Prove $K(\mathbb{Z}^n,2) = B\mathbb{T}^n$

Question

I want to proof statement $K(\mathbb{Z}^n,2) = B\mathbb{T}^n$, how can I do that? I need to use Chern classes or something like that? Hope for your help.

Answer 1

Let $X$ be a topological space, a map $f:X\rightarrow B\mathbb{T}^n$ defines a $\mathbb{T}^n$-bundle over $X$, defined by $1$-cocycle $c\in H^1(X,\mathbb{T}^n)=H^2(X,\mathbb{Z})$. This follow from the exact sequence $0\rightarrow \mathbb{Z}^n\rightarrow\mathbb{R}^n\rightarrow \mathbb{T}^n\rightarrow 1$; since $\mathbb{R}^n$ is contractible, the long exact sequence in cohomology gives $H^n(X,\mathbb{T^n})=H^{n+1}(X,\mathbb{Z}^n)$.

Since $K(\mathbb{Z}^n,2)$ represents the $2$-singular cohomology with coefficients in $\mathbb{Z}^n$; that is the set of homotopy classes of maps $[X,K(\mathbb{Z}^n,2)]=H^2(X,\mathbb{Z}^n)$ we deduce that $B\mathbb{T}^n=K(\mathbb{Z}^n,2)$.