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.