# Is the boundary of every 2-manifold the disjoint union of circles?

Note that in the title I am being sloppy and should say "Is the boundary of every 2-manifold with boundary the disjoint union of circles?".

Also when I say "disjoint union of circles" I mean up to homeomorphism -- otherwise counterexamples to the statement would be both trivial and uninteresting.

Anyway, consider the following quote from Manifold Atlas:

The boundary of a surface is a disjoint union (possibly empty) of circles. Surfaces with boundary can be constructed by removing open discs from surfaces without boundary.

This seems to be implying that the boundary of any 2-dimensional manifold with boundary is the (possibly empty) disjoint union of circles (up to homeomorphism), and that any 2-dimensional manifold with boundary is formed by removing circles from a 2-dimensional manifold (without boundary)? I know that there is a strong classification theorem for 2-dimensional manifolds (without boundary), but that such a strong a statement for 2-manifolds with boundary also holds was initially surprising to me.

However, now that I think about it, this result would follow from the classification for 1-manifolds as long as we assume that the boundary of a 2-manifold with boundary has to be compact.

Question:
Is this correct? And if it is correct, then why does the boundary of a 2-manifold with non-empty boundary have to be compact? (And thus a possibly empty disjoint union of circles?)

• One must put compactness into the hypothesis in some form, $\mathbb{R}\times [0,1]$ is a nice two-dimensional manifold with boundary whose boundary consists of two copies of $\mathbb{R}$. – Daniel Fischer Nov 18 '16 at 21:15
• They're only talking about compact manifolds. – user98602 Nov 18 '16 at 23:10
• Assuming compactness of the manifold, compactness of the boundary follows since it is a closed subspace of a compact space. – Moishe Kohan Nov 18 '16 at 23:28