Supposedly "trivial" implication that injective surfaces are incompressible My question is about a passage in Algorithmic Topology and Classification of 3-Manifolds by Sergei Matveev.
Let $F$ be a surface in some $3$-manifold $M$.
$F$ is called incompressible if for every embedded disc $D \subset M$ with $D \cap F = \partial D$, it is the case that $\partial D$ "bounds a disc" in $F$. 
I assume that "bounds a disc" means that there is an embedded disc $D' \subseteq F$ with $\partial D' = \partial D$. 
$F$ is called injective if it is connected and the homomorphism of fundamental groups induced by the inclusion $i: F \hookrightarrow M$ is injective.
It is supposed to be trivial that injective surfaces are incompressible. Trying to prove this, assume $F$ is injective and that $D \subset M$ is an embedded disc such that $D \cap F = \partial D$. $\partial D$ is obviously nullhomotopic in $M$, so by injectivity, $\partial D$ is nullhomotopic in $F$, so there is a continuous map $f : D^2 \to F$ such that $f(\partial D^2) = \partial D$. 
I don't see any easy way to get the required (required by my understanding of "bounds a disc") embedding with the same property.
The author explicitly uses the loop theorem for proving the reverse implication for two-sided $F$, so presumably, it's not necessary for the forward implication. 
 A: A simple closed curve in a surface bounds a disc in that surface if and only if the induced map on fundamental groups is trivial.
So if $C$ is an embedded circle in a surface $F$, I'm saying
$$\pi_1 C \to \pi_1 F$$
is the zero map if and only if $C$ bounds an embedded disc in $F$.  
You can break the proof up into steps. 
Step 1: the curve bounds a surface with boundary in $F$ if and only if the induced map $H_1 (C;\mathbb Z_2) \to H_1 (F;\mathbb Z_2)$ is zero. 
Step 2: In a surface with boundary $\partial F = C$, the induced map $\pi_1 C \to \pi_1 F$ is null if and only if $F$ is a disc.  
Step 1's proof can be done in many ways, the argument that leaps to mind uses simplicial or cellular homology.  i.e. triangulate the surface so that the curve $C$ is a subcomplex, and then look at the long exact homology sequence of the pair $(F,C)$. 
Step 2's proof needs the classification of compact surfaces.  The idea being that if the surface has non-zero genus, the boundary curve is a product of commutators (at least in the orientable case).  I guess minimally you need to know the classification of simply-connected surfaces for this argument -- that there's only one compact one and it's the 2-sphere, $S^2$. 
I don't think you can have a much simpler proof, as the theorem is basically equivalent to the classification of simply-connected surfaces.  
