[proofplan]
We cover $S^1$ by two contractible open arcs whose intersection has two connected components. After choosing product charts for a real line bundle over the two arcs, the bundle is encoded by a transition function into $\mathbb{R}^*$ on the overlap. Since each overlap component is connected, only the two signs of the transition function matter up to homotopy, and changes of product chart can alter the two signs simultaneously but cannot alter their ratio. The relative sign $+1$ gives the product bundle, while the relative sign $-1$ gives the Möbius bundle; these two relative signs are invariant under bundle isomorphism.
[/proofplan]
[step:Encode a real line bundle by two transition signs]
Let $\pi:E\to S^1$ be a real line bundle. Choose open connected arcs $U_0,U_1\subset S^1$ such that $S^1=U_0\cup U_1$ and $U_0\cap U_1=V_+\sqcup V_-$, where $V_+$ and $V_-$ are nonempty connected open arcs. Since each $U_i$ is homeomorphic to an open interval, it is paracompact and contractible. We apply the standard product-chart theorem for real vector bundles over contractible paracompact spaces: a topological real vector bundle over a contractible paracompact base is isomorphic to the product bundle of the same rank. Its hypotheses apply because $E|_{U_i}\to U_i$ is a topological real line bundle, each $U_i$ is paracompact, and each $U_i$ is contractible. Hence $E|_{U_i}$ is isomorphic to the product real line bundle over $U_i$. Choose bundle product charts
\begin{align*}
\Phi_i:E|_{U_i}&\to U_i\times \mathbb{R}
\end{align*}
for $i\in\{0,1\}$.
On $U_0\cap U_1$, the change of product chart
\begin{align*}
\Phi_1\circ \Phi_0^{-1}:(U_0\cap U_1)\times \mathbb{R}&\to (U_0\cap U_1)\times \mathbb{R}
\end{align*}
has the form
\begin{align*}
(x,t)&\mapsto (x,g(x)t)
\end{align*}
for a [continuous function](/page/Continuous%20Function)
\begin{align*}
g:U_0\cap U_1&\to \mathbb{R}^*.
\end{align*}
Define the sign map
\begin{align*}
\operatorname{sgn}:\mathbb{R}^*&\to\{\pm1\}
\end{align*}
by $\operatorname{sgn}(r)=+1$ for $r>0$ and $\operatorname{sgn}(r)=-1$ for $r<0$. Because $V_+$ and $V_-$ are connected and $\mathbb{R}^*=(0,\infty)\sqcup(-\infty,0)$ has exactly two connected components, the signs of $g$ are constant on each component. Define
\begin{align*}
\varepsilon_+&:=\operatorname{sgn}(g(x))\quad \text{for }x\in V_+.
\end{align*}
Define also
\begin{align*}
\varepsilon_-&:=\operatorname{sgn}(g(x))\quad \text{for }x\in V_-.
\end{align*}
Thus the chosen product charts determine a pair $(\varepsilon_+,\varepsilon_-)\in\{\pm1\}^2$.
[guided]
The point of using two arcs is that the overlap is not connected. This is the whole source of the classification.
Let $\pi:E\to S^1$ be a real line bundle. Choose open connected arcs $U_0,U_1\subset S^1$ with $S^1=U_0\cup U_1$ and with overlap
\begin{align*}
U_0\cap U_1=V_+\sqcup V_-,
\end{align*}
where $V_+$ and $V_-$ are nonempty connected open arcs. Since each $U_i$ is homeomorphic to an open interval, it is paracompact and contractible. We apply the standard product-chart theorem for real vector bundles over contractible paracompact spaces, which says that a topological real vector bundle over a contractible paracompact base is isomorphic to the product bundle of the same rank. The result applies here because $E|_{U_i}\to U_i$ is a topological real line bundle, each $U_i$ is paracompact, and each $U_i$ is contractible. Therefore $E|_{U_i}$ is isomorphic to the product real line bundle over $U_i$. Choose product charts
\begin{align*}
\Phi_i:E|_{U_i}&\to U_i\times \mathbb{R}
\end{align*}
for $i\in\{0,1\}$.
