# Smooth isotopy preserves orientation

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?

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.