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How can I show that a self-homeomorphism of an orientable surface with boundary that fixes identically a boundary component is orientation-preserving?

Amontillado
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  • What's your definition of orientation? For me, it's a choice of generator of $H^n(M;\partial M)$. In this case, use the long exact sequence associated to the pair $(M,\partial M)$. – Jason DeVito - on hiatus Jan 26 '19 at 20:00
  • That's my definition as well but I'm still not really sure how to use it. – Amontillado Jan 26 '19 at 21:17
  • Should the surface be connected? – Charlie Frohman Jan 26 '19 at 22:24
  • Amontillado: Here's a hint. I'll let you fill in the details. The tail end of the long exact sequence of a pair is $...\rightarrow H^{n-1}(\partial M)\rightarrow H^n(M;\partial M)\rightarrow H^n(M)\rightarrow ...$. You know each of these three groups (assuming, as @Charlie pointed out, that $M$ is connected). Now, if $f:M\rightarrow M$ is your homeo, it induces a self map of this long exact sequence. The self map $H^{n-1}(\partial M)\rightarrow H^{n-1}(\partial M)$ must be the identity (why?). Thus the map $H^n(M;\partial M)\rightarrow H^n(M;\partial M)$ is...? – Jason DeVito - on hiatus Jan 27 '19 at 01:08
  • I'm having some trouble with this. I don't see why $H_{1}(\partial M)\to H_{1}(\partial M)$ would be the identity cause if $M$ have more than one holes then $\partial M$ is not connected so $Η_{1}(\partial M)$ would be free abelian but not cyclic. Can't the induced map fix one generator and change the others? Let me ask you if an argument I've been thinking about is valid or not. – Amontillado Jan 28 '19 at 10:08
  • Since $M$ can be embedded in $\Bbb R^3$ then we can expand $f$ to a self homeomorphism $f^{'}$of an oriented surface $M^{'}$ with the same genus as $M$ but without the holes. In one expansion of the boundary $f^{'}$ would be the identity (since $f$ is the identity on that boundary component). This new homeomorphism will also induce an isomorphism of the homology groups. Is it safe to say that since $f'$ is the identity on some part of the surface then the induced map on $Η_{2}(Μ^{'})$ will be the identity as well? – Amontillado Jan 28 '19 at 10:09
  • Also, I just noticed, sorry I'm more used to homology groups than cohomology. – Amontillado Jan 28 '19 at 11:08

1 Answers1

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If you really want to work in the topological category then the answer of @JasonDeVito is the only one available.

If you are willing to work in the differentiable category, let me denote the surface and its diffeomorphism by $f : S \to S$.

Pick $p \in \partial S$ and pick a nonzero $v \in T_p \partial S$. By your assumption we have $D_pf(v)=v$.

The $1$-dimensional subspace $T_p \partial S \subset T_p S$ separates the 2-dimensional vector space $T_p S$ into two half spaces, one denoted $H_{in}$ consisting of vectors that point inward, and the other $H_{out}$ consisting of vector that point outward. Since $f : S \to S$ is a diffeomorphism it follows that $D_p f(H_{in}) = H_{in}$.

The two equations $D_p f(v)=v$ and $D_p f (H_{in}) = H_{in}$ imply that $D_p f$ preserves the orientation of $T_p S$. Therefore $f$ preserves orientation.

Lee Mosher
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  • Do you know if there is a straightforward proof that every homeomorphism that fixes a boundary component is isotopic to one that fixes a neighborhood of the boundary? It would follow if one could argue, eg, that the space of germs around $S^1$ of homeomorphisms of $S^1 \times [0,\infty)$ is connected, but this is not clear to me without differentiability (though it is of course true). –  Jan 27 '19 at 03:08
  • I think there probably is such a proof, similar to the Alexander Trick for proving that every homeomorphism of $D^2$ that restricts to the identity on $\partial D^2$ is isotopic to the identity. – Lee Mosher Jan 27 '19 at 05:01