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Let $p:E\to X$ be a universal covering map. Suppose that $E$ is compact and $X$ is path connected. Show that any continous $f:X\to S^1$ is homotopic to a constant.

Can you give me some hints?

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    $\begingroup$ This implies $X$ has finite fundamental group. Figure and why and see if with this knowledge you can get your result. $\endgroup$ – user98602 Jan 14 '16 at 17:41
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    $\begingroup$ Thank you for your hint! We know that $X$ is compact and we can take a finite covering of evenly covered open sets $\{U_i\}$, then $\{p^-1(U_i)\}$ is a open cover for $E$ and we can take a finite subcover $\{V_j\}$. For each $i$, $p^-1(U_i)$ is a union of disjoint open sets, so their number must be finite, and that number is equal to the cardinality of the fiber. Now since $E$ is simply connected, there is a bijection between the fiber $p^{-1}(X)$ and $\pi_1(X,x)$. $\endgroup$ – giulio Jan 14 '16 at 20:12

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