Universal cover of $(S^1\times S^2)\vee S^1$ How does the universal cover $X$ of $(S^1\times S^2)\vee S^1$ looks like? I know that the universal cover $Y$ of $S^1\vee S^1$ is as in the picture. First I thought that  $X$ would be the space obtained from $Y$ by attachin $S^2$ at  each vertex, but I soon found that it would be the universal cover of $S^1\vee S^1\vee S^2$. Then I thought that since $\Bbb R \times S^2$ is the universal cover of $S^1\times S^2$, $X$ would be obtained from $\Bbb R$ by attaching $\Bbb R \times S^2$ at each integer point, but I think this is also wrong, because the preimage of the wedge point will be not only the integers, but also infinitely many in each $\Bbb R\times S^2$. Any ideas?

 A: Choose a point $p \in S^1$. Consider the subset embedding
$$(S^1 \times \{p\}) \vee S^1 \hookrightarrow (S^1 \times S^2) \vee S^1
$$
On the left we have a wedge of two circles, with fundamental group free of rank 2. The induced homomorphism on fundamental groups that is induced by this embedding is an isomorphism between two rank 2 free groups:
$$\pi_1((S^1 \times \{p\}) \vee S^1) \approx \pi_1(S^1 \times S^2) \vee S^1)
$$
It follows that universal covering space of $(S^1 \times \{p\}) \vee S^1$, which is depicted in your post, embeds (equivariantly) into the universal covering space of $(S^1 \times S^2) \vee S^1$.
There are now two key questions to ask yourself. 
First, how does the universal cover of $S^1 \times \{p\}$ embed into the universal cover of $S^1 \times S^2$?
Answer: by including $\mathbb R \times \{p\}$ into $\mathbb R \times S^2$.
Second, how does one visualize this inclusion in the figure in your post?
Answer: Think of each vertical line in that figure as a copy of $\mathbb R \times \{p\}$. And now, to each of those lines, attach a separate copy of $\mathbb R \times S^2$.
