Universal covering space via fiber product

Question

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.

Answer 1

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.

(I think of this space you described like a man walking his dog through the streets (the 1-dimensional parts) and finding occasional dog parks (the spheres) where the dog can run free).