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Leary cover and good cover
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On most textbooks, the definition of Leray cover is: Let $displaystyle mathfrak U=U_i$ be an open cover of the topological space $X$, and $mathcal F$ a sheaf on $X$. We say that $mathfrak U$ is a Leray cover with respect to $mathcal F$ if, for every nonempty finite set $displaystyle i_1,ldots ,i_n$ of indices, and for all $k>0$, we have that $displaystyle H^k(U_i_1cap cdots cap U_i_n,mathcal F)=0$, in the derived functor cohomology.
But on Postnikov's Geometry VI: Riemannian Geometry, the definition of Leray cover is exactly the same definition of good cover.
I wonder the definition of good cover implies the usual definition of Leray cover.
Moreover, I wonder when a sheaf cohomology is homotopy invariant? And suppose $X$ is contractible, then if $displaystyle H^k(X,mathcal F)=0$ for $kge1$?
complex-geometry sheaf-theory sheaf-cohomology
asked Jul 19 at 15:02
Danny
784212
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up vote
1
down vote
favorite
On most textbooks, the definition of Leray cover is: Let $displaystyle mathfrak U=U_i$ be an open cover of the topological space $X$, and $mathcal F$ a sheaf on $X$. We say that $mathfrak U$ is a Leray cover with respect to $mathcal F$ if, for every nonempty finite set $displaystyle i_1,ldots ,i_n$ of indices, and for all $k>0$, we have that $displaystyle H^k(U_i_1cap cdots cap U_i_n,mathcal F)=0$, in the derived functor cohomology.
But on Postnikov's Geometry VI: Riemannian Geometry, the definition of Leray cover is exactly the same definition of good cover.
I wonder the definition of good cover implies the usual definition of Leray cover.
Moreover, I wonder when a sheaf cohomology is homotopy invariant? And suppose $X$ is contractible, then if $displaystyle H^k(X,mathcal F)=0$ for $kge1$?
complex-geometry sheaf-theory sheaf-cohomology
asked Jul 19 at 15:02
Danny
784212
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On most textbooks, the definition of Leray cover is: Let $displaystyle mathfrak U=U_i$ be an open cover of the topological space $X$, and $mathcal F$ a sheaf on $X$. We say that $mathfrak U$ is a Leray cover with respect to $mathcal F$ if, for every nonempty finite set $displaystyle i_1,ldots ,i_n$ of indices, and for all $k>0$, we have that $displaystyle H^k(U_i_1cap cdots cap U_i_n,mathcal F)=0$, in the derived functor cohomology.
But on Postnikov's Geometry VI: Riemannian Geometry, the definition of Leray cover is exactly the same definition of good cover.
I wonder the definition of good cover implies the usual definition of Leray cover.
Moreover, I wonder when a sheaf cohomology is homotopy invariant? And suppose $X$ is contractible, then if $displaystyle H^k(X,mathcal F)=0$ for $kge1$?
complex-geometry sheaf-theory sheaf-cohomology
asked Jul 19 at 15:02
Danny
784212
On most textbooks, the definition of Leray cover is: Let $displaystyle mathfrak U=U_i$ be an open cover of the topological space $X$, and $mathcal F$ a sheaf on $X$. We say that $mathfrak U$ is a Leray cover with respect to $mathcal F$ if, for every nonempty finite set $displaystyle i_1,ldots ,i_n$ of indices, and for all $k>0$, we have that $displaystyle H^k(U_i_1cap cdots cap U_i_n,mathcal F)=0$, in the derived functor cohomology.
But on Postnikov's Geometry VI: Riemannian Geometry, the definition of Leray cover is exactly the same definition of good cover.
I wonder the definition of good cover implies the usual definition of Leray cover.
Moreover, I wonder when a sheaf cohomology is homotopy invariant? And suppose $X$ is contractible, then if $displaystyle H^k(X,mathcal F)=0$ for $kge1$?
complex-geometry sheaf-theory sheaf-cohomology
asked Jul 19 at 15:02
Danny
784212
asked Jul 19 at 15:02
Danny
784212
asked Jul 19 at 15:02
Danny
784212
asked Jul 19 at 15:02
Danny
784212
784212
add a comment |Â
add a comment |Â
1 Answer
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Sheaf cohomology is not homotopy invariant, in particular there are sheaf whith non trivial $H^1$ on $[0,1]$. See here for a concrete example.
However, if $F$ is locally constant, i.e $underline A_X$ for $A$ a ring then its cohomology is homotopy invariant since it coincide with the usual singular cohomology with coefficients in $A$ (you might need some hypothesis on $X$, like locally contractible).
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Sheaf cohomology is not homotopy invariant, in particular there are sheaf whith non trivial $H^1$ on $[0,1]$. See here for a concrete example.
However, if $F$ is locally constant, i.e $underline A_X$ for $A$ a ring then its cohomology is homotopy invariant since it coincide with the usual singular cohomology with coefficients in $A$ (you might need some hypothesis on $X$, like locally contractible).
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
add a comment |Â
up vote
1
down vote
accepted
Sheaf cohomology is not homotopy invariant, in particular there are sheaf whith non trivial $H^1$ on $[0,1]$. See here for a concrete example.
However, if $F$ is locally constant, i.e $underline A_X$ for $A$ a ring then its cohomology is homotopy invariant since it coincide with the usual singular cohomology with coefficients in $A$ (you might need some hypothesis on $X$, like locally contractible).
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Sheaf cohomology is not homotopy invariant, in particular there are sheaf whith non trivial $H^1$ on $[0,1]$. See here for a concrete example.
However, if $F$ is locally constant, i.e $underline A_X$ for $A$ a ring then its cohomology is homotopy invariant since it coincide with the usual singular cohomology with coefficients in $A$ (you might need some hypothesis on $X$, like locally contractible).
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
Sheaf cohomology is not homotopy invariant, in particular there are sheaf whith non trivial $H^1$ on $[0,1]$. See here for a concrete example.
However, if $F$ is locally constant, i.e $underline A_X$ for $A$ a ring then its cohomology is homotopy invariant since it coincide with the usual singular cohomology with coefficients in $A$ (you might need some hypothesis on $X$, like locally contractible).
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
answered Jul 19 at 17:05
Nicolas Hemelsoet
4,890316
4,890316
add a comment |Â
add a comment |Â
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