Here are two problems that look trivial, but I could not prove.

i) If $p:E \to B$ and $j:B \to Z$ are covering maps, and $j$ is such that the preimages of points are finite sets, then the composite is a covering map. I suppose that for this, the neighborhood $U$ that will be eventually covered by the composite will be the same that is eventually covered by $j$, but I can´t prove that the preimage can be written as a disjoint union of open sets homeomorphic to $U$.

ii) For this I have no idea what to do, but if I prove that it´s injective, I'm done. Let $p:E \to B$ be a covering map, with $E$ path connected and $B$ simply connected; prove that $p$ is a homeomorphism.

  • $\begingroup$ I think you need extra conditions on $j$, otherwise $j \circ p$ might even fail to be surjective. $\endgroup$ – Zhen Lin Oct 31 '11 at 7:27
  • $\begingroup$ @Zhen Lin: By definition both $p$ and $j$ are surjections, so their composition must be as well. $\endgroup$ – Brian M. Scott Oct 31 '11 at 7:35
  • $\begingroup$ I didn’t change it when I edited the question, but I’m reasonably sure that eventually covered should be evenly covered. $\endgroup$ – Brian M. Scott Oct 31 '11 at 7:36
  • $\begingroup$ Ah, now that the question has been rephrased it is much clearer that $j$ is also a covering map. (I thought it was arbitrary with finite fibres.) $\endgroup$ – Zhen Lin Oct 31 '11 at 8:41

Lets call an open neighborhood $U$ of a point $y$ principal (wrt. a covering projection $p: X \to Y$), if it's pre image $p^{-1}(U)$ is a disjoint union of open sets, which are mapped homeomorphically onto $U$ by $p$.

By definition a covering projection is a surjection $p: X \to Y$, such that every point has a principal neighborhood. It is easy to see, that if $U$ is a principal neighborhood of a point $y$, then any open neighborhood $U'$ of $y$ with $U' \subset U$ is again principal.

i) Let $p: X \to Y$ and $q: Y \to Z$ be covering projections, where $q^{-1}(\{z\})$ is finite for every $z \in Z$. Let $z \in Z$ and $U$ a principal neighborhood of $z$. For every point $y \in q^{-1}(\{z\})$ choose a principal neighborhood $V_y$. We can assume that $V_y$ is a subset of the component of $q^{-1}(U)$ corresponding to $y$, possibly replacing $V_y$ with its intersection with that component.

Now let $$U' = \bigcap_{y \in q^{-1}(\{z\})}q(V_y),$$ then $U'$ is an open (being the intersection of finitely many open subsets) neighborhood of $z$. It should be easy to verify that $U'$ is principal.

ii) Let $e,e' \in E$ with $p(e) = p(e')$ and $\gamma: I \to E$ be a path from $e$ to $e'$. Now $p \circ \gamma$ is a closed path, and therefore nullhomotopic. Lifting such a homotopy shows that $e=e'$.


on the first one: Take $z\in Z$ and let $U$ be an evenly covered neighborhood of $z$. Then $j^{-1}(U)=\cup_{i\leq k} V_i$ (dj union) and $j^{-1}(z)= \{b_1,...b_k\}$ where $b_i \in V_i$. Now choose $b_i \in X_i \subset V_i$ where now the $X_i$'s are evenly covered by $p$. Then let $Y= \cap_{i\leq k} j(X_i)$. $Y$ is open since this is a finite intersection. Now $Y$ evenly covered by the composite.

on the second one: Suppose $p$ were not injective. let $p(x_1)=p(x_2)=y$ Then take a path $\gamma$ from $x_1$ to $x_2$. $p\gamma$ is then a loop at $y$ homotopic to the trivial loop. So when the trivial loop and $p\gamma$ lift to paths starting at $x_1$ the two lifts must end at the same point. But the lift of the trivial loop ends at $x_1$ and the lift of $p\gamma$ which is $\gamma$ ends at $x_2$.

  • $\begingroup$ Sorry )=!!! But i did not understand you did, in the second problem, using the lifting to conclude the injectivity :/? $\endgroup$ – Daniel Nov 1 '11 at 21:49

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