$H^1(X) = [X,\mathbb{T}]$?

Question

This is a stupid question, but here goes.

I have a compact Hausdorff space $X$, and I am talking about $[X,\mathbb{T}]$, the group of homotopy classes of maps $X \to \mathbb{T}$, where $\mathbb{T}$ is the circle (viewed as a topological group).

Is it true, in some sense, that $[X,\mathbb{T}] = H^1(X)$, the 1st cohomology group of $X$?

My completely shallow motivation for this question is that I want to know whether it is OK to refer to $[X,\mathbb{T}]$ as a cohomology group.

Now, I realize that there are many cohomology theories out there: singular, de Rahm, Čech, and so on. However, I have never studied any of them. In addition, from the bit of poking around I've done, it seems I should be specifying a coefficient ring $R$ too. Anyway, it seems a bit much for me to digest the general definitions just to decide whether what I'm using is the $n=1$ case of some general theory.

Answer 1

The circle is an Eilenburg-MacLane space, $(K(\mathbb{Z},1))$, and so the answer is yes: Link