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Let $f:X\rightarrow S^1$ be a continuous function for $X$ connected and $x^* \in X$. Given that $\hat{f}_1, \hat{f}_2:X \rightarrow \mathbb{R}$ are liftings of $f$ such that $\hat{f}_1(x^*) = \hat{f}_2(x^*)$, I am trying to prove that $\hat{f}_1(x) = \hat{f}_2(x)$ for all $x \in X$.

$\hat{f}_1 - \hat{f}_2$ is a continuous function, and $(\hat{f}_1 - \hat{f}_2)(X)$ is a connected set in $\mathbb{R}$. I'm trying to prove the aforementioned result, but I'm having a bit of difficulty. It would suffice to show that $(\hat{f}_1 - \hat{f}_2)(x) = 0$ for all $x$, but I don't necessarily know how to show the image of $X$ under this function can't be a connected interval in $\mathbb{R}$. Any suggestions?

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Hint: consider the set of points in $X$ where the two lifts coincide. It is nonempty and obviously closed. To show it is open, use the definition of covering space.

Eduardo Longa
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  • I'm afraid I don't see why it is obviously closed. The covering space is just the real line, correct? –  Mar 27 '19 at 02:34
  • "obviously closed": Consider the continuous map $F:X\rightarrow \mathbb{R}\times \mathbb{R}$ given by $F(x) = (\hat{f}_1(x), \hat{f}_2(x))$. Let $\Delta \mathbb{R}\subseteq \mathbb{R}\times \mathbb{R}$ denote the diagonal - that is $\Delta \mathbb{R} = {(t,t): t\in \mathbb{R}}$. Prove that $\Delta \mathbb{R}$ is a closed subset of $\mathbb{R}\times \mathbb{R}$. Then $F^{-1}(\Delta \mathbb{R})$ is the subset of $X$ for which $\tilde{f}_1 = \tilde{f}_2$. – Jason DeVito - on hiatus Mar 27 '19 at 02:44