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I've seen that the universal covering of $S^1 \vee S^1 \vee S^2$ is infinitely many $S^2$'s chain together by line segments where the endpoints are identified. But when we are wedge two copies of $S^1$, I'm not sure how to make the two line segments distinguishable, yet still keeping the cover simply connected.

Any help will be appeciated!

nekodesu
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  • What do you mean by distinguishable? The projection map distinguishes the two segments. – Michael Burr May 03 '18 at 01:08
  • I meant the two line segments cannot just be merged as one line segments right? But there endpoints can be identified which will make the covering not simply connected. – nekodesu May 03 '18 at 01:28
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    Think about the universal cover of $S^1\vee S^1$, each of the four cardinal directions represents a path along an edge and they never duplicate. Just because points are identified in the covering map, they don't give loops in the cover. In your case, the universal cover of $S^1\vee S^1\vee S^2$ should be the infinite graph of the universal cover of $S^1\vee S^1$ with a sphere at every intersection. – Michael Burr May 03 '18 at 01:39
  • Ah! That makes sense! Do you mind transplant your comment to an answer so I can accept it? – nekodesu May 03 '18 at 01:46
  • I added it as community wiki – Michael Burr May 03 '18 at 12:29

1 Answers1

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Since the sphere is simply connected, the universal cover of $S^1\vee S^1\vee S^2$ is the universal cover of $S^1\vee S^1$ with a sphere at every intersection point. Since distinct paths in the universal cover of $S^1\vee S^1$ never intersect, the issue that the OP brings up about two paths coming together never occurs.

Michael Burr
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