# Given a local diffeomorphism $f: N \to M$ with $M$ orientable, then $N$ is orientable.

Given a local diffeomorphism $f: N \to M$ with $M$ orientable. Why is $N$ orientable? My professor wrote this in class without giving a proof and said “you should try to prove this for fun :)”. I am clueless, please help 🙁

#### Solutions Collecting From Web of "Given a local diffeomorphism $f: N \to M$ with $M$ orientable, then $N$ is orientable."

Do you know about the connection between differential forms and orientability? That $M$ is orientable means precisely that $M$ has a non-vanishing volume form $\omega$. To show $N$ is orientable, we must show $N$ also admits a non-vanishing volume form. It seems natural to consider the pullback of $\omega$, and you can check directly that it is non-vanishing at each point because $f$ is a local diffeomorphism.

In order to have a more complete answer (especially having the algebraic details) this proof is added.

Since $N$ is locally diffeomorphic to $M$, at each point $p$ of the manifold $N$ there exists an open set $U$ around $p$ and a function $F$ such that:

$F:U\to F(U)$ is a diffeomorphism. Having this diffeomorphism in mind one would have the following isomorphism:

$$g:\bigwedge^{n}F(U)\to\bigwedge^{n}U$$

where $g$ is defined as follows:

$$g(\phi)(X_{1},\dots, X_{n})(p):= \phi(F_{\star}(X_{1\ p}),\dots, F_{\star}(X_{n\ p}))$$

Now since $M$ is orientable there exists a non-vanishing $C^{\infty}$ $n$-form $\Omega_{M}$ on $M$.

Now by defining

$$\Omega_{N}:=g(\Omega_{M})$$

We’ll have a non-vanishing $C^{\infty}$ $n$-form on N as follows:

$$\Omega_{N}(X_{1},\dots, X_{n})(p)=g(\Omega_{M})(X_{1},\dots, X_{n})(p)=\Omega_{M}(F_{\star}(X_{1\ p}),\dots, F_{\star}(X_{n\ p}))\neq 0$$