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Let $F, G : M \to N$ be diffeomorphisms of compact, connected, oriented, $n$-manifolds. If $F$ and $G$ are smoothly homotopic, prove that $F$ and $G$ are either both orientation-preserving or both orientation reversing.

Degree is homotopic invariant, hence $F$ and $G$ has the same degree, but how can I proceed?

To complete the proof with Henry's help in the answer as well as in the comment:

Lemma. If $f: M \longrightarrow N$ is a diffeomorphism of smooth manifolds, then $\deg(f) = 1$ if $f$ is orientation preserving, and $\deg(f) = -1$ if $f$ is orientation reversing.

Proof: Because $f$ is a diffeomorphism, for each $y$, there is only one $x$ such that $f(x) = y$. Hence, by $$\deg(f) = \sum_{x \in f^{-1}(y)} \operatorname{sgn} (\det(df_x)),$$ where $y \in N$ is a regular value of $f$, we know that $\deg(f) = \pm 1.$

We know that degree is homotopic invariant, hence $F$ and $G$ has the same degree. Therefore, $F,G$ are both either orientation preserving or both orientation reversing.

Thanks very much, Henry!!!

1LiterTears
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  • Hint: use that the determinant is continuous. – Berci Aug 17 '13 at 23:07
  • I'm not sure about either of your proofs. For the lemma, I would use the formula $$\deg(f) = \sum_{x \in f^{-1}(y)} \mathrm{sgn}(\det(df_x)),$$ where $y \in N$ is a regular value of $f$. Since $f$ is a diffeomorphism, this formula will be very easy to work with. Now for the proof of your main result, I don't know why you introduce $H$. You already know degree is homotopy invariant, so by the lemma the degree of both $F$ and $G$ will be $1$ (both orientation preserving) or $-1$ (both orientation reversing). – Henry T. Horton Aug 18 '13 at 00:12
  • As for $H$, I thought there was a diffeomorphism between the function $G$ and $F$!! Laugh, my English is very bad.. – 1LiterTears Aug 18 '13 at 01:00
  • But how about my proof of lemma? Not elegant or wrong? – 1LiterTears Aug 18 '13 at 01:04
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    It's good now. I would just say $\deg(f) = 1$ corresponds to $f$ being orientation-preserving and $\deg(f) = -1$ corresponds to $f$ be orientation reversing. – Henry T. Horton Aug 18 '13 at 01:27
  • Get it, thank you @HenryT.Horton :-) You have been very helpful! – 1LiterTears Aug 18 '13 at 01:28

1 Answers1

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To finish the problem, prove the following simple lemma:

If $f: M \longrightarrow N$ is a diffeomorphism of smooth manifolds, then $\deg(f) = 1$ if $f$ is orientation preserving, and $\deg(f) = -1$ if $f$ is orientation reversing.

Combining this with your observation about homotopy invariance of degrees, you see that $\deg(F) = \pm 1 = \deg(G)$, so they are both either orientation preserving or both orientation reversing.

Henry T. Horton
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  • That's wonderful! Thanks Henry. That's exactly what I need - I was not quite confident with the lemma. Thank you! – 1LiterTears Aug 17 '13 at 23:31