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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