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Let $M$ be a non-orientable smooth $d$-dimensional manifold. Let $\tilde{M}$ be the oriented double cover for $M$; and let $f\in Diff^{1+\beta}(M)$. Define $\tilde{f}:\tilde{M}\rightarrow \tilde{M}$ the following way: $\tilde{f}((x,o_x)):=(f(x),o(d_xf(b)))$ where $b$ is some basis for $T_xM$ from the equivalence class $o_x$, $o(b')$ is the equivalence class of a basis $b'$.

My question is, how can one show that also $\tilde{f}\in Diff^{1+\beta}(\tilde{M})$ ?

Thanks ahead for any heplers

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  • $\begingroup$ Short answer: The projection map and $f$ are local diffeos, so near any point, you can invert them. Now, locally, write $\tilde{f}$ as a composition of diffeos. $\endgroup$ Feb 8, 2016 at 15:42
  • $\begingroup$ Well the important part is to show that the differential is $\beta$-H\"older, so how do I write the differential in this case? I understand I can identify it with $d_\cdot f$, but have no idea how to approach it rigorously.. $\endgroup$ Feb 8, 2016 at 15:58
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    $\begingroup$ I think my suggestion still works. If you compose a $\beta$-H\"older function with a smooth $(C^\infty)$ function, the result is $\beta$-Ho"older, isn't it? If so, that's enough because you can (locally) write $\tilde{f}$ as $\pi^{-1}\circ f\circ \pi$ and $\pi$ is $C^\infty$. $\endgroup$ Feb 8, 2016 at 18:16
  • $\begingroup$ And to be clear what I mean by writing $\tilde{f}$ like that locally, pick $p\in \tilde{M}$ and let $\pi:\tilde{M}\rightarrow M$ be the covering map. Consider $\pi(\tilde{f}(p))\in M$. Since $\pi$ is a covering, there is an open set $U$ around $\pi(\tilde{f}(p)$ for which $\pi^{-1}(U)$ is 2 disjoint pieces, if of which is taken by $\pi$, diffeomorphically, onto $U$. Now, $\tilde{f}(x)$ is in one of those copies, call it $V$. Then $\pi|_V:V\rightarrow U$ is a diffeo and on $\tilde{f}(U)$, we have $\tilde f = (\pi|_V)^{-1} \circ f \circ \pi$. $\endgroup$ Feb 8, 2016 at 18:19
  • $\begingroup$ Okay I think I can take it from here, thanks! $\endgroup$ Feb 9, 2016 at 21:01

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