Equivalence of different cohomology groups

Question

Let $X$ be a topological space (may be assumed to be locally compact). Let $A$ be either a field or $\mathbb{Z}$. One can consider various cohomology groups:

(1) singular cohomology $H_{sing}^*(X,A)$;

(2) cohomology with coefficients in the constant sheaf $\underline{A}$ (defined e.g. using the machinery of derived functors);

(3) Chech cohomology $\check{H}^*(X,A)$.

I am aware that if $X$ is a smooth manifold then all cohomology groups are canonically isomorphic.

Question. Under what conditions on $X$ all three cohomology groups are canonically isomorphic? I am particularly interested in equivalence of (1) and (2).

I am not an expert, but I suspect that this question has been investigated in literature long ago; a reference will be helpful.

Answer 1

It's interesting to think about the case where $X$ is compact Hausdorff but not locally contractible. We have a short exact sequence of sheaves $\mathbb{Z}\to\mathbb{R}\to S^1$ (where $\mathbb{R}$ really means the sheaf of continuous $\mathbb{R}$-valued functions, and so on). This gives a long exact sequence of cohomology groups. We have partitions of unity, which we can use to prove that the higher cohomology of $\mathbb{R}$ is trivial. One can also check that a map $u\colon X\to S^1$ lifts to $\mathbb{R}$ if and only if it is nullhomotopic. Using this we get $$ H^1_{\text{sheaf}}(X) = [X,S^1] = \{\text{homotopy classes of maps from $X$ to $S^1$}\}. $$ On the other hand, one can prove the Hurewicz theorem and universal coefficient theorem for singular (co)homology by simplicial methods, so they do not depend on any special assumptions about the topology of $X$. If $X$ is path-connected, this gives $$ H^1_{\text{sing}}(X) = \text{Hom}(H_1(X),\mathbb{Z}) = \text{Hom}(\pi_1(X),\mathbb{Z}). $$ There is a natural map $$ \phi_X\colon H^1_{\text{sheaf}}(X) \to H^1_{\text{sing}}(X), $$ but it is not obviously bijective.

However, it seems that $\phi_X$ is an isomorphism in at least one interesting case. Let $F$ be a free abelian group of countable rank, and take $$ X = \text{Hom}(F,S^1) \simeq\prod_{i=0}^\infty S^1. $$ As this is compact, any map to a CW complex lands in a finite subcomplex. Using this, it is not hard to see that $X$ does not have the homotopy type of a CW complex. Also, it is not locally simply connected.

There is an evident evaluation map $$\chi\colon F\to\text{Map}(X,S^1)\to [X,S^1]=H^1_{\text{sheaf}}(X).$$
We also have $$ \pi_1(X)=\prod_{i=0}^\infty\pi_1(S^1)=\prod_{i=0}^\infty\mathbb{Z}, $$ or more naturally $\pi_1(X)=F^*$, where $A^*$ means $\text{Hom}(A,\mathbb{Z})$. Using this we get $H^1_{\text{sing}}(X)=H_1(X)^*=F^{**}$. The composite $\phi\chi$ is the usual map from $F$ to $F^{**}$. It is a nontrivial algebraic fact that this map is an isomorphism, even though $F$ is infinitely generated. (This is quite different to the situation with vector spaces over a field, of course.) This means that $\chi$ is injective and $\phi$ is surjective. I think that in fact they are both isomorpphisms, but I do not instantly see a proof of that.

Answer 2

I think this holds if $X$ is a CW complex. I don't have a reference, but try Bredon or Iversen.

Edit: Bredon, "Sheaf Theory", Chapter III "Comparison with Other Cohomology Theories". Perhaps this contains what you are looking for.