Let $M$ be an oriented surface. If a diffeomorphism $f:M \to M$ is $C^2$-homotopic to the identity, then $f$ preserves orientation
The only question I found related to this is Map isotopic to identity is orientation preserving
I couldn't understand the provided answer. My book's definition of an oriented surface is "M is orientable if it admits an oriented atlas.", but it looks like this answer is using something about volume forms, which also can be used to define orientabiity, as I found: https://en.wikipedia.org/wiki/Orientability#Orientability_of_manifolds
I'm trying to prove without using the concepts of volume forms, but if they're needed, then I'd be glad if someone could explain better to me the given answer.
Since it involves diffeomorphisms and homotopy, could it possibly be related to homotopy equivalences?
Using the classic definitions, $f$ being homotopic to the identity, then it can be deformed smoothly to the identity function. How it affects the atlas? I guess that since $f$ is from $M$ to itself and it's homotopic to the identity, it's intuitive that it'll preserve orientation, but I don't have any idea in how to prove it.
HINT: You're confusing orientability of $M$ (which is given) and the question of whether a diffeomorphism is orientation-preserving. Without loss of generality, assume $M$ is connected, and start with compatible orientations on all the tangent spaces. Fix any point $x\in M$, and consider the invertible linear maps $d(f_t)_x\colon T_x M\to T_{f_t(x)}M$ as $t$ varies from $0$ to $1$. What can you say? (Then why does it work for all $x\in M$?)
EDIT: Here's an alternative argument. (It's always difficult to guess what you know and feel comfortable with.) Since $M$ is orientable, there is a volume form, i.e., a nowhere-zero $n$-form $\omega$; so, in particular, we may take $\int_M \omega>0$. Now, homotopic maps $f,g\colon M\to M$ have the property that $\int_M f^*\omega = \int_M g^*\omega$. In particular, if $f$ is homotopic to the identity, then we have $\int_M f^*\omega = \int_M \omega > 0$. But by the degree formula (see, e.g., Guillemin and Pollack), $\int_M f^*\omega = \deg(f)\int_M \omega$, so $\deg (f) = 1$, which says that $f$ is orientation-preserving.