Why does a map from simply connected space to U(1) factors through R?

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SUMMARY

A smooth map s from a simply connected manifold M to U(1) can be expressed as s = e^(iu) for some smooth function u from M to R. This conclusion is based on the fact that R serves as the universal cover of U(1), and the exponential function e^ix acts as a covering map. The property that every map from a simply connected space lifts to its universal cover is crucial in establishing this relationship.

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eok20
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I'm having trouble seeing why the following is true: let M be a simply connected manifold and s a smooth map from M to U(1). Then why does it follow that s = e^(iu) for some smooth function u from M to R?

Thanks!
 
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Because R is the universal cover of U(1), and e^ix is a covering map.
 
Tinyboss said:
Because R is the universal cover of U(1), and e^ix is a covering map.

Ahh...right, every map from a simply connected space lifts to the universal cover.

Thanks!
 

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