# Intuitive Approach to Sheaf and Cech Cohomology [closed]

Sheaf and Cech cohomology $H^*(X,\mathcal{F})$ (which give the same result when applied to good enough topological spaces) are a useful generalisation of the concepts of de Rham and Dolbeault cohomology, just by putting $\mathcal{F}=\mathbb{R}$ or the sheaf of holomorphic functions $\Omega^p(M)$. But each generalisation involves losing intuition about the measured geometric properties.

My question is: is it possible to understand the geometric intuition behind sheaf and Cech cohomology in the same way one can understand the geometry behind de Rham cohomology?

## closed as unclear what you're asking by abx, Stefan Kohl, José Figueroa-O'Farrill, Peter Crooks, Vidit NandaFeb 18 '15 at 15:28

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• If you put $G=\mathbb{C}$ you do not get Dolbeault cohomology. You just get ordinary cohomology with coefficients in $\mathbb{C}$. – Steven Gubkin Feb 18 '15 at 12:38
• If you work just a little bit with sheaf cohomology, you'll get very soon the intuition of what it is. – abx Feb 18 '15 at 13:19
• Here is a computation which helped my intuition. Take a combinatorially triangulated manifold $X$. For each vertex $v$ of the triangulation, let $U(v)$ be the union of the interiors of all simplices (of all dimensions) containing $v$. So $U(v)$ is an open ball. Let $\underline{\mathbb{Z}}$ be the sheaf of locally constant $\mathbb{Z}$-valued functions on $X$. Compute $H^{\ast}(X, \underline{\mathbb{Z}})$ with respect to the Cech cover $U(v)$. You should get the cochain complex of the triangulation. – David E Speyer Feb 18 '15 at 15:05
• Dear Jjm, I have cast the final vote to close since your question is more suited for math stackexchange, where I hope it will receive the comprehensive answer that it deserves. I also suggest that you modify it there by removing the erroneous reference to Dolbeault cohomology. – Vidit Nanda Feb 18 '15 at 15:29
• @,Jjm Nothing! We can just leave it here and the stackexchange automatons will deal with it :) – Vidit Nanda Feb 18 '15 at 15:58