TITLE: Complement of $S^1$ in a simply connected oriented surface
QUESTION [1 upvotes]: The complement of any $C^1$-embedded circle $S^1$ in an oriented simply connected surface of regularity $C^1$ has exactly two components by the theorem of Stokes. Now it seems evident to me that there is always one component which is both simply connected and relatively compact. 
I do not see the cause although it seems "obvious" to me that one of them has to be covered by the image of any zero homotopy of the embedded circle. 
Are there any good references which would permit a hint?

REPLY [1 votes]: If we suppose one of the two components to be relatively compact, then it is also simply connected by Poincare duality between De Rham cohomology with and without compact supports (cf. Bott-Tu) and between De Rham cohomology and simplicial homology alias abelianised fundamental group. 
At least in the proof of Riemann uniformisation theorem, relative compactness of one component of complements of occuring cercles can be achieved by means of an appropriate exhaustion function.
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