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

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?

Reference:

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.

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.

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.