Let $\Sigma \subset \mathbb{R}^3$ be a compact embedded surface with boundary $\partial \Sigma$ and $i:\Sigma\setminus \partial\Sigma \to \mathbb{R}^3 \setminus \partial\Sigma$ the inclusion.

Is the following true?

If $i_*(\pi_1(\Sigma\setminus \partial \Sigma))=0$, then $\Sigma$ is orientable.


1 Answer 1


Yes, the surface is orientable. To simplify the LaTex and the exposition, I will change the notation and setting a small amount.

Suppose that $F$ is a compact connected embedded surface in three-sphere. Suppose that the image of $\pi_1(F)$ in $\pi_1(S^3 - \partial F)$ is trivial. We must show that $F$ is orientable.

In the base case, where $F$ has no boundary, this follows from Alexander's theorem.

Suppose that $\alpha$ is a boundary component of $F$. Let $\alpha'$ be a curve embedded in $F$ which is disjoint from, but is isotopic to (in $F$), the boundary component $\alpha$. Thus $\alpha$ and $\alpha'$ cobound an annulus in $F$. Since the image of $\pi_1(F)$ is trivial in $\pi_1(S^3 - \partial F)$, we have that $\alpha'$ bounds an immersed disk in $S^3 - \partial F$. By Dehn's lemma (that is, by a version of the Disk Theorem) $\alpha'$ bounds an embedded disk $D$ in $S^3 - \partial F$. Thus $\alpha$ is an unknot.

This holds for all boundary components of $F$. In fact, since all of the boundary components bound disks in $S^3 - \partial F$, we deduce that $\partial F$ is a split link. Surgering along separating spheres reduces us to the case where $F$ has only one boundary component. [There is some work here.]

Now consider the annulus $A$ between $\alpha$ and $\alpha'$. Note that $A$ is two-sided. So we may and do isotope $D$ slightly to make it transverse to $F$ and disjoint from the interior of $A$. Suppose that $\beta$ (perhaps equal to $\alpha'$) is an innermost curve of $F \cap D$. Let $D' \subset D$ be the subdisk bounded by $\beta$. Let $B \subset F$ be a small annulus neighbourhood of $\beta$.

We surger $F$ along $D'$ to obtain a new surface $F'$. That is, we form $F - B$ and glue on a pair of disks, both parallel to $D'$. If $F'$ is non-orientable, then it still has trivial $\pi_1$-image and has lower complexity (either has no boundary, has lower genus, or meets $D$ in a simpler way) than $F$. This is a contradiction.

We deduce that $F'$ is orientable and so is two-sided. We also deduce that the two boundaries of $B$ are attached to opposite sides of $F'$. Thus there is a (orientation reversing) curve $\gamma$ in $F$ that meets $\beta$ exactly once. We deduce that $\gamma$ has linking number one with $\alpha$. Thus $\gamma$ is non-trivial in the image of $\pi_1(F)$, a contradiction.

I think that the condition can be reduced to "$H_1$-image is trivial". The above argument does not immediately work (because surgery along an orientable surface can cause genus to increase).


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