$\mathbb{R}^n$ minus a simple closed curve is simply connected for $n\geq 4$? I would like to prove the statement in my question, at least for the case $n=4$. Here is some context for it; in the study of the general Jordan curve theorem, one confronts with the following result (36.2 in Munkres' book, 19.4 in Bredon's, 2B.1 b) in Hatcher's)

Theorem. For an embedding $h:\mathbb{S}^r\rightarrow\mathbb{S}^n$ with $r<n$, one has that
$$\tilde H_i(\mathbb{S}^n-h(\mathbb{S}^r))\cong\begin{cases}
\mathbb{Z},&\text{ if }i=n-r-1\\
0&\text{ otherwise}
\end{cases}$$

This completely chatacterizes the homology of the space $\mathbb{S}^n-h(\mathbb{S}^r)$, but it does not say anything about its fundamental group; a very well-known example of this wild behaviour is given by Alexander's Horned sphere, for it is an embedding of $\mathbb{S}^2$ in $\mathbb{S}^3$ whose complement is not simply connected.
However, I think this kind of behaviour does not happen when $r=1$ and $n\geq 4$; in other words, the complement of any simple closed curve in $\mathbb{R}^n$ for $n\geq 4$ must be simply connected.
The inuition is clear in simple cases, but how can we prove it from the previous facts, if possible?
 A: Your intuition is false unless the embedding is smooth, in which case this is a standard result of differential topology, using the tool of transversality. It is probably sufficient to assume that the sphere has codimension at least 3 and is locally flat but I do not know anything about locally flat embeddings.
The Double suspension theorem of Cannon (expanding on work of Edwards) shows that if $P$ is a homology sphere of dimension $n$, then $\Sigma(\Sigma(P)) = \Sigma^2(P) \cong S^{n+2}$.
Just as there is a natural map $S^0 \to \Sigma P$ whose complement is homeomorphic to $P \times \Bbb R$, there is a map $h: S^1 = \Sigma(S^0) \to \Sigma^2 P$ whose complement is homeomorphic to $P \times \Bbb R^2$.
Corollary. If $G$ is the fundamental group of a homology n-sphere, then $G$ is the fundamental group of the complement of a wild embedding of $S^1$ into $S^{n+2}$.
Thus you can get, for instance, a group of order 120 (called the binary icosahedral group) as the complement of a wild 1-knot in $S^5$.
In fact work or Kervaire shows that any finitely presented superperfect group, a group whose first and second homology is zero, arises as the fundamental group of a homology n-sphere for every $n \geq 5$. Possibly $n \geq 4$: I did not check.
So the situation is more or less as bad as it can be.
