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The results are listed here: http://topospaces.subwiki.org/wiki/Homotopy_of_torus

Is there an intuitive way to understand these results?

In particular, why would the higher homotopy group be the trivial group?

Taiben
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    Homotopy groups of a product of two spaces are the direct products of the respective homotopy groups of the factors. Torus is a product of two circles. If you know the homotopy groups of a circle you know those of a torus. – Jyrki Lahtonen Sep 01 '14 at 07:31
  • Thanks! I just realized that n-torus is the product of n circles. I thought it was the product of n-spheres. – Taiben Sep 01 '14 at 07:36
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    In addition, recall that the higher homotopy groups of a space $X$ that admits a universal cover are the same as the higher homotopy groups of $X$ itself. The universal cover of the $n$-torus is $\Bbb R^n$. –  Sep 01 '14 at 07:38
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    To make @MikeMiller's comment slightly more precise; the induced homomorphisms $\pi_k(p)\colon \pi_k(Y)\to\pi_k(X)$ of a covering map $p\colon Y\to X$ is injective for $k=1$ and are isomorphisms for $k\geq 2$. – Dan Rust Sep 01 '14 at 12:29

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Let $\mathbb{T}^n$ denote the $n$-torus. This is the product of $n$ circles, i.e., $\mathbb{T}^n=(S^1)^n$. Thus, $\pi_1(\mathbb{T}^n)=\prod^{n}\pi_1(S^1)=\mathbb{Z}^n$. Since $\mathbb{T}^n$ admits a universal cover $\alpha:\mathbb{R}^n\to\mathbb{T}^n$, for $i\geq 2$, the homomorphisms $\pi_i(\alpha):\pi_i(\mathbb{R}^n)\to\pi_i(\mathbb{T}^n)$ are isomorphisms. Since $\mathbb{R}^n$ is homotopy equivalent to the point, $\pi_i(\mathbb{T}^n)$ is the trivial group for $i\geq 2$.