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I was following the proof that the lift of any path is constant posted in this thread: Is every lift of a constant path necessarily constant? , specifically the final post on the thread. I was left confused at the final step, which states by continuity and connectedness, we are done, because we only seem to have proven that it is locally constant at 0. I was hoping someone could explain where I was going wrong (sorry if it is really obvious!).

J.Jones5552
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  • That post could have a more rounded finish of its argument. You can finish it up, for example, by continuous induction. They proved that the lift is constant in a neighborhood of $0$. By continuity it is also constant in the closure of that neighborhood. Repeat the argument with the infimum of the set of points in which the lift is not equal to the same constant and conclude that the set in which it is not equal to the same constant is empty. – olsen5 Oct 30 '17 at 00:11
  • Can you quickly outline how one would go about showing it is empty? – J.Jones5552 Oct 30 '17 at 00:19
  • Take the infimum of it. That infimum doesn't belong to it, and by the same argument done at $0$ it has a neighborhood in which the lift is constant. That neighborhood would intersect the set, contradiction. – olsen5 Oct 30 '17 at 00:35
  • how do we know that the infinimum does not belong to the set? – J.Jones5552 Oct 30 '17 at 00:44
  • The part where it was equal to the value at $0$ was closed. In other words, by continuity of the path. – olsen5 Oct 30 '17 at 00:50
  • https://math.stackexchange.com/questions/44850/locally-constant-functions-on-connected-spaces-are-constant – Andres Mejia Oct 30 '17 at 13:39

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A locally constant map $f:X \to Y$ for which $X$ is connected and $Y$ is a metric space ($T_1$, really), must be everywhere constant.

Suppose that there exist $x,y \in X$ so that $f(x) \neq f(y)=y_0$.

note that $f^{-1}(\{y_0\})$ is open (since for every $u \in f^{-1}(\{y_0\})$ is a neighborhood of $u$ where the function is locally constant, and hence also in the preimage of $y_0$)

Let $U=f^{-1}(\{y_0\})$, $V=U^{c}=f^{-1}(Y\setminus \{y_0\})$. Their intersection is clearly trivial, they together cover $X$, and they are both nonempty by assumption, so we have a contradiction.

method 2: since $Y$ is $T_1$, $\{y_0\}$ is closed, so the preimage is clopen, and hence all of $X$.

We can drop the hypotheses on $Y$ in the following way: since $f$ is locally constant, we can take any $v \in V$, then as above, there is a neighborhood of the point that does not map to $y_0$, so the complement is open as well.

Andres Mejia
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