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Let $X$ be a compact, path-connected subset of $\mathbb{R}^n$. I need a reference for the fact that the $n$-th homotopy group $\pi_n(X)$ is trivial.

EDIT: Quite embarrassing. Indeed this is false, as the comment below indicates ($\pi_3(S^2)=\mathbb{Z}$). Should have thought more before posting the question.

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This is not true (so it's hard to give a reference). For example, $S^2$ is a compact path-connected subset of $\mathbb{R}^3$ and $\pi_3(S^2) = \mathbb{Z}$. See also here. –  t.b. Nov 5 '11 at 9:22

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Unless $n \leq 2$ this is not true and simple counterexamples are provided by the $(n-1)$-spheres $S^{n-1} \subset \mathbb{R}^{n}$ — which are path-connected and compact, of course.

Indeed, $\pi_{3}(S^2) = \mathbb{Z}$ and $\pi_{n}(S^{n-1}) = \mathbb{Z}/2\mathbb{Z}$ for $n \geq 3$.

The cases $n=0,1$ are trivial and for the case $n=2$ there's the following fact:

It turns out that for all subsets $X$ of the plane we have $\pi_{k}(X) = 0$ for $k \geq 2$ by a theorem of Andreas Zastrow (1998), available here.

That reference is distinctly non-trivial and I don't know if there is a simple proof under the hypotheses you're interested in. The statement you ask about does not seem obvious even for relatively nice planar sets, such as the Hawaiian earring, for example.

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