Leary cover and good cover

153 Views Asked by At

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 ${\displaystyle H^{k}(U_{i_{1}}\cap \cdots \cap U_{i_{n}},{\mathcal {F}})=0}$, 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 $k\ge1$?

1

There are 1 best solutions below

0
On BEST ANSWER

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).