Maps of discs into surfaces

Question

Let $f:D \to S $ be a continuous map from the closed unit disc in $R^2$ into a closed surface of genus $g \geq 1$ such that $f|_{\partial D}$ is an embedding (homeomorphism onto its image) with Image $\gamma$ a simple closed curve. Then, $\gamma$ bounds an embedded disc $D* \subset S$ and I wonder if it is true that $D* \subset f(D)$.

Answer 1

This is the answer for not necessarily compact manifolds of arbitrary dimension. It uses significantly more machinery than the other answer.

Let $M$ be a simply connected manifold with sphere boundary $S$, and let $M'$ be the manifold obtained by filling in the boundary with a disc. If the inclusion $S \hookrightarrow M$ is null-homotopic, then $M = D^n$. A null-homotopy gives me a map $(D^n,S) \to (M,S)$, and I can complete this to a map $S^n \to M'$ of degree 1. By Poincare duality this is an isomorphism on cohomology (if $M'$ has a nontrivial cohomology class $\alpha \in H^k(M')$, there is some $\beta \in H^{n-k}(M')$ with $\alpha \smile \beta \neq 0$; invoke that we have a nontrivial homomorphism $H^*(M') \to H^*(S^n)$. Do this with $\Bbb Q$ and $\Bbb Z/p$ coefficients to kill all torsion and non-torsion cohomology.) The assumption that $M$ is simply connected implies by Whitehead that it's a homotopy equivalence. Hence by the generalized topological Poincare conjecture $M'$ is homeomorphic to $S^n$ and $M$ is homeomorphic to $D^n$ as desired.

(This is even easier if $M$ is not simply connected; just lift the map from the disc to the universal cover of $M$ and fill in the other boundary components to get a non-surjective map of degree 1.)

Now suppose I had an embedded sphere $S \subset M$ (with a tubular neighborhood) that bounded a ball on one side. Call the other side (with $S$) $M'$. Suppose I had an extension to the disc $D \to M$ of this map that did not surject onto the ball. Then by puncturing I would see that this implies that the inclusion of the boundary $S \subset M'$ is null-homotopic. The above then implies that $M' = D^n$ as desired.

I never actually needed that $M$ was compact here!

Answer 2

Cute question. Supposing the genus is at least zero, yes. (If you were working on the sphere you could take your map-from-disc to be the top hemisphere and the chosen embedded disc to be the bottom hemisphere.)

For suppose not. Then there's some point $x$ in the interior of $D^*$ that the image of the disc misses. Then the boundary curve $\gamma$ is a loop around $x$ that's null-homotopic in the surface with that point deleted. But the punctured genus $g$ surface deformation retracts onto the wedge of $2g$ circles (draw a picture of the fundamental domain with the puncture in the middle!) and this loop $\gamma$ maps to the (nontrivial) word $[a_1,b_1]\cdots [a_g,b_g]$ in $\pi_1(\vee_{2g} S^1) = F_{2g}$, the free group on $2g$ generators.

So $\gamma$ is not null-homotopic in $\Sigma_g \setminus \{x\}$, and this is a contradiction. So $D^* \subset f(D)$.