On the overlap $U_0\cap U_1$, both $\Phi_0$ and $\Phi_1$ identify the same fiber $E_x$ with $\mathbb{R}$. Therefore their transition map is fiberwise multiplication by a nonzero scalar. Thus there is a continuous map
\begin{align*}
g:U_0\cap U_1&\to \mathbb{R}^*
\end{align*}
such that
\begin{align*}
\Phi_1\circ\Phi_0^{-1}(x,t)=(x,g(x)t)
\end{align*}
for every $x\in U_0\cap U_1$ and every $t\in\mathbb{R}$.
Define the sign map
\begin{align*}
\operatorname{sgn}:\mathbb{R}^*&\to\{\pm1\}
\end{align*}
by $\operatorname{sgn}(r)=+1$ for $r>0$ and $\operatorname{sgn}(r)=-1$ for $r<0$. Now each of $V_+$ and $V_-$ is connected. The function $g$ is continuous and never vanishes, so it cannot change sign on a connected component of its domain. Hence the two signs
\begin{align*}
\varepsilon_+&:=\operatorname{sgn}(g(x))\quad \text{for }x\in V_+
\end{align*}
and
\begin{align*}
\varepsilon_-&:=\operatorname{sgn}(g(x))\quad \text{for }x\in V_-
\end{align*}
are well-defined. These two signs are the only discrete information carried by the transition function on the two overlap components.
[/guided]
[/step]
[step:Show that only the relative sign is independent of the product charts]
Changing the product chart over $U_i$ amounts to multiplying it by a continuous function
\begin{align*}
a_i:U_i&\to \mathbb{R}^*
\end{align*}
for $i\in\{0,1\}$. Since each $U_i$ is connected, $\operatorname{sgn}(a_i)$ is constant on $U_i$; write this sign as $\delta_i\in\{\pm1\}$. Define the new transition function
\begin{align*}
g':U_0\cap U_1&\to\mathbb{R}^*
\end{align*}
by
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}.
\end{align*}
Therefore its signs on $V_+$ and $V_-$ are
\begin{align*}
\varepsilon_+'&=\delta_1\varepsilon_+\delta_0^{-1}.
\end{align*}
Similarly,
\begin{align*}
\varepsilon_-'&=\delta_1\varepsilon_-\delta_0^{-1}.
\end{align*}
Consequently
\begin{align*}
\varepsilon_+'\varepsilon_-'=\varepsilon_+\varepsilon_-,
\end{align*}
because $\delta_0^2=\delta_1^2=1$. Thus the relative sign
\begin{align*}
\rho(E):=\varepsilon_+\varepsilon_-\in\{\pm1\}
\end{align*}
is independent of the chosen local product charts.
[guided]
We must check that the number $\rho(E)$ is attached to the bundle, not to the auxiliary product charts. A different product chart over $U_i$ is obtained by composing with fiberwise multiplication by a continuous nonzero function
\begin{align*}
a_i:U_i&\to \mathbb{R}^*
\end{align*}
for $i\in\{0,1\}$. Since $U_i$ is connected and $a_i$ never vanishes, the sign of $a_i$ is constant on $U_i$; denote this sign by $\delta_i\in\{\pm1\}$.
With the new charts, define the transition function
\begin{align*}
g':U_0\cap U_1&\to\mathbb{R}^*
\end{align*}
by
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}.
\end{align*}
Therefore the signs on the two overlap components are
\begin{align*}
\varepsilon_+'&=\delta_1\varepsilon_+\delta_0^{-1}.
\end{align*}
Similarly,
\begin{align*}
\varepsilon_-'&=\delta_1\varepsilon_-\delta_0^{-1}.
\end{align*}
Multiplying these two equations and using $\delta_0^2=\delta_1^2=1$, we get
\begin{align*}
\varepsilon_+'\varepsilon_-'=\varepsilon_+\varepsilon_-.
