I'm following Spivak's comprehensive guide to differential geometry and I'm getting a bit stuck on a calculation about orientability. The example is showing that the Mobius strip considered as a line bundle over $S^1$ is non-orientable. Intuitively I have no issues, but I would like to construct things a bit more rigorously and am struggling.
Here are some definitions I'm paraphrasing from the book.
We can equip the trivial bundle $\epsilon^n(X)=X\times \mathbb{R}^n$ with the standard orientation on each fibre ${x}\times \mathbb{R}^n$. We write this as $[(x,e_1),...,(x,e_n)]$. Equivalences are fibre-wise isomorphisms, so we can consider the following case. Let $f:\epsilon^n(X)\rightarrow \epsilon^n(X)$ be an equivalence, with $X$ connected. In this case, $f$ either preserves or reverses the orientation on each fibre. To see this, we define the functions $a_{ij}:X\rightarrow \mathbb{R}$ by \begin{align*} f(x,e_i)=\sum_{i=1}^na_{ji}(x)\cdot (x,e_j), \end{align*} which requires that $\text{det}(a_{ij}):X\rightarrow \mathbb{R}$ is continuous and non-zero.
If $\pi:E\rightarrow B$ is a non-trivial $n$-plane bundle, we just need to amend this definition to be local. An orientation of $E$ is defined to be a collection of orientations $\mu_p$ for $\pi^{-1}(p)$ which satisfy the following compatibility condition for any open connected sets $U\subset B$: If $t:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ is an equivalence, and the fibres of $U\times\mathbb{R}^n$ are equipped with the standard orientation, then $t$ is either orientation preserving or reversing on all fibres. Note that if a single equivalence $t$ satisfies this relation, then so must all equivalences $t'$, since $t'\circ t^{-1}:U\times\mathbb{R}^n\rightarrow U\times\mathbb{R}^n$ is also an equivalence. Then as long as we have compatible orientations $\mu_p$ over a collection of sets $U$ that cover $B$, then we obtain an orientation of $E$, $\mu=\{\mu_p\}$.
Here is how I'm trying to do this. What I am imagining is constructing sections on $S^1$; I construct two on each neighborhood, $+1$ and $-1$. I need to do this on a collection of open sets $U$ that cover $S^1$. There is a simple isomorphism $\pi^{-1}(U)\rightarrow U\times \mathbb{R}$ for the problem here, which I can just take to be the identity. Then I want to show that such equivalences cannot define an orientation. Here is my attempt.
We take the bundle \begin{align*} [0,1]\times \mathbb{R} \Big/((0,a)\sim (1,-a)). \end{align*} Consider the following sections on $S^1$: \begin{align*} \sigma^1_\pm:[0,0.4)&\rightarrow \pi^{-1}([0,0.4) )\\ s&\mapsto (s,\pm 1) \\ \sigma^2_\pm:(0.3, 0.7)&\rightarrow \pi^{-1}((0.3,0.7))\\ s&\mapsto (s,\pm 1)\\ \sigma^3_\pm:(0.6,1]&\rightarrow \pi^{-1}((0.6,1] )\\ s&\mapsto (s,\pm 1) \end{align*} I take the equivalences on each neighborhood to be $t_U = \text{Identity}_U$, which either preserves or reverses the orientation on each neighbourhood as required.
Now note the following conditions on each overlapping region (obviously doesn't change choosing a different point):
\begin{align} &t_{[0,0.4)}\circ \sigma^1_\pm(3.5)=(3.5,\pm 1)\\ &t_{(0.3,0.7)}\circ \sigma^2_\pm(3.5)=(3.5,\pm 1)\\ &t_{(0.3,0.7)}\circ \sigma^2_\pm(6.5)=(6.5,\pm 1)\\ &t_{(0.6,1]}\circ \sigma^3_\pm(6.5)=(6.5,\pm 1)\\ \end{align}
So this means that we have $t_U\circ t^{-1}_{U'}(a, 1)=(a, \pm 1)$, and $t_U\circ t^{-1}_{U'}(a, -1)=(a, \pm 1)$, where the sign is changed only if we take different sections (meaning one is $\sigma^i_+$ and the other is $\sigma^j_-$) on the overlapping region $U\cap U'$. But we have the identification $(0,a)\sim (1,-a)$, and so $\sigma^1_+(0)=\sigma^3_-(1)$. But $t_{[0,0.4)} \circ \sigma^1_+(0) = t_{(0.6,1]} \circ \sigma^3_-(1)$, and since these sections are the differently signed ones, this contradicts what we have above, and there is no orientation.
I guess I'm fundamentally unsure that this proves what I want it to. I would appreciate any guidance, but please be as explicit as you can, because it's the construction that I'm really struggling with rather than the concept.