A covering space of a Hausdorff space is Hausdorff

Question

Let $p:Y\to X$ be a covering space. If $X$ is Hausdorff, so is $Y$.

Hello,

I have a question to this task. I want to show that $Y$ is a Hausdorff space. Hence for $y_1, y_2\in Y$ with $y_1\neq y_2$ have disjoint neighbourhoods.

I want to prove this:

Let $p:Y\to X$ be a covering space. $y_1, y_2\in Y$, with $y_1\neq y_2$. $x_1:=p(y_1)$ and $x_2:=p(y_2)$. Then exist open sets $x_1\in U_1\subseteq X$ and $x_2\in U_2\subseteq X$, such that $p^{-1}(U_i)$ with $i=1,2$ is the disjoint union of open sets.

Hence there are open sets $y_1\in V_1\subseteq p^{-1}(U_1)$ and $y_2\in V_2\subseteq p^{-1}(U_2)$.

Claim: $V_1\cap V_2=\emptyset$.

I do not think, that this works... Is this approach any good? Do you have tips on how I can do better? Thanks in advance.

Answer 1

Hint:

First case $x_1\neq x_2$, since $X$ is separated, there exists $U_1,U_2$ open neighborhoods of respectively $x_1$ and $x_2$ such that $U_1\cap U_2$ is empty. $p^{-1}(U_1)\cap p^{-1}(U_2)$ is empty and $y_1\in p^{-1}(U_1), y_2\in p^{-1}(U_2)$.

Second case $x_1=x_2=x$, since $p$ is a covering, there exists a neighborhoo $U$ of $x$ such that for every $y\in p^{-1}(x)$, there exists a neighborhood $V_y$ such that the restriction of $p$ to $V_y$ is an homeomorphism $p_{\mid V_y}:V_y\rightarrow U$, and $V_{y_1}\cap V_{y_2}$ is empty by definition.

Answer 2

You need to break your proof up into two parts.

Suppose $x_1 \neq x_2$. We can choose $U_1$ and $U_2$ to be disjoint, and so $p^{-1}(U_1)$ and $p^{-1}(U_2)$ will be disjoint (if they had a point $z$ in common, $p(z) \in U_1\cap U_2 = \emptyset$ which is impossible). Now you can take $V_1 = p^{-1}(U)$ and $V_2 = p^{-1}(U_2)$.

Suppose $x_1 = x_2$; call this point $x$. Let $U$ be an evenly covered open neighbourhood of $x \in X$, then $p^{-1}(U) \cong \bigsqcup_{\alpha} U_{\alpha}$ where $p|_{U_{\alpha}} : U_{\alpha} \to U$ is a homeomorphism. Suppose $y_1 \in U_{\beta_1}$ and $y_2 \in U_{\beta_2}$, then it follows that $\beta_1 \neq \beta_2$. To see this, suppose $\beta_1 = \beta_2$; call this value $\beta$. As $p|_{U_{\beta}}$ is a homeomorphism and $p|_{U_{\beta}}(y_1) = p|_{U_{\beta}}(y_2)$, we have $y_1 = y_2$; this is a contradiction as $y_1$ and $y_2$ were assumed to be distinct. As $\beta_1 \neq \beta_2$, we can take $V_1 = U_{\beta_1}$ and $V_2 = U_{\beta_2}$.