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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 $\mathbb{S}^3 \setminus M$?

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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.

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  • $\begingroup$ Why does it extend? $\endgroup$
    – Eduardo Longa
    Apr 9, 2019 at 0:20
  • $\begingroup$ @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.) $\endgroup$
    – Kyle Miller
    Apr 9, 2019 at 0:28

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