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The following problem gives me a bit of trouble:

Let $p:E\to X$ be a covering map. Let $g_1,g_2$ be two lifts of the continuous map $f:Y\to X$. Show that $T:=\{y\in Y:g_1(y)=g_2(y)\}\subseteq Y$ is clopen. Also show that for $Y$ connected, $g_1$ and $g_2$ are equal if $g_1(y)=g_2(y)$ for one $y\in Y$.

The second part is clear, since $Y$ connected is equivalent to $\emptyset,Y$ being the only clopen sets in $Y$, thus $y\in T\implies T=Y$.
For the first part: It's clear that the only non-trivial case is where $g_1,g_2$ are lifts of $f$ having a different starting point (since if they had the same starting point, they would be identical by the uniqueness property). But I don't know where to go from here.
I would appreciate any hints!

Kyle
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blst
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1 Answers1

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Let $y\in Y$. By the definition of covering map, there exists a neighbourhood $U$ of $f(y)$ such that $p^{-1}(U)$ is a disjoint union of open sets, each one homeomorphic to $U$ (via $p$). Let $V_1$ be the one where $g_1(y)$ lies, and $V_2$ the one where $g_2(y)$ lies.

Notice that $V_1$ and $V_2$ are either equal or disjoint, and which one of the two options depends only on whether $g_1(y)=g_2(y)$ or not. Additionally, for any $x\in U$ there is exactly one $e'\in V_1$ and exactly one $e''\in V_2$ such that $p(e')=p(e'')=x$.

Let $W=g_1^{-1}(V_1)\cap g_2^{-1}(V_2)$. This is obviously an open neighbourhood of $y$ and $g_1(W)\subset V_1$ and $g_2(W)\subset V_2$. Now, this, together with the thing stated in the previous paragraph implies that if $y\in T$ then actually $W\subset T$, and if $y\notin T$ then also $W\subset Y\setminus T$. This proves that $T$ is clopen.

user0820
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