Can Every Smooth Map from a Manifold to the Complex Numbers be Lifted to the Exponential Map?

[Euge]

Here is this week's POTW:

-----
Suppose $M$ is a smooth path-connected manifold. Consider the differential form

$$\nu = \Re\left\{\frac{1}{2\pi i} \frac{dz}{z}\right\}$$

which generates $H^1_{dR}(\Bbb C^\times)$, the first de Rham cohomology of $\Bbb C^\times$. Show that every smooth map $f : M \to \Bbb C^\times$ can be lifted to smooth map $M\to \Bbb C$ via the exponential map, provided that the image of $\nu$ under $f^* : H^1_{dR}(\Bbb C^\times) \to H^1_{dR}(M)$ is zero.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Here we use the fact that the exponential map $\Bbb C\xrightarrow{\exp}\Bbb C^*$ is a covering space. Suppose $f_*(\pi_1M)$ is nonzero. Then for some loop $\gamma$ in $M$, $f_*(\gamma)$ is (up to homotopy) a circular loop $c$ about $0$. Hence $\int_c\nu = \int_{f_*(\gamma)} \nu = \int_\gamma f^*\nu = \int_\gamma 0 = 0$. On the other hand, Cauchy's integral formula yields $\int_c \nu = 1$. This is a contradiction. Thus $f_*(\pi_1M) = 0$. In particular, $f_*(\pi_1M)$ is a subgroup of $\exp_*(\pi_1\Bbb C)$ (in fact, this subgroup is also zero as $\Bbb C$ is simply connected). By the path lifting criterion for covering spaces of smooth manifolds, $f$ has a smooth lift to a map $M \to \Bbb C$ via $\exp$.