Let $M$ be an $n$ dimensional oriented riemannian spin manifold, then the bundle of oriented orthonormal frames admits a spin structures. That is a principal $\text{Spin}(n)$ bundle denoted $\text{Spin}(M)$ such that there exists a double covering $\Lambda:\text{Spin}(M)\rightarrow SO(M)$ compatible with the double covering Lie group homomorphism $\lambda:\text{Spin}(n)\rightarrow SO(n)$.
I am trying to prove the following:
Let $e=(e_1,\cdots, e_n)$ be an oriented orthonormal frame for some contractible open set $U\subset M$, then there exists precisely two local sections $\epsilon_{\pm}$ of $\text{Spin}(M)$ satisfying $\Lambda\circ \epsilon_{\pm}=e$.
The book I am using proves this by first asserting that the image of $e$ is an open set of $SO(M)$ diffeomorphic to $U$, however I do not believe this is true. It is easy to see that $e$ is essentially a local section of $SO(M)$, and as such is embedding of $U$ into $SO(M)$. Indeed let $\pi:SO(M)\rightarrow M$ be the projection map, and $V=\text{im }e$, then $\pi|_{V}$ is a continuous (in the subspace topology) inverse of $e$, so $e$ is an homeomorphism onto its image. It is easy to see that the differential $D_xe$ is injective for all $x\in U$, so $e$ is an immersion which is also a homeomorphism onto it's image and thus an embedding. However, the dimension of $V$ is $n$, while the dimension of $SO(M)$ is $n+\frac{n^2-n}{2}$, and an embedded submanifold is open if and only if the dimensions of the submanifold is equal to the dimension of the total space, so $V$ can't be open in $SO(M)$.
With this in mind, I thought I could try something like this:
The image of $e$ is a contractible embedded submanifold of $SO(M)$, denote it by $V$, and let $V'=\pi^{-1}(U)$. We use the principal bundle chart defined by: $$ \begin{align} \psi^{-1}:U\times SO(n)&\longrightarrow V'\\ (x,g)&\longmapsto e(x)\cdot g \end{align}$$ It follows $\psi(V)=U\times\{I_n\}$, where $I_n$ is the identity matrix. Examining the smooth map $\text{Id}_U\times\lambda:U\times \text{Spin}(n)\rightarrow U\times SO(n)$, we see that $(\text{Id}_U\times\lambda)^{-1}(\psi(V))=U\times\{\pm 1\}$. Let $\delta_{\pm}:U\rightarrow U\times \{\pm 1\}\subset U\times \text{Spin}(n)$, these are sections of the trivial principal bundle $U\times\text{Spin}(n)$, and satisfy: $$\psi^{-1}\circ (\text{Id}\times \lambda)\circ \delta_{\pm}=e$$
After this I got stuck, as I don't know how to turn $\delta_{\pm}$ into sections of $\text{Spin}(M)$. Is there anyway to do it this way? The author originally used that $\Lambda|_{\Lambda^{-1}(V)}$ is a trivial two sheeted covering of $V$, but I don't think I can use that argument if $V$ is not open, which I strongly believe it not to be.
Any help would be appreciated.
You are certainly right that the image $e(U)$ cannot be open in $SO(M)$, but this is not really a problem. You should keep in mind that a local section of a principal fiber bundle is equivalent to a local trivialization (basically as you suggest in the question). Take a contractible neighborhood $V$ of $e$ in $SO(n)$ and consider the map $U\times V\to SO(M)$ defined by $(x,g)\mapsto e(x)\cdot g$, where the dot denotes the principal right action. This is the restriction of the inverse $\psi^{-1}$ that you mention in the question, so it is a diffeomorphism onto an open neighborhood $W$ of $e(U)$ in $SO(M)$. Since $\Lambda:\Lambda^{-1}(W)\to W$ is a covering and $W$ is contractible, $\Lambda^{-1}(W)$ must be the disjoint union of two open subsets on which $\Lambda$ restricts to a diffeomorphism. Composing the inverses of these diffeomorphisms with $e$, you get two sections covering $e$.
Conversely, it is easy to see that for a connected manifold $N$, there are at most two maps $f,g:N\to Spin(M)$ such that $\Lambda\circ f=\Lambda\circ g$.