about covering space and compact base space

Question

Let $p: \tilde{X} \rightarrow X$ be a covering space with $p^{-1}(x)$ finite and nonempty for all $x \in X .$ Show that $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.

Proof :

(1) If $\tilde{X}$ is compact. since $p$ is continuous and $X$ is the continuous image of a compact set hence is compact.

(2) If $\tilde{X}$ is Hausdorff, show that $X$ is Hausdorff. For $x, y \in X .$ Suppose there is an evenly covered open set $U$ containing both $x, y .$ Then $p^{-1}(U)$ contains $p^{-1}(x)$ and $p^{-1}(y) .$ Take one of the preimage of this open set, say, $p^{-1}(U)$ since $\tilde{X}$ is Hausdorff, there exist open sets $U_{x}$ and $U_{y}$ in $p^{-1}(U)$ containing $p^{-1}(x)$ and $p^{-1}(y)$ and $U_{x} \cap U_{y}=\phi .$ i.e. $p^{-1}(x) \in U_{x}$ and $p^{-1}(y) \in U_{y} .$ So we have $x \in p\left(U_{x}\right)$ and $y \in P\left(U_{y}\right)$ and $p\left(U_{x}\right) \cap p\left(U_{y}\right)=\phi$ .

Now, suppose there is no such open set exists. Let $V_{x}$ be an open set containing $x$ but not $y$ and $V_{y}$ be an open set containing $y$ but not $x .$ Then $p^{-1}\left(V_{x}\right)=\bigcup \tilde{U}_{\tilde{x}}$ and $p^{-1}\left(V_{y}\right)=\bigcup \tilde{V}_{\tilde{y}}$ $\operatorname{Fix} \tilde{x} \in p^{-1}\left(V_{x}\right) \subset U_{\alpha}$ for some $\alpha .$ since $\tilde{X}$ is Hausdorff. Pick an open set $\tilde{V}_{x} \subset p^{-1}\left(V_{x}\right)$ containing $\tilde{x}$ and $\tilde{V}_{\tilde{y}} \subset p^{-1}\left(V_{y}\right)$ containing $\tilde{y}$ for each $\tilde{y} \in p^{-1}(y)$ such that $\tilde{V}_{\tilde{x}} \cap \tilde{V}_{\tilde{y}}=\phi .$ Then $\left(\left(\cap \tilde{V}_{x}\right) \cap \tilde{V}_{y}=\phi \text { for all } y . \text { since } p\left(\left(\cap \tilde{V}_{\tilde{x}}\right) \cap \tilde{V}_{\tilde{y}}\right)=p\left(\cap \tilde{V}_{\tilde{x}}\right) \cap p\left(\tilde{V}_{\tilde{y}}\right) \text { and } x \in p\left(\cap \tilde{V}_{\tilde{x}}\right) \text { and } y \in p\left(\hat{V}_{\tilde{y}}\right)\right.$ Claim that $p\left(\cap \tilde{V}_{\tilde{x}}\right) \cap p\left(\tilde{V}_{\tilde{y}}\right)=\phi$ Suppose $p\left(\cap \tilde{V}_{\tilde{x}}\right) \cap p\left(\tilde{V}_{\tilde{y}}\right)=\{z\} .$ Then $p^{-1}(z) \subset \cap \tilde{V}_{\tilde{x}}$ and $p^{-1}(z) \subset \cap \tilde{V}_{\tilde{y}}$ for all $y .$ This is a contradiction to the fact that $\left(\cap \tilde{V}_{\tilde{x}}\right) \cap \tilde{V}_{\tilde{y}}=\phi .$ Hence $X$ is Hausdorff. (3) If $X$ is Hausdorff, prove that $\tilde{X}$ is Hausdorff. For $x, y \in \tilde{X},$ if $p(x) \neq p(y) .$ since $X$ is Hausdorff, there is an open set $U_{x}$ containing $p(x)$ and open set $U_{y}$ containing $p(y)$ such that $U_{x} \cap U_{y}=\phi .$ So their preimages $p^{-1}\left(U_{x}\right)$ and $p^{-1}\left(U_{y}\right)$ are also disjoint. If not, suppose $p^{-1}\left(U_{x}\right) \cap p^{-1}\left(U_{y}\right)=\{z\} .$ Then $p(z) \in U_{x}$ and $p(z) \in U_{y}$ which is a contradiction. Now, suppose $p(x)=p(y)$ in $X,$ take the open set $U_{x}$ containing $p(x) .$ Then $x \in P^{-1}\left(U_{\alpha}\right)$ and $y \in p^{-1}\left(U_{\alpha}\right) .$ Hence $x$ and $y$ are either the same point or on different sheet. Therefore, $\tilde{X}$ is Hausdorff.

