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Let $N$ a $n$-dimensional connected manifold and let $h: N \rightarrow N$ a diffeomorphism such that $h$ is smoothly isotopic to the identity map $\text{id}_N : N \rightarrow N$.

It's clair that the identity map $\text{id}_N$ is an orientation preserving map. Using the smooth isotopy relation, how can I prove that $h$ is also an orientation preserving map?

Thanks in advance for help!

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Let $\omega$ be the volume form, $f_t^*\omega=h_t\omega, h_t:N\rightarrow \mathbb{R}$. For every $x\in N$, $h_x:[0,1]\rightarrow\mathbb{R}$ define by $h_x(t)=h_t(x)$ is continuous, this implies that $h_x([0,1])$ is connected and an interval. $h_x(0)=1$, this implies that $h_x>0$, if not there exists $t$ such that $h_x(t)<0$, and IVT implies the existence of $t_0$ such that $h_x(t_0)=0$. Contradiction, since $h_{t_0}\omega$ is a volume form.

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  • $\begingroup$ Sorry, can you maybe add some details of your proof? I never worked with the volume form and I'm actually a little bit lost... $\endgroup$ – userr777 Oct 2 '18 at 18:10

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