Isomorphism between second Cech cohomology group and third cohomology

Question

In some notes I came across the following equality (or to be more precise: isomorphism) that $\check{H}^2(M,\mathbb{T}) \cong H^3(M,\mathbb{Z})$ where $\check{H}$ is Cech cohomology and $\mathbb{T}$ is the circle. I would like to understand why is it true. EDIT: Here $M$ is assumed to be smooth manifold (this is enough for my purposes).

Answer 1

Here is an alternative approach to the one given in the comments. I will blithely assume the following fact, although I am not sure in what generality it is true: if $A$ is a topological abelian group and $X$ is reasonable, then the Cech cohomology $\check{H}^n(X, A)$ should agree with the representable cohomology given by homotopy classes of maps from $X$ to the classifying space $B^n A$.

Assuming this, it now suffices to observe that the classifying space $B^2 S^1$ can be identified with the classifying space $B^3 \mathbb{Z}$, since $S^1 = B \mathbb{Z}$.

In this language, the long exact sequence associated to the short exact sequence of topological groups

$$0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$$

deloops to the fiber sequence of classifying spaces

$$B^2 \mathbb{Z} \to B^2 \mathbb{R} \to B^2 S^1$$

and since $\mathbb{R}$ is contractible, so is $B^2 \mathbb{R}$. This fiber sequence therefore identifies $B^2 \mathbb{Z}$ with the loop space of $B^2 S^1$, and hence identifies $B^2 S^1$ with the delooping $B(B^2 \mathbb{Z}) \cong B^3 \mathbb{Z}$.