(This is not meant to be a duplicate of this post.)
I would like to show that
Let $M,N$ be oriented, compact, connected, smooth $n$-manifolds, and let $F,G:M\to N$ be diffeomorphisms that are homotopic. Then $F,G$ are either both orientation-preserving or both orientation-reversing.
I'm aware that this post asked exactly the same question. However, the proof there relies on the notion of degree of a map. In the proof I attempted below I wish to adopt an alternative approach.
Let $H:M\times I\to N$ be a homotopy from $F$ to $G$. It is known that (by Whitney approximation theorem) such $H$ can be chosen to also be smooth.
It is not hard to show that a map (says F) is orientation-preserving if and only if for any positively-oriented orientation form $\Omega$ on $N$, the pullback $F^*\Omega$ is also a positively-oriented orientation form on $M$.
Using this, we can apply the Stokes' theorem on the differential $n$-form $H^*\Omega$, which is closed: \begin{align} d(H^*\Omega)=H^*(d\Omega)=H^*(0)=0 \end{align} where $d\Omega=0$ because $\Omega$ is a top-degree form on $N$. Thus \begin{align} 0&=\int_{M\times I}d(H^*\Omega) \\ &=\int_{\partial(M\times I)}H^*\Omega \\ &=\int_{\partial M\times I}H^*\Omega +\int_{M\times\{0\}}H^*\Omega +\int_{M\times\{1\}}H^*\Omega \end{align} Now $M\times\{0\}$ and $M\times\{1\}$ are just two copies of $M$. However, since in the statement of Stokes' theorem, the orientation on the boundary must be induced from the ambient orientation, these two copies must have opposite orientation (one can recall the low-dimensional picture encountered in multivariable calculus). Without loss of generality, let us assume that $M\times\{1\}$ is positively-oriented. Then we have \begin{align} \int_{M\times\{0\}}H^*\Omega=-\int_MF^*\Omega,\qquad \int_{M\times\{1\}}H^*\Omega=+\int_MG^*\Omega \end{align} The proof is completed if one can show that $\displaystyle\int_{\partial M\times I}H^*\Omega=0$, for then we will have \begin{align} \int_MF^*\Omega=\int_MG^*\Omega \end{align} Since $F^*\Omega$ and $G^*\Omega$ are smooth non-vanishing forms, the two integrals are nonzero, so for the equality to hold, they must have the same sign, which is true if and only if $F,G$ are either both orientation-preserving or both orientation-reversing. Hence, my question boils down to
How to show that $\displaystyle\int_{\partial M\times I}H^*\Omega=0$?
Note that this is certainly true if $M$ is already closed (compact without boundary). What about in general?
Any comment, hint or answer is greatly welcomed and appreciated.
Clearly false in general. Let $F$ be the identity on the disc and $G$ be reflection in one coordinate. Then $\int F^*\Omega = - \int G^*\Omega$ and in any $G$ is orientation reversing.
You need your homotopy between diffeomorphisms to be boundary-preserving, that is, $H(\partial M \times I) \subset \partial N$. Then choose a form $\Omega$ which integrates to one and vanishes on the boundary. Then $\int F^*\Omega$ is an invariant of maps $F: (M,\partial M) \to (N, \partial N)$, giving the same result for any two maps which are homotopic through boundary-preserving maps.
(This reflects the algebraic fact that degree is defined in terms of $H_n(M,\partial M)$, or equivalently in terms of $H^n(M,\partial M)$.)