# Any two non-separating curves on a surface are equivalent

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.

• Hint: find a compact connected subsurface with boundary in $S$ which contains $C\cup D$ and is not disconnected by either one of them. Commented Dec 30, 2020 at 10:46
• 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? Commented Dec 30, 2020 at 13:18
• 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. Commented Dec 30, 2020 at 13:37
• Yes, that's right. Commented Dec 30, 2020 at 13:47
• The fact that your extension might not exist (if you swap wrong boundary components). Commented Dec 30, 2020 at 13:51

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.