Covering Maps are Open Maps

Recall from the Covering Spaces page that if $X$ is a topological space then a covering space of $X$ is a pair $(\tilde{X}, p)$ where $\tilde{X}$ is a path connected and locally path connected topological space and $p : \tilde{X} \to X$ is a continuous map such that for every $x \in X$ there exists a path connected open neighbourhood $U$ of $x$ such that $p$ restricted to every path component of $p^{-1}(U)$ is a homeomorphism onto $U$.

We will now state a nice result which tells us that if $X$ is a topological space and $(\tilde{X}, p)$ is a covering space of $X$ then $p$ is an open map.

Theorem 1: Let $X$ be a topological space. If $(\tilde{X}, p)$ is a covering space of $X$ then $p : \tilde{X} \to X$ is an open map.

Recall that a map between topological spaces is said to be open if the image of any open set in the domain is an open set in the range.

• Proof: Let $U$ be an open set in $\tilde{X}$. If $U = \emptyset$ then trivially $p(U) = p(\emptyset) = \emptyset$ which is open in $X$. So assume that $U \neq \emptyset$.

• Let $x \in p(U)$. We want to show that $x$ is an interior point of $U$. Let $V$ be an elementary neighbourhood of $x$. Let $C$ be a path component of $p^{-1}(V)$. Then $p$ restricted to $C$ is a homeomorphism onto $V$.

• Since $C$ is a path component it is open in $C \subset \tilde{X}$, and since $U$ is open in $\tilde{X}$ we have that $C \cap U$ is open in $C$. Since $p$ is a homeomorphism from $C$ onto $V$ we have that $p(C \cap U)$ is open in $V$ and is also open in $X$.

• But we have that:

(1)

\begin{align} \quad x \in p(C \cap U) \subseteq p(U) \end{align}

• So $x \in \mathrm{int} (p(U))$ which shows that $p(U)$ is open. Hence $p : \tilde{X} \to X$ is an open map.