Proving that $S^1 \vee S^1$'s covering space is universal

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In Hatcher, they give the following example of a simply connected covering space of $S^1 \vee S^1$ enter image description here. I am having trouble convincing myself that it is simply connected. I suspect it is contractible. However, I am getting lost with the infinity of of steps it would take to contract it.

Might anyone have any hints? Thank you in advance,

Raphaël Fua

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Don't worry about showing that the cover is contractible: just show that every loop is homotopic to the constant loop.

For that, let $E$ be the supposed universal cover in your post and let $\alpha$ be a loop in $E$. Since the image of $\alpha$ must be compact, its image must live in a finite sub-tree $T$ of $E$. The point is that $\alpha$ can't be wild and run off to all these strange infinite corners: it's contained in a finite sub-tree. Now, Hatcher proves that finite trees are contractible, so use that to give a homotopy of $\alpha$ down to the constant map. That same homotopy in $T$ is a homotopy within $E$ by inclusions, so you're done.