Null-homotopic embedded circle in surface bounds disk

Question

The following is stated all over, for instance in Hatcher's Notes on 3-Manifold Topology: A null-homotopic embedded circle in a surface bounds a disk in the surface. I can't seem to find or come up with a proof of this. Any help would be greatly appreciated.

Answer 1

Let $S$ be the surface, assumed to be connected. Let $\gamma \subset S$ be the null-homotopic embedded circle. We must find an embedded disc in $S$ with boundary $\gamma$.

If $S$ is the sphere or the plane, this is just the Schönflies Theorem.

Otherwise, consider a universal covering map $f : X \to S$ and let $\tilde\gamma \subset X$ be a homeomorphic lift of $\gamma$. From the classification of surfaces we see that $X$ itself is the sphere or the plane, and so $\tilde\gamma = \partial D$ for some embedded disc $D \subset X$. The idea is now to prove that the restriction $f \mid D : D \to S$ is injective, and it will then follow that $f(D)$ is an embedded disc with $\partial(f(D)) = \gamma$.

Consider the deck transformation action of $\pi_1 S$ on $X$. To prove that $f \mid D$ is injective, it is equivalent to prove that for each deck transformation $T$, if $T$ is not the identity then $D \cap (T\cdot D) = \emptyset$.

Suppose that $D \cap (T\cdot D) \ne \emptyset$. There are two cases.

In the first case, suppose that $\partial D \cap (T \cdot \partial D) = \emptyset$. The Schönflies Theorem tells us that $D$ is one of the two components of $X - \partial D = X - \tilde\gamma$, and similarly $T \cdot D$ is one of the two components of $X - \partial (T \cdot D) = X - (T \cdot \tilde\gamma)$. It follows that either $D \subset T \cdot D$ or $T \cdot D \subset D$. In either case, the Brouwer Fixed Point Theorem implies that $T$ has a fixed point. Since $T$ is a deck transformation, it follows that $T$ is the identity.

In the second case, suppose that $\partial D \cap (T \cdot \partial D) \ne \emptyset$, and so for some $x,y \in \partial D = \tilde\gamma$ we have $T \cdot x = y$. If $x=y$ then $T$ is the identity, as in the previous case. If $x \ne y$ then we have $f(x)=f(y)$ and so $f \mid \tilde\gamma$ is not injective, contradicting that $\tilde\gamma$ is a homeomorphic lift of $\gamma$.