Suppose $Z\to X,\ Z\to Y$ are covering spaces.

Does there exist a space $W$ s.t. $X\to W$ and $Y\to W$ are covering spaces?


Exercise 1.3.11 in page 80 from Allen Hatcher's book Algebraic topology gives an answer.

Exercise 1.3.11

Construct finite graphs $X_1$ and $X_2$ having a common finite-sheeted covering space $\widetilde X_1 = \widetilde X_2$, but such that there is no space having both $X_1$ and $X_2$ as covering spaces.


2 Answers 2


Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).

Interesting side note: The covering spaces $Z\to X$ and $Z\to Y$ are defined by free actions of ${\mathbb Z}_2$ on $Z$. These generate a ${\mathbb Z}_2\times{\mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.

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    $\begingroup$ wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction $\endgroup$ Feb 11, 2015 at 23:20
  • $\begingroup$ I originally had a similar idea but in the end gave up on it because I concluded that the graph with 2 vertices , one loop at each vertex, and a straight edge between the vertices, is a covering space of the graph with 2 vertices and 3 straight edges by seeing that it’s homeomorphic to the graph with 4 vertices and the edges {1,2} twice, {2,3} once and {3,4} twice $\endgroup$ Aug 21, 2022 at 10:11

No, it is not true.

The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 \times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 \# F_1$ (that symbol $\#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $\mathbb{R}^2 \to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $\mathbb{H}^2 \to F_2$. And, of course, each of $\mathbb{R}^2$, $\mathbb{H}^2$ is homeomorphic to $D^2$.

But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 \mapsto W$ and $F_2 \mapsto W$ are each of finite degree. The fundamental group $\pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $\pi_1F_1$, $\pi_1F_2$ respectively. Since $\pi_1F_1 = \mathbb{Z} \oplus \mathbb{Z}$ is abelian, $A_1 \cap A_2$ is a finite index abelian subgroup of $A_2$. But $\pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.

  • $\begingroup$ $\pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me? $\endgroup$ Feb 1, 2015 at 18:33
  • $\begingroup$ @AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $\mathbb{H}^2$ fixes some line in $\mathbb{H}^2$, but the action of the deck transformation group on $\mathbb{H}^2$ has no finite index subgroup fixing a line. $\endgroup$
    – Lee Mosher
    Feb 1, 2015 at 21:06
  • $\begingroup$ here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping $\endgroup$ Feb 6, 2015 at 14:07
  • $\begingroup$ @AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces). $\endgroup$
    – Lee Mosher
    Feb 6, 2015 at 20:51
  • $\begingroup$ thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions $\endgroup$ Feb 6, 2015 at 21:56

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