutions and long exact sequences arising there I recommend the book on homological algebra by Weibel, but there you can see precisely where those maps in the long exact sequence actually come from (injective resolutions).






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            $begingroup$

            They are literally the same, since your long exact sequence actually comes from injective resolutions for your sheaves, in particular this long exact sequence coincides everywhere, with the left exact sequence $$0 to mathcal{F}(X) to mathcal{I}(X)to mathcal{G}(X).$$
            And now you can finish your argument, since your short exact sequence makes said left exact sequence into a sequence itself.
            About the precise mechanics about injective resolutions and long exact sequences arising there I recommend the book on homological algebra by Weibel, but there you can see precisely where those maps in the long exact sequence actually come from (injective resolutions).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              They are literally the same, since your long exact sequence actually comes from injective resolutions for your sheaves, in particular this long exact sequence coincides everywhere, with the left exact sequence $$0 to mathcal{F}(X) to mathcal{I}(X)to mathcal{G}(X).$$
              And now you can finish your argument, since your short exact sequence makes said left exact sequence into a sequence itself.
              About the precise mechanics about injective resolutions and long exact sequences arising there I recommend the book on homological algebra by Weibel, but there you can see precisely where those maps in the long exact sequence actually come from (injective resolutions).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                They are literally the same, since your long exact sequence actually comes from injective resolutions for your sheaves, in particular this long exact sequence coincides everywhere, with the left exact sequence $$0 to mathcal{F}(X) to mathcal{I}(X)to mathcal{G}(X).$$
                And now you can finish your argument, since your short exact sequence makes said left exact sequence into a sequence itself.
                About the precise mechanics about injective resolutions and long exact sequences arising there I recommend the book on homological algebra by Weibel, but there you can see precisely where those maps in the long exact sequence actually come from (injective resolutions).






                share|cite|improve this answer









                $endgroup$



                They are literally the same, since your long exact sequence actually comes from injective resolutions for your sheaves, in particular this long exact sequence coincides everywhere, with the left exact sequence $$0 to mathcal{F}(X) to mathcal{I}(X)to mathcal{G}(X).$$
                And now you can finish your argument, since your short exact sequence makes said left exact sequence into a sequence itself.
                About the precise mechanics about injective resolutions and long exact sequences arising there I recommend the book on homological algebra by Weibel, but there you can see precisely where those maps in the long exact sequence actually come from (injective resolutions).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 14:05









                EnkiduEnkidu

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