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I am stuck in problem 53.4 from Munkres: 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 map.

This is my work so far:

First, $ p $ is both continuous and surjective, because the composition of continuous/surjective maps is again continuous/surjective (note that both $ r $ and $ q $ are continuous and surjective, since they are covering maps).

Let $ z \in Z $. Since $ r $ is a covering map, there exists an open neighborhood $ U $ of $ z $ that is evenly covered by $ r $. For all $ y \in r^{-1}(z) $, there exists an open neighborhood $ V_{y} $ of $ y $ that is evenly covered by $ q $, since $ q $ is a covering map. We now define \begin{align*} U' := \bigcap_{y \in r^{-1}(z)} r(V_{y}) \end{align*} We claim that $ U' $ is both open and evenly covered by $ p $, making $ p $ a covering map (since for all elements of $ Z $, a corresponding $ U' $ can be constructed).

$ U' $ is open in $ Z $ because it is a finite intersection of finitely many open subsets $ r(V_{y}) $, as $ r^{-1}(z) $ is finite. Each $ r(V_{y}) $ is open in $ Z $ because the covering map $ r $ is open. Since each $ V_{y} $ is open, it follows that $ r(V_{y}) $ is open. We now show that $ U' $ is evenly covered my $ p $, that is, $ p^{-1}(U') $ is a union of disjoint open sets in $ X $.

I am stuck in showing whether $p^{-1}(U')$ can be expressed as a disjoint union of open sets in $X$, each of which is homeomorphic to $U'$. Is it true that my $U'$ is evenly covered by $p$?

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  • $\begingroup$ not necessarily, you should start by an evenly covered neighbourhood of $z$, see my answer. $\endgroup$ – Henno Brandsma Apr 14 '18 at 11:25
  • $\begingroup$ Your $U'$ should be intersected with $U$ as well. You need to keep the properties of $U$. Then you get the same set I did. $\endgroup$ – Henno Brandsma Apr 14 '18 at 12:06
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Continuity and surjectivity are clear, as you state. The work is in the evenly covered part of the definition:

Take a fixed $z \in Z$. This has the points $y_1, \ldots y_n$ (for some finite $n$ by assumption) as preimages in $Y$ and an evenly covered open neighbourhood $U_z$ such that for every $i=1,\ldots,n$ we have open $V_i$ pairwise disjoint and containing $y_i$ such that $r|_{V_i}$ is a homeomorphism between $V_i$ and $U_z$.

Now each $y_i$ likewise has an evenly covered (for $q$) neighbourhood $W_{y_i}$. We can now replace $U_z$ by the better neighbourhood $U'_z:= \cap_{i=1}^n r[W_{y_i} \cap V_i]$ (which is open as a finite intersection of open sets). $U'_z$ is still evenly covered by $r$ but now the $y_i$ disjoint neighbourhoods are $W_{y_i} \cap V_i$ (and these in turn are still evenly covered by $q$).

You can now easily check that $U'_z$ is the required evenly covered neighbourhood for $p$. See this answer for this same idea with slightly more details, which I found later.

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  • $\begingroup$ Is it then true that the preimage under $q$ of the disjoint neighbourhoods $W_{y_{i}} \cap V_{i}$ is a disjoint union of subsets of $X$? If yes, why? $\endgroup$ – fcm Apr 14 '18 at 11:45
  • $\begingroup$ @fcm Yes, because this holds for $W_{y_i}$. Say, $q^{-1}[W_{y_i}]$ is a disjoint union of $O_j, j \in J$ then $q^{-1}[W_{y_i} \cap V_i]$ is a disjoint union of $O_j \cap q^{-1}[V_i]$ etc. $\endgroup$ – Henno Brandsma Apr 14 '18 at 11:48
  • $\begingroup$ I apologize for my wording, but I meant to ask if the preimage of the collection of disjoint sets $W_{y_{i}}∩V_{i}$ is a disjoint union of subsets of $X$. In other words, why is $ (q^{-1} \circ r^{-1})(U_{z}') $ a disjoint union of subsets of $ X $? $\endgroup$ – fcm Apr 14 '18 at 11:53
  • $\begingroup$ @fcm that's clear. They're just all the sets $q^{-1}[V_i] \cap O_j$ where $i=1\ldots,n$ and $j \in I_i$. $\endgroup$ – Henno Brandsma Apr 14 '18 at 11:57
  • $\begingroup$ where $O_j, j \in I_i$ is the disjoint decomposition of $q^{-1}[W_{y_i}]$. plus a composition of homeomorphisms is a homeomorpism etc. $\endgroup$ – Henno Brandsma Apr 14 '18 at 12:03

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