Suppose $M$ is a smooth manifold, and $N \subset M$ is a closed embedded submanifold of codimension $\geq 3$. Is is true that $$ \pi_1(M) = \pi_1(M \setminus N)?$$ If $N$ is a finite collection of points, then this follows inductively from this question. The proof there uses the Seifert-van-Kampen theorem, but I don't know how to adopt it for higher dimensional $N$.
I was able to show the claim for $M = \mathbb{R}^n$, if $N = \mathbb{R}^{n-3}$ is a linear (or affine) subspace, by projecting to the quotient $\mathbb{R}^n / \mathbb{R}^{n-3} \cong \mathbb{R}^3$. Using this, I had the following idea to show the injectivity of $\pi_1(M \setminus N) \to \pi_1(M)$, but I'm not sure if it is correct:
Let $\gamma$ and $\delta$ be two paths in $M \setminus N$, which are homotopic in $M$, and choose a homotopy $H: I \times I \to M$. We can cover $M$ by open sets $U_i \cong \mathbb{R}^n$, such that $N \cap U_i \cong \mathbb{R}^{n-3}$ is identified with a linear subspace of $\mathbb{R}^n$. By the Lemma of Lebesgue, we can partition $I \times I$ into squares $[k/N, (k+1)/N] \times [l/N, (l+1)/N]$, such that each square maps into (at least) one $U_i$. Strictly speaking, if we restrict $H$ to such a square we do not get a homotopy, because $H$ is not constant at the left and right border. But we can interpret it as a homotopy between two paths from $H(k/N,l/N)$ to $H((k+1)/N, (l+1)/N)$. And because $U_i \cong \mathbb{R}^n$, and $N \cap U_i \cong \mathbb{R}^{n-3}$, we can replace $H$ on that square by a homotopy which avoids $N$. Gluing everything together we obtain a homotopy between $\gamma$ and $\delta$ which avoids $N$.
There is one gap, which might not be too wild: What if the corner points lie on $N$? Then the new homotopy can never avoid $N$.
And I don't have a clue how to show that $\pi_1(M \setminus N) \to \pi_1(M)$ is surjective.
Any help or references are welcome :)
I think a cleaner argument is to use the idea of transversality. As a reference, see chapter 2 of Guillemin-Pollack.
Suppose we're given an element of $\pi_1(M, m)$, $m \notin N$, represented by a function $\gamma: S^1 \to M$. Without loss of generality, we may suppose that $\gamma$ is smooth. Then there is another map $\gamma'$ that is homotopic rel basepoint to $\gamma$ and is transverse to $N$. Since $\dim S^1 + \dim N < \dim M$, the tranversality condition just means that the image of $\gamma'$ is disjoint from $N$. Therefore $\gamma' \in \pi_1(M \setminus N)$. This shows that the map $\pi_1(M \setminus N) \to \pi_1(M)$ is surjective.
To show injectivity, we just apply the same argument to the homotopies. Suppose $\gamma_0$, $\gamma_1: S^1 \to M$ are homotopic via a homotopy $H: S^1 \times I \to M$. Without loss of generality, we may assume we have chosen smooth representatives, and that $\gamma_0$ and $\gamma_1$ are disjoint to $N$ by the previous paragraph. But again there is another map $H'$ that is homotopic rel $\gamma_0, \gamma_1$ and is transverse to $N$. We still have $\dim (S^1 \times I) + \dim N < \dim M$, so tranversality means that the image of $H'$ is disjoint from $N$. In other words, $H': S^1 \times I \to M \setminus N$ is a homotopy between $\gamma_0$ and $\gamma_1$, so $\gamma_0$ and $\gamma_1$ represent the same element of $\pi_1(M \setminus N)$. This shows that $\pi_1(M \setminus N) \to \pi_1(M)$ is injective.
Finally, a word about the transversality theorem: instead of relying on constructions based on the Lebesgue number lemma, the more nitty-gritty parts of the analysis is handled by Sard's theorem, which is arguably more complicated but exploits the smooth structure available in the problem.