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Let $\pi_1: M \rightarrow M_1$ and $\pi_2: N \rightarrow M_2$ be two smooth covering maps. Now $\phi: M \rightarrow N$ is a smooth diffeomorphism. Does this induce a smooth diffeomorphism $f: M_1 \rightarrow M_2$ such that the diagram commutes ($ f \circ \pi_1 = \pi_2 \circ \phi$)?

I stumbled over this question in the context of classical mechanics and since I am not very familiar with topology, I thought I might give this question a chance here.

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  • $\begingroup$ The issue here is set theoretical: the map $f\colon M_1 \to M_2$ may not exist, due to some overdetermination. Take the case of $M\to M_1$ covering and $M\to M$ the identity. However, if you do have a map $f$ making the diagram commutative, then $f$ is smooth. $\endgroup$ – Orest Bucicovschi Jul 2 '15 at 21:52
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Generally, no. $\phi$ must be fibre-preserving to induce a smooth map of the base spaces,

$$\pi_1(x) = \pi_1(y) \implies \pi_2(\phi(x)) = \pi_2(\phi(y)).$$

If a smooth map $\psi \colon M \to N$ of the covering spaces is fibre-preserving, it induces a smooth map of the base spaces. Locally, around each $x \in M_1$, we can define a smooth map via $\pi_2 \circ \psi \circ (\pi_1\lvert_V)^{-1}$. The choice of the local inverse of $\pi_1$ doesn't matter since $\psi$ is fibre-preserving, hence all the local smooth maps coincide on the overlap of their domains, and fit together to make a global smooth map $\tilde{\psi}\colon M_1 \to M_2$. If we have a fibre-preserving diffeomorphism $\phi$ of the covering spaces, the induced map is a diffeomorphism of the base spaces if and only if also the inverse $\phi^{-1}$ is fibre-preserving.

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