# Universal covering space via fiber product

Given 3 topological spaces $X,Y,Z$ and 2 functions $f:X \rightarrow Z$, $g:Y \rightarrow Z$, I define the fiber product between $X$ and $Y$ over $Z$ by: $$X\times_Z Y := \{(x,y) \in X \times Y | f(x) = g(y) \}.$$ My problem is to find the universal covering space of $X :=\mathbb{P}^2 \vee \mathbb{S}^1$. Here I call $\mathbb{P}^2$ the real projective plane, while $\mathbb{S}^1$ is the circle, and by $X \vee Y$ I mean the wedge sum between $X$ and $Y$. My idea was the following: I call $Y$ the 2-sheeted covering space of $X$ given by the wedge sum of $\mathbb{S}^2$ (the 2-sphere, which is the universal covering space of the proj plane) and 2 copies of the circle, each of those attached to the lifted basepoint; and I call $Z$ the wedge sum of 2 copies of the circle. Now, it is clear that there exists a onto map $f:Y \twoheadrightarrow Z$, which sends the whole sphere into the basepoint of $Z$. On the other hand, the universal covering space $\tilde{Z}$ of $Z$ is well known (cfr. Hatcher, p. 77, it is a fractal tree). So, If I consider the fiber product $$Y \times_Z \tilde{Z},$$ defined as above, this is a covering space of $Y$. If this covering space of $Y$ were simply connected, then it would be the universal one, and it would then be the universal covering space of $X$. I finally turn to my question: how can I visualize the space $Y \times_Z \tilde{Z}$ graphically? It should be $\tilde{Z}$ with a sphere (instead of a point) placed at each segment intersection, but I cannot convince myself about this. If this were correct, than it would be clear that $Y \times_Z \tilde{Z}$ turns out to be simply connected.

Imagine traveling through your fiber product. You are represented by a pair of dots traveling in both spaces; the dots move over time, but they must map to the same point of $Z$ at all times.
While the $X$-dot is not in a sphere, it's position is uniquely determined by the $Y$-dot's position. When the $Y$-dot is at a vertex of the tree, the $X$ dot is free to move around the sphere. So the effect of it all is that you out a sphere at each vertex of the tree, separating the edges coming into the vertex into two groups.