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My guess is that the fundamental group should be $\mathbb{Z}$, but other than my intuition, I can't seem to find a way to show this.

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- Thread starter mathsq
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- #1

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My guess is that the fundamental group should be $\mathbb{Z}$, but other than my intuition, I can't seem to find a way to show this.

- #2

mathwonk

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look at the universal cover, an infinite string of spheres like beads on a long string.

- #3

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- #4

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If you wanted to use Van Kampen, couldn't you take the "left of the sphere with the string" along with the "right of the sphere with the string"? Each piece will be homotopic to a circle and their overlap will be homotopic to a figure 8. It seems that the amalgamation will identify the two generators and you indeed get the integers, which is the "obvious" answer.

- #5

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Oh, I was thinking of your bead incorrectly, I see why it is the universal cover, never mind!

- #6

mathwonk

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well you are right i ignored the question. but the cover makes it obvious the group is Z. (i think?)

- #7

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I've just thought again about your answer. Do I have this right- in your bead, I imagine your sphere (and interval) being cut in halves along the equator, and then the interval is made to point out in the opposite direction. You then glue these all back together in the obvious way. It will have trivial fundamental group since it is homotopic to a wedge of spheres and the covering map sends a point to where it came from in the above construction.

The set of deck transformations is Z, because it just matters where you send some particular sphere (I was worried about "flipping" everything being a self-homeomorphism- but thinking about it, this won't be a deck transformation).

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