# Showing that the orientable double covering of a manifold is an orientable manifold (do Carmo Differential Forms ex. 3.16)

I am trying to solve exercise 3.16 of do Carmo's Differential Forms and Applications. The problem is as follows:

Let $M$ be a connected differentiable manifold. For each $p\in M$, denote by $\mathcal{O}_p$ the quotient space of the set of all bases of $T_pM$ under the following equivalence relation: tow bases are equivalent if they are related by a matrix with positive determinant. Clearly $\mathcal{O}_p$ has two elements, and each element $O_p$ of $\mathcal{O}_p$ is called an orientation at p. Now let $$\tilde{M}=\{(p,O_p)\mid p\in M,O_p\in\mathcal{O}\},$$ and let $f_\alpha:U_\alpha\rightarrow M$ be a parametrization of $M$ with $p\in f_\alpha(U_\alpha)$. Define $\tilde{f}_\alpha:U_\alpha\rightarrow \tilde{M}$ by $\tilde{f}_\alpha(x_1,\ldots,x_n)=(f_\alpha(x_1,\ldots,x_n),[\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}])$ where $(x_1,\ldots,x_n)\in U_\alpha$, and $[\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}]$ denotes the element of $\mathcal{O}_p$ determined by this basis. Show that if $\{(U_\alpha,f_\alpha)\}$ is a differentiable structure in $M$, then $\{(\tilde{U}_\alpha,\tilde{f}_\alpha)\}$ is a differentiable structure in $\tilde{M}$ which is orientable (even if $M$ is not).

What I am having trouble understanding is how $\{(U_\alpha,f_\alpha)\}$ is a differentiable structure on $\tilde{M}$. In particular, I can't see why $\bigcup_\alpha\tilde{f}_\alpha(U_\alpha)=\tilde{M}$. The basis $\{\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_1}\}$ should be determined at each $p$ by $f_\alpha$ so that given some $p\in U_\alpha$, $\tilde{f}_\alpha(p)$ only maps $p$ to one component of $\tilde{M}$. Thus, unless $p$ happens to be included in some $U_\beta$ with $f_\beta$ that provides an oppositely oriented basis for $T_pM$, I see no reason why $(f_\alpha(p),+)$ and $(f_\alpha(p),-)$ (where by $+$ and $-$ I denote the two elements of $\mathcal{O}_p$) should appear in the image of the coordinate charts.

• Just a guess: I've seen differentiable structures defined where we assume that any parametrization which is compatible with the parametrizations we already have is included in the differentiable structure. Perhaps the author intended the collection $\{(\tilde{U}_\alpha,\tilde{f}_\alpha)\}$ to generate a differentiable structure in this way? Of course, it would then be left to you to show that if $(f_\alpha(p),+)$ is in the image of the charts, then so is $(f_\alpha(p),-)$. – 211792 Sep 1 '16 at 22:22
• Note that if $f_\alpha\colon U_\alpha\to M$ is a parametrization, so is $f_\alpha^-\colon U_\alpha^-\to M$, where we use coordinates $(x_2,x_1,\dots,x_n)$ on $U_\alpha^-$. (I'm assuming $n\ge 2$. Any $1$-manifold is orientable.) – Ted Shifrin Sep 1 '16 at 22:27
• Ah so assuming we expand our differentiable structure to include all necessary charts then the resulting manifold will be orientable because a transition map $f_\beta^{-1}\circ f_\alpha$ will only be defined on sets where the orientations of $f_\beta$ and $f_\alpha$ agree? – user293794 Sep 1 '16 at 22:35