Let $q:X\to Y$ and $r:Y \to Z$ be covering maps;let $p=r\circ q$. Show that if $r^{-1}(z)$ is finite for each $z\in Z$, then $p$ is a covering space.

I'm confused that $p=r\circ q$ is obvious surjective and each open set $U\subset Z$, $r^{-1}(U)=\cup_\alpha V_\alpha\subset Y$,and each $q^{-1}(V_\alpha)=\cup_\beta W_\beta \subset Z$ and each $W_\beta$ is homeomorphic to $U$ by $p$. So how do we use the condition $r^{-1}(z)$ is finite for each $z\in Z$? Thank you!

  • 3
    $\begingroup$ I think you have the definition of "covering map" wrong. The fact that $r$ is a covering map does not imply that for every open $U \subset Z$, $r^{-1}(u)$ is a disjoint union of open sets mapped homeomorphically onto $U$ by $r$. $\endgroup$ – Chris Eagle May 14 '12 at 10:43
  • $\begingroup$ To illustrate what @Chris said: consider $U = Z = \mathbb{S}^1$ the unit circle and $Y = \mathbb{R}$ the real line in your argument. $\endgroup$ – Willie Wong May 14 '12 at 11:02
  • $\begingroup$ @ChrisEagle Thank you for your correction on my mistake.It should be for any point $z\in Z$,there is a neighbourhood $z\in U\subset Z$ which is covered by $p$ evently:) $\endgroup$ – Jiangnan Yu May 14 '12 at 13:52

First let me give you a counterexample (modified from Allan Hatcher's Algebraic Topology, p79 exercise 6) when $r^{-1}(z)$ is not finite.

The following are all subsets of $\mathbb{R}^2$ with the induced topology.

Defn Let $C_{n,m}^\pm$ denote the circle of radius $1/n$ centered at $(\pm 1 \pm 1/n, 2m)$ for $n,m \in \mathbb{N}$.

Defn Let $\gamma^+_m$ denote a smooth curve contained in $[-1,1]\times[2m, 2m+1)$ and similarly $\gamma^-_m$ in $[-1,1]\times (2m-1,2m]$ such that $\gamma^+_m \cap \gamma^-_m = \{ (1,2m), (-1,2m)\}$.

Defn Let $Z = \cup_{n\in \mathbb{N}} C^+_{n,1}$. Let $Y = \cup_{n,m\in\mathbb{N}} C^+_{n,m}$ and let $$X = \left(\cup_{n,m\in\mathbb{N}}( C^+_{n,m}\cup C^-_{n,m} )\right) \cup\left(\cup_{m\in\mathbb{N}}(\gamma^+_m\cup\gamma^-_m)\right) $$

We let $r:Y\to Z$ be the obvious map taking $C^+_{n,m}\to C^+_{n,1}$. This is clearly a covering map.

Let $q:X\to Y$ be the following:

  • $q: C^\pm_{n,m} \to C^+_{n,m}$ in the obvious way if $n < m$.
  • $q: C^\pm_{n,m} \to C^+_{n+1,m}$ if $n \geq m$.
  • Noticing that $\gamma^+\cup \gamma^-$ is a circle topologically, let $q: \gamma^+_m\cup \gamma^-_m \to C^+_{m,m}$ be a double cover, so that $(\pm 1,2m) \mapsto (1,2m)$.

$q$ clearly gives a double cover of $Y$ by $X$.

Now, let $B$ be an arbitrary open neighborhood of $(1,1)$ in $Z$. Since the topology is induced from that of $\mathbb{R}^2$, $B$ contains $C_{N,1}^+$ for $N \geq N_0$ sufficiently large. But for $M > N_0$, the connected component of $(1,2M)$ in the inverse image $p^{-1}(C_{M,1}^+)$ contains both $\gamma^\pm_M$, and hence the same connected component of $p^{-1}(B)$ double covers $B$ under the map $p$, and hence cannot be a homeomorphism. This shows that $p$ is not a covering map.

Now, how do we use the information that $r^{-1}(z)$ is finite?

Sketch of proof:

  • For any $z$ in $Z$, let $U_z$ be a neighborhood of $z$ such that it is evenly covered by $r$. By the finiteness assumption we can write $r^{-1}U_z$ as a finite disjoint union $\sqcup_{k = 1}^{n_z} V_{k,z}$, and let $y_k \in V_{k,z}$ be the corresponding preimage of $z$. Note that $V_{k,z}$ are homeomorphic to $U_z$.

  • Each $y_k$ has an open neighborhood $V'_k$, which we can assume to lie inside $V_{k,z}$, such that $q^{-1}V'_k$ can be written as disjoint unions of open sets each homeomorphic to $V'_k$.

  • Now consider the set $U'_z \subset Z$ defined as $$ U'_z = \cap_{k= 1}^{n_z} r(V'_k) $$ I claim that $U'_z$ is open (this is where finiteness of $n_z$ is used!!) and its preimage $p^{-1}(U'_z)$ can be written as a disjoint union of open subsets of $X$ each of which homeomorphic to $U'_z$.

Going back to the counterexample, you see that roughly speaking each of the $V'_m$ has to be a neighborhood of $(1,2m)$ that does not contain the entire circle $C_{m,m}^+$. Hence necessarily $\cap_{m} r(V'_m)$ cannot be open, and the above procedure cannot work.

  • $\begingroup$ Thank you Willie Wong,now I know the proof.I will learn your counter example carefully tomorrow:)) $\endgroup$ – Jiangnan Yu May 14 '12 at 15:38

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