\end{align*}
Thus changing product charts may flip both signs together, but it cannot change their product. Hence
\begin{align*}
\rho(E):=\varepsilon_+\varepsilon_-\in\{\pm1\}
\end{align*}
is a well-defined invariant of the real line bundle $E\to S^1$ relative to the chosen two-arc cover.
[/guided]
[/step]
[step:Identify relative sign $+1$ with the product line bundle]
Suppose first that $\rho(E)=+1$. Then $\varepsilon_+=\varepsilon_-$. Multiplying one product chart by $-1$ if necessary, we may assume both signs are $+1$. Hence the transition function $g$ takes values in $(0,\infty)$ on both components of $U_0\cap U_1$.
Define
\begin{align*}
h:U_0\cap U_1&\to (0,\infty)
\end{align*}
by $h(x)=g(x)$. Let
\begin{align*}
\ell:U_0\cap U_1&\to \mathbb{R}
\end{align*}
be the continuous function $\ell(x)=\log h(x)$. By the [Existence of Partitions of Unity](/theorems/57), choose a continuous [partition of unity](/page/Partition%20of%20Unity) $\varphi_0,\varphi_1:S^1\to[0,1]$ subordinate to the open cover $\{U_0,U_1\}$, so the closed support $\operatorname{supp}(\varphi_i)$ is contained in $U_i$ and $\varphi_0+\varphi_1=1$ on $S^1$. Define continuous functions
\begin{align*}
b_0:U_0&\to\mathbb{R}
\end{align*}
and
\begin{align*}
b_1:U_1&\to\mathbb{R}
\end{align*}
as follows: on $U_0\cap U_1$, set $b_0(x)=\varphi_1(x)\ell(x)$ and $b_1(x)=-\varphi_0(x)\ell(x)$; away from the overlap, extend each by $0$. These extensions are continuous because $\operatorname{supp}(\varphi_1)\subset U_1$ and $\operatorname{supp}(\varphi_0)\subset U_0$. On $U_0\cap U_1$ we have
\begin{align*}
b_1(x)-b_0(x)=-(\varphi_0(x)+\varphi_1(x))\ell(x)=-\ell(x).
\end{align*}
Now define positive change-of-product-chart functions
\begin{align*}
a_i:U_i&\to(0,\infty)
\end{align*}
by $a_i(x)=e^{b_i(x)}$. The new transition function is
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}=e^{b_1(x)}e^{\ell(x)}e^{-b_0(x)}=1.
\end{align*}
With transition function $1$, the two local products glue to the global product bundle $S^1\times\mathbb{R}\to S^1$. Therefore $E$ is isomorphic to the product real line bundle.
[guided]
Assume $\rho(E)=+1$. Then the two transition signs agree. Multiplying one product chart by $-1$ if necessary changes both signs at once, so we may arrange that $g$ is positive on both $V_+$ and $V_-$. Thus
\begin{align*}
h:U_0\cap U_1&\to (0,\infty), & h(x)&=g(x)
\end{align*}
is a continuous positive function.
Because $h$ is positive, the logarithm gives a continuous function
\begin{align*}
\ell:U_0\cap U_1&\to \mathbb{R}, & \ell(x)&=\log h(x).
\end{align*}
Choose a continuous partition of unity $\varphi_0,\varphi_1:S^1\to[0,1]$ subordinate to $\{U_0,U_1\}$, with $\operatorname{supp}(\varphi_i)\subset U_i$ and $\varphi_0+\varphi_1=1$. Define continuous functions $b_0:U_0\to\mathbb{R}$ and $b_1:U_1\to\mathbb{R}$ by setting, on $U_0\cap U_1$,
\begin{align*}
b_0(x)&=\varphi_1(x)\ell(x), & b_1(x)&=-\varphi_0(x)\ell(x),
\end{align*}
and extending by $0$ away from the overlap. These extensions are continuous because the relevant supports lie inside the corresponding open sets.
On the overlap,
\begin{align*}
b_1(x)-b_0(x)=-(\varphi_0(x)+\varphi_1(x))\ell(x)=-\ell(x).
