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.