# Showing whether a certain subset is evenly covered by a map

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 $p$?

• not necessarily, you should start by an evenly covered neighbourhood of $z$, see my answer. – Henno Brandsma Apr 14 '18 at 11:25
• 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. – 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.

• 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? – fcm Apr 14 '18 at 11:45
• @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. – Henno Brandsma Apr 14 '18 at 11:48
• 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 $X$? – fcm Apr 14 '18 at 11:53
• @fcm that's clear. They're just all the sets $q^{-1}[V_i] \cap O_j$ where $i=1\ldots,n$ and $j \in I_i$. – Henno Brandsma Apr 14 '18 at 11:57
• 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. – Henno Brandsma Apr 14 '18 at 12:03