# The three definitions of Cech cohomology (simplical complex vs. presheaf vs. sheaf)

I came across the following three definitions of Cech cohomology group of a topological space $X$:

1. [Source: Munkres, Elements of Algebraic Topology, pp. 437]. The Cech cohomology group of $X$ in dimension $k$, with coefficients in the abelian group $G$ is $$\check{H}^k(X,G) =\lim_{\substack{\longrightarrow\\\mathcal{U}\in A}} H^k(N(\mathcal{U}), G)$$ where $A$ is the directed set consisting of all open coverings of the space $X$, directed by letting $\mathcal{U} < \mathcal{V}$ if $\mathcal{V}$ is a refinement of $\mathcal{U}$, and $N(\mathcal{U})$ is an abstract simplicial complex called nerve of $\mathcal{U}$.
2. [Source: Bott and Tu, Differential Forms in Algebraic Topology, pp. 112] The Cech cohomology of $X$ with values in the presheaf $\mathscr{F}$ is $$\check{H}^k(X,\mathscr{F}) =\lim_{\substack{\longrightarrow\\\mathscr{U}\in A}} H^k(\mathcal{U}, \mathscr{F})$$ where $A$ is the directed set consisting of all open coverings of the space $X$, directed by letting $\mathscr{U} < \mathscr{V}$ if $\mathscr{V}$ is a refinement of $\mathscr{U}$.
3. [Source: Miranda, Algebraic Curves and Riemann Surfaces, pp. 296] The $k^{th}$ Cech cohomology group of a sheaf $\mathscr{F}$ on $X$, for $k\geq 0$ is $$\check{H}^k(X,\mathscr{F}) =\lim_{\substack{\longrightarrow\\\mathscr{U}\in A}} H^n(\mathcal{U}, \mathscr{F})$$ where $A$ is the directed set consisting of all open coverings of the space $X$, directed by letting $\mathscr{U} < \mathscr{V}$ if $\mathscr{V}$ is a refinement of $\mathscr{U}$.

I have seen the proof of isomorphism between Cech and de Rham cohomology using all these definitions of Cech cohomology. Depending on the definition, we have to use different kinds of tools for the proof.

How are all these definitions equivalent? Is there any reference which discusses the equivalence of these definitions? Are all these definitions special case of a general definition of Cech cohomology group?

• Every sheaf is a presheaf, hence 2 imples 3. The first one seems to be equivalent to the second using the constant presheaf $\underline{G}$. – Alan Muniz Sep 5 '18 at 14:47
• In general, "presheaf cohomology" differs from "sheaf cohomology", but there is a Grothendieck spectral sequence relating them. – JHF Sep 5 '18 at 14:56
• @AlanMuniz So what exactly do we mean by the isomorphism between Cech and de Rham cohomology of a smooth manifold? If we use definition 1 then we can prove using Weil's method (cf. Mortia, Geometry of Differential Forms, Theorem 3.19). Moreover, the proof using definition 2 is much simpler as compared to the one using definition 3. This is because a short exact sequence of presheaves always leads to a long exact sequence of cohomology, but in the case of sheaves we need the condition of paracompactness (cf. Hirzebruch,Topological methods in algebraic geometry, Lemma 2.7.1, Theorem 2.10.1) – rationalbeing Sep 5 '18 at 16:18
• @JHF Can you please give a reference where I can read more about it? In Bott & Tu (proposition 10.6), Cech cohomology of manifold $M$ with values in constant presheaf is proved isomorphic to the de Rham cohomology without any mention of long exact sequence in cohomology. Whereas, Griffith & Harris argument (pp. 44, Principles of Algebraic Geometry) for the isomorphism between Cech cohomology of sheaves and de Rham cohomology hinges on the fact that manifolds are paracompact and hence we can get a long exact sequence of cohomology form a short exact sequence of sheaves. – rationalbeing Sep 5 '18 at 16:32
• As mentioned by Alan Muniz, definition 2 is more general than definition 3. But as I checked in both books, for a sheaf both definitions agree (although Bott and Tu are not very precise). If you apply 2 /3 to the constant sheaf $\underline{G}$, it remains to verify that the result agrees with 1. I do not know a reference, but it is absolutely plausible.The $n$-cochains in the sense of 2/3 are based on the intersections of $n$ members of $\mathcal{U}$, and precisely these are the $n$-simplices of $N(\mathcal{U})$. I guess you can find your own proof for the equivalence of 2/3/ and 1. – Paul Frost Sep 5 '18 at 17:06