Let $S_1$ and $S_2$ be two surfaces with compact boundary and the same number of boundary components. Let $M_1$ and $M_2$ be the surfaces obtained by glueing closed disks to the boundary of $S_1$ and $S_2$, respectively. Is there a direct proof that if $M_1$ and $M_2$ are homeomorphic then $S_1$ and $S_2$ are also homeomorphic? I know that this can be done in the compact case using normal forms, as in Massey's book. And if the surfaces are not compact?
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2$\begingroup$ One can always isotope the homemorphism such that it is identity on a disk. $\endgroup$– Anubhav MukherjeeCommented Feb 21, 2021 at 23:39
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1$\begingroup$ You need equal number of boundary components for this to hold. $\endgroup$– Moishe KohanCommented Feb 22, 2021 at 0:53
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1$\begingroup$ What you need ultimately is the "unique disc lemma": if $M$ is a surface without boundary, any two embeddings $i,j: D^2 \to M$ are nearly isotopic: there is a family of homomorphisms $F_t: M \to M$ so that $F_0$ is the identity and $i F_1 = j$ or $i F_1 = jr$, where r is reflection of the disc. In particular there is an ambient isotopy taking the image of one to the image of the other. You can then extend this to embeddings of k disjoint discs. Now you recover S from M by deleting the interiors of these discs, and the isotopies above justify that the result is unique up to homeomorphism. $\endgroup$– mmeCommented Feb 22, 2021 at 11:40
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1$\begingroup$ The unique disc lemma is basically a form of the Schoenflies theorem (or more accurately a form of the annulus theorem in dimension 2). $\endgroup$– mmeCommented Feb 22, 2021 at 11:42
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1$\begingroup$ For a single disc, the question is answered in all dimensions here (see the "For the rest..." paragraph), as an application of the Annulus Theorem. That proof should easily generalize to a finite disjoint union of discs. $\endgroup$– Lee MosherCommented Feb 26, 2021 at 14:44
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