\end{align*}
Now define positive product-chart changes
\begin{align*}
a_i:U_i&\to(0,\infty), & a_i(x)&=e^{b_i(x)}.
\end{align*}
The changed transition function is
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}=e^{b_1(x)}e^{\ell(x)}e^{-b_0(x)}=1.
\end{align*}
A transition function equal to $1$ means the two local products identify fibers without twisting on the overlap, so they glue to $S^1\times\mathbb{R}\to S^1$. Hence $E$ is isomorphic to the product real line bundle.
[/guided]
[/step]
[step:Identify relative sign $-1$ with the Möbius line bundle]
Suppose now that $\rho(E)=-1$. After possibly multiplying one product chart by $-1$, we may assume
\begin{align*}
\varepsilon_+=+1,\qquad \varepsilon_-=-1.
\end{align*}
Let
\begin{align*}
s:U_0\cap U_1&\to\{\pm1\}
\end{align*}
be the locally constant sign function with $s(x)=+1$ for $x\in V_+$ and $s(x)=-1$ for $x\in V_-$. Then $h(x):=s(x)g(x)$ defines a continuous map $h:U_0\cap U_1\to(0,\infty)$. Applying the logarithm and partition-of-unity construction from the previous step to this positive function $h$, choose positive functions $a_i:U_i\to(0,\infty)$ such that
\begin{align*}
a_1(x)h(x)a_0(x)^{-1}=1
\end{align*}
for $x\in U_0\cap U_1$. Since $g(x)=s(x)h(x)$, the corresponding changed transition function is
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}=s(x).
\end{align*}
Thus $E$ is isomorphic to the bundle obtained by gluing $U_0\times\mathbb{R}$ and $U_1\times\mathbb{R}$ using transition function $+1$ on $V_+$ and transition function $-1$ on $V_-$.
We now compare this two-overlap gluing with the usual quotient model of the Möbius line bundle. Identify $S^1$ with $[0,1]/(0\sim1)$, choose $U_0$ to be the image of a neighbourhood of $[0,1/2]$, and choose $U_1$ to be the image of a neighbourhood of $[1/2,1]$. Then one component of $U_0\cap U_1$ lies near the interior point $1/2$, while the other lies near the identified endpoint $0\sim1$. In the quotient bundle
\begin{align*}
M:=([0,1]\times\mathbb{R})/((0,t)\sim(1,-t)),
\end{align*}
the local products over $U_0$ and $U_1$ are obtained from the evident products over the corresponding subintervals. On the overlap near $1/2$, these two products come from the same representative in $[0,1]\times\mathbb{R}$, so the transition function is $+1$. On the overlap near $0\sim1$, passing from the representative near $0$ to the representative near $1$ uses the defining identification $(0,t)\sim(1,-t)$, so the transition function is $-1$. Therefore the two-overlap gluing model with signs $+1$ and $-1$ is isomorphic to $M$, the Möbius line bundle. Hence every real line bundle with relative sign $-1$ is isomorphic to the Möbius line bundle.
[guided]
Assume $\rho(E)=-1$. After multiplying one product chart by $-1$ if necessary, we may arrange the signs as
\begin{align*}
\varepsilon_+=+1,\qquad \varepsilon_-=-1.
\end{align*}
Define the locally constant sign function
\begin{align*}
s:U_0\cap U_1&\to\{\pm1\}
\end{align*}
by $s(x)=+1$ for $x\in V_+$ and $s(x)=-1$ for $x\in V_-$. Define
\begin{align*}
h:U_0\cap U_1&\to(0,\infty)
\end{align*}
by $h(x)=s(x)g(x)$. This function is continuous because $s$ is constant on each connected component of $U_0\cap U_1$ and the sign of $g$ agrees with $s$ on each component.
