|
6
|
|
edited Jul 7 2011 at 9:16 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the dual of the cohomology with compact support of $X$.
$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
EDIT : to answer precisely the question I include a summary of the comments.
there is a typo in the wiki: $[F,X[k]]$ should be understood as $[F,F[k]]$. for any sheaf of $F$-modules $S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$. Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.
|
|
|
|
|
5
|
|
edited Jun 24 2011 at 13:39 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the cohomology with compact support of $X$.
$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
EDIT : to answer precisely the question I include a summary of the comments.
there is a typo in the wiki: $$[F,X[k]]$ [F,X[k]]$ should be understood as $[F,F[k]]$. for any sheaf of $F$-modules $S$, S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$. Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.
|
|
|
|
|
4
|
|
edited Jun 24 2011 at 6:40 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the cohomology with compact support of $X$.
$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
EDIT : to answer precisely the question I include a summary of the comments.
there is a typo in the wiki: $$[F,X[k]]$ should be understood as $[F,F[k]]$. for any sheaf of $F$-modules $S$, $[F[−k],S]=H^k(X,S)$. Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.
|
|
|
|
3
|
|
edited Jun 24 2011 at 6:27 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to*$. f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_*(f_!F,F)=f_*Hom_X(F,f^!FHom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the cohomology with compact support of $X$.
$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
|
|
|
|
|
2
|
|
edited Jun 23 2011 at 21:09 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to*$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_*(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the cohomology with compact support of $X$.
$Hom_X(F,f^!F)=f^!F$ Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(f^!F)=\Gamma(X,f^!F)$, f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.
|
|
|
|
|
1
|
|
answered Jun 23 2011 at 20:59 |
I think here is how one should understand the last paragraph of the wiki.
Consider $f:X\to*$. We have (all functors are derived and my $Hom$ are sheaf $Hom$)
$$
Hom_*(f_!F,F)=f_*Hom_X(F,f^!F)
$$
$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the cohomology with compact support of $X$.
$Hom_X(F,f^!F)=f^!F$ and thus the r.h.s. is $f_*(f^!F)=\Gamma(X,f^!F)$, the homology of $X$.
|
|
|
