# Complement of tubular neighborhood

Let $$M$$ be a closed, connected, orientable and embedded surface inside the unit 3-sphere $$\mathbb{S}^3$$ and consider a small tubular neighborhood $$U$$ of $$M$$:

$$U = \{ x \in \mathbb{S}^3 : d(x, M) \leq \varepsilon \},$$

(for small $$\varepsilon > 0$$). I know that $$U$$ has the same homotopy type of $$M$$. Is it true that $$\mathbb{S}^3 \setminus U$$ has the same homotopy type of $$\mathbb{S}^3 \setminus M$$?

A tubular neighborhood $$U$$ of $$M$$ has a homeomorphism $$M\times (-1,1)\to U$$, where $$M$$ is the image of $$M\times 0$$. By, say, restricting $$(-1,1)$$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $$M\times [-1,1]\to \overline{U}$$, where $$\overline{U}$$ is the closure of $$U$$ in $$M$$. The homeomorphism can be used to deformation retract $$\overline{U}-M$$ to $$\partial \overline{U}$$, the image of $$M\times\{-1,1\}$$. This extends to a deformation retract of $$S^3-M$$ onto $$S^3-U$$. Thus $$S^3-U \hookrightarrow S^3-M$$ is a homotopy equivalence.
• @EduardoLonga Extend each step of the deformation retract by $\operatorname{id}_{M-U}$. The maps agree on $\partial \overline{U}$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.) Apr 9, 2019 at 0:28