Does every real line bundle admit a flat connection?

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Consider a three-fold intersection $U_{i} \cap U_{j} \cap U_{k}$ of a trivialising Leray cover $\{\mathcal{U}\}$ for a real line bundle L over a $C^k$ manifold $M$. We have $g_{ij}g_{jk}g_{ki} = 1$ and thus \begin{align} \ln|g_{ij}| + \ln|g_{jk}| + \ln|g_{ki}| = 0 \end{align} Thus $\{\ln|g_{ij}|\} \in Z^{1}(\mathcal{U}, C^k)$. That is, this array defines a $1$-cocycle with coefficients in the sheaf of $C^{k}$ functions on the overlaps. Now it is a theorem that if the sheaf that the cochains take values in, admits partitions of unity, then the cohomology vanishes. That is, every cocycle is a coboundary so we can write \begin{align} \ln |g_{ij}| = f_{i} - f_{j} = \text{ln}(e^{f_{j}}e^{-f_{i}}) \end{align} for a $1$-coycle $\{f_{i}\}$ on the open cover. Hence \begin{align} \ln|e^{-f_{j}}g_{ij}e^{f_{i}}| = 0 \end{align} Now if $\{s_{i}\}$ is a section which gives a trivialisation over $U_{i}$ define $s'_{i} = e^{f_{i}}s_{i}$. Doing this in every coordinate patch we get that the transition functions with respect to this new trivialisation can be found like \begin{align} e^{f_{i}}s_{i} = e^{f_{i}}g_{ij}e^{-f_{j}}e^{-f_{j}s_{j}} \implies s'_{i}= e^{f_{i}}g_{ij}e^{-f_{j}}s'_{j} \end{align} but $|e^{-f_{j}}g_{ij}e^{f_{i}}| = 1$. So the transition functions of any line bundle can be taken to be in $\{-1,1\}$.

Since the transition functions of this bundle can be taken to be constants it is a theorem that the bundle admits a flat connection.

I don't have any intuition for why this result would be true. Is this result true, and if not, where is the mistake in my working?

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This result is true and your approach is correct. Not an intuition, but the obstruction of the existence of a flat connection is the curvature which vanishes here because the dimension is $1$.

https://en.wikipedia.org/wiki/Connection_(vector_bundle)#Curvature