(4) If $X$ is compact, how we can prove $\tilde{X}$ is compact ?.

i think we have :

theorem : Let $X$ be a Hausdorff locally compact in $x \in X .$ then for each open nbd $U$ of $x$ there exists an open nbd $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subset U$.

so we can write $ X= \bar{V_1} \cup \bar{V_2} \cup \bar{V_3} ... \cup \bar{V_n} $ then

$ \tilde{X} = p^{-1}(X)= p^{-1}(\bar{V_1} \cup \bar{V_2} \cup \bar{V_3} ... \cup \bar{V_n} ) =p^{-1}(\bar{V_1} ) \cup p^{-1}(\bar{V_2}) \cup p^{-1}(\bar{V_3}) ... \cup p^{-1}(\bar{V_n} ) $ now is every $p^{-1}(\bar{V_n})$ compact ? is my proof true ?

Answer 1

Your proof is essentially correct. For (4) observe that by compactness of $X$ you can find finitely many open $V_i \subset X$ such that $\overline V_i$ is compact and contained in an open $U_i \subset X$ which is evenly covered by $p$. This means that $p^{-1}(U_i)$ is the union of finitely many (recall that fibers are finite) pairwise disjoint open $W_i^k \subset \tilde X$ such that all $p_i^k : W_i^k \stackrel{p}{\to} U_i$ are homeomorphisms. Then $p^{-1}(\overline V_i) = \bigcup_k (p_i^k)^{-1}(\overline V_i)$. The latter is a finite union of compact sets, hence $p^{-1}(\overline V_i)$ is compact. We conclude that $\tilde X = \bigcup_i p^{-1}(\overline V_i)$ is compact.

Answer 2

A covering map $p: X \to Y$ with finite fibres is a closed map: if $y \in Y$, and $p^{-1}[\{y\}]= \{x_1,\ldots x_n\} \subseteq O$, then let $O_i, i =1, \ldots, n$ be an open neighbourhood of $x_i$ obeying $O_i \subseteq O$, such that $p: O_i \to p[O_i]$ is a homeomorphism between open sets (we only use the local homeomorphism property of $p$), then $U:=\bigcap_{i=1}^n p[O_i]$ is an open neighbourhood of $y$ such that $p^{-1}[U] \subseteq O$.

Next use that this means that $p$ is a proper map (fibres are compact and $p$ is closed) (e.g. my recent answer here) an in particular $Y$ is compact implies $X=p^{-1}[Y]$ is compact. Or in your context: $X$ compact implies $\tilde{X}=p^{-1}[X]$ is compact.

So we can reduce it to a more general fact, which is useful to know in itself.

That $\tilde{X}$ compact implies $X$ compact is a consequence of surjectivity and continuity of $p$, as you state.

And if $x \neq y$ in $X$ and $\tilde{X}$ Hausdorff, then $F_x=p^{-1}[\{x\}]$ and $F_y=p^{-1}[\{y\}]$ are disjoint finite (compact would also work) sets so that in a Hausdorff space $\tilde{X}$ we have open disjoint sets $F_x \subseteq O_x, F_y \subseteq O_y$ and then the argument from the first paragraph (for showing $p$ closed) gives us that we have open $x \in O'_x$ with $p^{-1}[O'_x] \subseteq O_x$ and ditto for $O'_y \ni y$ and then it's easy to see that $O'_x \cap O'_y = \emptyset$ (as $p$ is onto) showing $X$ to be Hausdorff. So here too we only need closedness of $p$ and compact fibres really.

And if $u \neq v$ in $\tilde{X}$ then either $x:=p(u) =p(v)$ and we use the evenly covered neighbourhood $U$ of $x$ ($u$ and $v$ are in different open parts of $p^{-1}[U]$ etc.), or $p(u) \neq p(v)$ and they have disjoint open neighbourhoods in $X$ and we pull them back (using continuity of $p$) to conclude again that $u$ and $v$ have disjoint open neighbourhoods in $\tilde{X}$.