Problem: Let $\Sigma$ be a orientable surface possibly non-compact with boundary. Let $C,$ and $D$ be two simple-closed curves(smooth embedding) on $\Sigma\backslash \partial \Sigma$ such that both $\Sigma\backslash C$, and $\Sigma\backslash D$ are connected. Does there always exist a homeomorphism $f:\Sigma\to \Sigma$ such that $f(C)=D$.

This is clear if our surface $\Sigma$ is compact: considering tubular-neighborhoods we have two compact subsurface $M:=\Sigma\backslash\big((-1,1)\times C\big)$, and $N:=\Sigma\backslash\big((-1,1)\times D\big)$ with $\chi(M)=\chi(\Sigma)=\chi(N)$. Note that we are using the following fact: If a finite CW-complex $X$ can be written as union of two subcomplexes $X=Y\cup Z$, then we have the inclusion-exclusion formula: $\chi(X)=\chi(Y)+\chi(Z)-\chi(Y\cap Z)$. Now, we have the

Classification theorem(compact surfaces): Two compact orientable surfaces having the same number of boundary components and same Euler-characteristic, are homeomorphic.

Now, after boundary rearrangement(which is again a variation of classification theorem for compact surfaces), we have a homeomorphism $\widetilde f:M\to N$ with $\widetilde f(-1\times C)=-1\times D$, and $\widetilde f(1\times C)=1\times D$. Now, extends the $\widetilde f$ in the whole tubular-neighborhoods.

So, the crucial step is to use Euler-characteristic, and for the non-compact case we have compactly supported Euler-characteristics(considering the rank of compactly supported cohomology), and the inclusion-exclusion formula does hold.

But, the problem is in the classification theorem(compact surface) and its relation with compactly supported cohomology. Also, the boundary rearrangement of a compact surface is another one.

Any help will be appreciated.

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    $\begingroup$ Hint: find a compact connected subsurface with boundary in $S$ which contains $C\cup D$ and is not disconnected by either one of them. $\endgroup$ Dec 30, 2020 at 10:46
  • $\begingroup$ So, after your hint(countable compact exhaustion of a surface by subsurfaces), one has to use Isotopy Extension Theorem to extend the homeomorphism of the sub-surface to a homeomorphism of the original surface. Am I right? $\endgroup$
    – Sumanta
    Dec 30, 2020 at 13:18
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    $\begingroup$ No: Find a homeomorphism of the subsurface sending $C$ to $D$ and fixing the boundary pointwise. Then extend to the rest of $S$ by the identity map. $\endgroup$ Dec 30, 2020 at 13:37
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$ Dec 30, 2020 at 13:47
  • 2
    $\begingroup$ The fact that your extension might not exist (if you swap wrong boundary components). $\endgroup$ Dec 30, 2020 at 13:51

1 Answer 1


Here are the steps of the proof.

  1. Suppose that $S$ is a compact connected oriented surface with boundary, $C_1, C_2$ are distinct boundary components of $S$. Then there exists a pair of pants $P\subset S$ such that $\partial P= C_1\cup C_2\cup C$, where $C$ is a loop in the interior of $S$.

  2. Given a pair of pants $P$ as in 1, there is an orientation-preserving homeomorphism $h: P\to P$ which swaps $C_1$ and $C_2$.

  3. For a homeomorphism $h$ as in 2, there exists an extension of $h$ to the rest of $S$, $\tilde{h}: S\to S$, which preserves all the boundary components of $S$ except for $C_1, C_2$, of course.

  4. If $F$ is a connected oriented surface with boundary, $A_1, A_2$ are simple nonseparating oriented loops in the interior of $F$. Then there exists an orientation-preserving homeomorphism $f: F\to F$ which sends $A_1$ to $A_2$ and preserves each boundary component of $F$.

  5. Suppose that $\Sigma$ is a connected oriented surface, possibly with boundary and $A_1, A_2$ are oriented simple loops in the interior of $\Sigma$. Then there exists a compact subsurface $F\subset int(\Sigma)$ containing $A_1\cup A_2$.

  6. Suppose that $F$ is compact subsurface of the interior of a surface $\Sigma$ and $f: F\to F$ is an orientation-preserving homeomorphism preserving each boundary component. Then $f$ extends to an (orientation-preserving) homeomorphism $\Sigma\to \Sigma$.

  7. Combining steps 4, 5, and 6, you obtain the desired result.


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