Map isotopic to identity is orientation preserving

Question

Let $M$ be an $n$-dimensional orientable and compact smooth manifold and $f:M\to M$ be a smooth map isotopic to the identity map. Is it true that $f$ is orientation preserving?

Answer 1

Orientations on a smooth $n$-manifold are equivalent to (equivalence classes of) nonvanishing $n$-forms, where $\omega \sim \eta$ if there's a positive smooth function $f$ with $f\omega = \eta$. Equivalently, because $C^\infty(M,\Bbb R^+)$ is connected, $\omega$ and $\eta$ are equivalent iff they are in the same path component of the space of nonvanishing $n$-forms.

Now pick an orientation $[\omega]$. If $f_t$ is a homotopy through diffeomorphisms between the identity and $f$, then $f_t^*\omega$ is a path in the space of $n$-forms between $\omega$ and $f^*\omega$; and hence, they're equivalent orientations.

Note that there is no good way of talking about whether or not a smooth map is orientation-preserving or reversing, because $d_pf$ needn't be an isomorphism - which is the condition you need to make sense of "$f$ is orientation-(blah) at $p$". This is why we restrict to diffeomorphisms here.

Answer 2

Whatever your definition of orientation is, it gives a continuous homomorphism $\operatorname{Diffeo}(M)\to\{\pm 1\}$. So an isotopy gives a continuous map $[0;1]\to\{\pm 1\}$ which is constant (by intermediate value thm).

(Of course, one have to check that the map $\operatorname{Diffeo}(M)\to\{\pm 1\}$ is indeed continuous — and the details depend on the definition of 'orientation-preserving' you're using.)