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 $p$?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.