Orientable surface

Question

Suppose that two smooth surfaces $S$ and $\tilde{S}$ are diffeomorphic and that $S$ is orientable.

I want to prove that $\tilde{S}$ is orientable.

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Since $S$ and $\tilde{S}$ are diffeomorphic, we have that there is a smooth map between these two surfaces, which is a bijection and the inverse is also smooth.

We have that $S$ is orientable, then there is a smooth choice of unit normal at any point of $S$.

Do we conclude then that $\tilde{S}$ is also orientable because of the fact that there is a bijective map between $S$ and $\tilde{S}$ ?

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EDIT:

A surface $S$ is orientable if there exists an atlas $A$ for $S$ with the property that, if $\Phi$ is the transition map between any two surface patches in $A$, then $\det (J(\Phi )) > 0$ where $\Phi$ is defined.

An oriented surface is a surface $S$ together with a smooth choice of unit normal $N$ at each point, i.e., a smooth map $N : S \rightarrow \mathbb{R}^3$ (meaning that each of the three components of $N$ is a smooth function $S \rightarrow \mathbb{R}$) such that, for all $p \in S, \ N(p)$ is a unit vector perpendicular to $T_pS$. Any oriented surface is orientable!

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Answer 1

In general, an orientation is not exactly a ``smooth choice of unit normal''. This definition is only really helpful for 2D surfaces embedded in 3D. I'm guessing that you need to consider more general manifolds. The best way to generalise is as follows. The normal vector $\mathbf{n}$ to a 2D surface in 3D is really a machine that maps pairs of tangent vectors to the triple product, $(\mathbf{u},\mathbf{v}) \to \mathbf{n}\cdot(\mathbf{u}\times\mathbf{v})$. If this is positive, $(\mathbf{u},\mathbf{v},\mathbf{n})$ is a right-handed set. Otherwise left-handed. In this spirit, we have the following definition.

Defn. An n-manifold $M$ is orientable if it admits a global, non-vanishing n-form $\omega$, called the volume form.

So an orientation is a recipe for turning $n$ tangent vectors into a number. If the number is positive, your vectors are positively oriented. Otherwise negatively oriented.

Now suppose that $S$ is orientable with volume form $\omega$ and $\phi: S \to \tilde{S}$ is your diffeomorphism. Then $\phi^{-1}: \tilde{S}\to S$ is also a diffeomorphism. Moreover the pullback of $\omega$ to $\tilde{S}$, $$\omega'= \phi^{-1 *} \omega$$ defines a global, non-vanishing n-form on $\tilde{S}$ (if it vanished anywhere, $\phi^{-1}_{*}$ would be taking a tangent vector to $0$, which contradicts $\phi^{-1}$ being a diffeomorphism) We conclude that $\tilde{S}$ is orientable.

Alternatively, using your vector intuition, we could note that $\phi_{*}$ has full rank on every tangent space and varies smoothly over $S$, so a smooth choice of $n$ tangent vectors on S (another definition of ``orientation'') maps to another such choice on $\tilde{S}$. Hope that helps!

Answer 2

Using your first definition of orientable, let $$\{(U_i,\phi_i)\}_{i \in I} $$ be an atlas for $S$ satisfying the orientation requirement. To be clear on the notation, $U_i \subset S$ is open and $\phi_i : U_i \to \mathbb{R}^2$ is a homeomorphism, and the orientation requirement says that each overlap map $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j) $$ is a diffeomorphism having positive Jacobian determinant at each point of its domain.

Suppose that $f : S \to \tilde S$ is a differeomorphism. Then it's pretty straightforward to show that $$\{(f(U_i),\phi_i \circ f^{-1})\}_{i \in I} $$ is an atlas for $\tilde S$ satisfying the orientation requirement.

Answer 3

A surface is not orientable if and only if it contains a Möbius band embedded. This property is invariant by diffeomorphims or homeomorphism.