Since $h$ is positive, define
\begin{align*}
\ell:U_0\cap U_1&\to\mathbb{R}
\end{align*}
by $\ell(x)=\log h(x)$. Choose a continuous partition of unity $\varphi_0,\varphi_1:S^1\to[0,1]$ subordinate to $\{U_0,U_1\}$, with $\operatorname{supp}(\varphi_i)\subset U_i$ and $\varphi_0+\varphi_1=1$. Define continuous functions $b_0:U_0\to\mathbb{R}$ and $b_1:U_1\to\mathbb{R}$ by $b_0(x)=\varphi_1(x)\ell(x)$ and $b_1(x)=-\varphi_0(x)\ell(x)$ on $U_0\cap U_1$, and by $0$ away from the overlap. These extensions are continuous because the relevant supports lie inside the corresponding open sets. On $U_0\cap U_1$,
\begin{align*}
b_1(x)-b_0(x)=-(\varphi_0(x)+\varphi_1(x))\ell(x)=-\ell(x).
\end{align*}
Define positive functions $a_i:U_i\to(0,\infty)$ by $a_i(x)=e^{b_i(x)}$. Then
\begin{align*}
a_1(x)h(x)a_0(x)^{-1}=e^{b_1(x)}e^{\ell(x)}e^{-b_0(x)}=1
\end{align*}
for $x\in U_0\cap U_1$. Since $g(x)=s(x)h(x)$, the changed transition function is
\begin{align*}
g'(x)=a_1(x)g(x)a_0(x)^{-1}=s(x).
\end{align*}
Thus $E$ is isomorphic to the two-overlap gluing model whose transition is $+1$ on $V_+$ and $-1$ on $V_-$.
It remains to identify this gluing model with the usual Möbius line bundle. View $S^1$ as $[0,1]/(0\sim1)$, choose $U_0$ around the image of $[0,1/2]$, and choose $U_1$ around the image of $[1/2,1]$. The overlap has one component near $1/2$ and one component near the identified endpoint $0\sim1$. In
\begin{align*}
M:=([0,1]\times\mathbb{R})/((0,t)\sim(1,-t)),
\end{align*}
the product charts over these two arcs are inherited from the product over subintervals of $[0,1]$. Near $1/2$, the two charts use the same representative and hence have transition $+1$. Near $0\sim1$, the passage from the representative at $0$ to the representative at $1$ uses $(0,t)\sim(1,-t)$ and hence has transition $-1$. Therefore the two-overlap gluing model with signs $+1$ and $-1$ is precisely the Möbius line bundle up to bundle isomorphism.
[/guided]
[/step]
[step:Separate the two resulting isomorphism classes]
The invariant $\rho(E)\in\{\pm1\}$ is preserved by bundle isomorphism, because a bundle isomorphism written in local product charts changes transition functions only by multiplication by nonvanishing functions on $U_0$ and $U_1$, and the preceding computation shows that such changes preserve $\varepsilon_+\varepsilon_-$. The product line bundle has transition function $1$ on both overlap components, so its relative sign is $+1$. The Möbius line bundle has transition signs $+1$ and $-1$, so its relative sign is $-1$.
Therefore the product line bundle and the Möbius line bundle are not isomorphic, and every real line bundle over $S^1$ is isomorphic to exactly one of them. This proves that there are exactly two isomorphism classes of real line bundles over $S^1$.
[guided]
The invariant
\begin{align*}
\rho(E)\in\{\pm1\}
\end{align*}
is preserved by bundle isomorphism. Indeed, after writing a bundle isomorphism in local product charts over $U_0$ and $U_1$, the transition functions on the source and target differ only by multiplication by nonvanishing functions on $U_0$ and $U_1$. The computation in the chart-independence step shows that such multiplication leaves the product of the two signs unchanged.
The product line bundle has transition function $1$ on both overlap components, so its relative sign is $+1$. The Möbius line bundle has transition signs $+1$ and $-1$, so its relative sign is $-1$. Since these invariant values are different, the two bundles are not isomorphic. The previous two steps show conversely that every real line bundle over $S^1$ has one of these two invariant values and is isomorphic to the corresponding model. Hence there are exactly two isomorphism classes of real line bundles over $S^1$: the product line bundle and the Möbius line bundle.
[/guided]
[/step]
