I have the following proposition (taken from Klingenberg's Lectures on Closed Geodesics):
Let $\pi: E \rightarrow S$ and $\mathcal{O} \subset E$ be a finite dimensional fibre bundle over the circle with a riemannian metric and a riemannian connection, and an open set that intersects every fibre respectively. Let $\phi: F \rightarrow S$ be another fibre bundle of the same kind. Assume that
$$f: \mathcal{O} \rightarrow F$$
is a differentiable fibre map ($\phi \circ f=\pi$).
So, we have that the induced mapping:
$$\tilde{f}: H^1(\mathcal{O}) \rightarrow H^1(F); \quad (\xi(t)) \mapsto \left(f\circ \xi(t)\right)$$
is continuous
My question is: why does $$\tilde{f}: H^1(\mathcal{O}) \rightarrow H^1(F)$$
?? More explicitly, why is the image of a $H^1$ section (an absolutely continuous with square-integrable derivative, which is defined a.e.) a $H^1$ section?
It follows from compactness of $S^1$ and the fact that $f$ is differentiable ($C^{\infty}$) that the derivative is bounded. The chain rule then implies the facts needed.