Sections on Möbius bundle correspond to $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+n)=(-1)^nf(x)$ I want to show that sections of the Möbius bundle correspond to functions 
 $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x+n)=(-1)^nf(x)$. 
Given a section of the Möbius bundle I want to define such an $f$, this will give a map from sections of the Möbius bundle into the set of desired functions.
We view $S^1$ as $\mathbb{R}/\mathbb{Z}$ where $\mathbb{Z}$ acts on $\mathbb{R}$ by $n\cdot x=x+n$.
We get the Mobius bundle over $S^1$ by considering the $\mathbb{Z}$ action on $\mathbb{R}^2$ that is $n\cdot(x,y)=(x+n,(-1)^n y)$. Denote the Möbius bundle by $L$.
The projection $\pi:L\rightarrow S^1$ is then $\pi(x,y)=(x)$. 
Recall that a section of a line bundle over $S^1$ is a map $s:S^1\rightarrow L$ such that $\pi \circ s= id_{S^1}$.
Intuitively this makes sense but I am not sure how to rigoursly identify these things.
Given a section $s:S^1\rightarrow L$, we want to define an $f:\mathbb{R}\rightarrow \mathbb{R}$ such that  $f(x+n)=(-1)^nf(x)$. The $(-1)^n$ term appears when we act on $(x,y)\in L$ by $n$. So if for a section $s(x)=(s_1(x),s_2(x))$ I think we will need to define $f$ in terms of $s_2$. But this only gives $(s_1(x),s_2(x))=n\cdot(s_1(x),s_2(x))=(s_1(x)+n,(-1)^ns_2(x))$. 
Do we do something like $f(x)=\pi_2(n\cdot s(\overline{x}))$? where $\overline{x}$ is the $x$ minus the nearest integer to $x$ and $n$ is the nearest integer to $x$? 
This feels wrong, it seems like there should be a neater way to do this since $x+n$ and $(-1)^n$ both appear in the action of $\mathbb{Z}$/
 A: The space $L$ is the orbit space obtained by quotienting $\mathbb{R}^{2}$ by the indicated $\mathbb{Z}$-action. That is, points of $L$ are $\mathbb{Z}$-orbits of points $(x,y)\in\mathbb{R}^{2}.$ Let the orbit of $(x,y)$ be denoted $(x:y),$ so that
$$(x:y)=(x+n:(-1)^{n}y)$$
holds for all $x,y\in\mathbb{R}$ and $n\in\mathbb{Z}.$
Lemma. If $x,y_{1},y_{2}\in\mathbb{R}$ are such that $(x:y_{1})=(x:y_{2}),$ then $y_{1}=y_{2}.$
Proof of lemma. By definition, if $(x:y_{1})=(x:y_{2})$ then there must exist $n\in\mathbb{Z}$ such that $n\cdot(x,y_{1})=(x,y_{2}).$ But $n\cdot(x,y_{1})=(x+n,(-1)^{n}y_{1}).$ Thus, we must have $n=0,$ and $y_{2}=(-1)^{n}y_{1}=y_{1}.$ $\qquad\square$
The bundle map $p\colon L\to S^{1}$ is defined a few stages. First, note that the $x$-axis in $\mathbb{R}^{2}$ is invariant under the $\mathbb{Z}$-action, so that the action on $\mathbb{R}^{2}$ restricts to an action on the $x$-axis. Furthermore, on the $x$-axis the action is  by translations, i.e., $\mathbb{Z}$ acts on $\mathbb{R}\times\{0\}$ in a way which everyone knows gives $S^{1}$ as the orbit space; so we make the identification $S^{1}\cong(\mathbb{R}\times\{0\})/\mathbb{Z}.$ Finally, introduce the orthogonal projection $\pi\colon\mathbb{R}^{2}\to\mathbb{R}\times\{0\},$ and define $p=q\circ\pi.$ This gives a commuting square, in which the vertical maps are both the quotient map $q\colon\mathbb{R}^{2}\to L$ given by $q(x,y)=(x:y).$
$$
\begin{array}{ccc}
\mathbb{R}^{2} & \xrightarrow{\pi} & \mathbb{R}\times\{0\}\\
\downarrow & & \downarrow\\
L & \xrightarrow{p} & (\mathbb{R}\times\{0\})/\mathbb{Z}
\end{array}
$$
A section $s\colon(\mathbb{R}\times\{0\})/\mathbb{Z}\to L$ is just a map which satisfies $p(s(x:0))=(x:0)$ for all $x\in\mathbb{R}.$ By definition of $p,$ this means that if $x\in\mathbb{R},$ then there exist a $y\in\mathbb{R}$ such that $s(x:0)=(x:y).$ We know such a $y$ is unique because of the lemma proved above. With this in mind, define $f_{s}\colon\mathbb{R}\to\mathbb{R}$ by the formula $s(x:0)=(x:f_{s}(x)).$
All that remains is to check that the function $f_{s}$ so-defined satisfies the functional equation. Given $x\in\mathbb{R}$ and $n\in\mathbb{Z},$ we have
$$(x+n:f_{s}(x+n))=s(x+n:0)=s(x:0)=(x:f_{s}(x))=(x+n:(-1)^{n}f_{s}(x)).$$
Therefore $f_{s}(x+n)=(-1)^{n}f_{s}(x).$ 
So a (global) section $s$ of the bundle gives a function $f_{s}\colon\mathbb{R}\to\mathbb{R}$ satisfying the given equation.
