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Let $X$ be the subspace of $\mathbb R^3$ which is union of the spheres of radius $1/n$ and centered at $(1/n,0,0)$. Then $X$ is simply connected.

I had thought for it in this way to attach $2$-cells to a single point, namely the origin, but then I realized the space I will get is a wedge sum of spheres and not the space given in the question.

Please help to figure out the fundamental group of the space in question.

Not a homework problem

  • 1
    But $\pi_{1} (S^{2})=0$,so how are you getting $\mathbb Z$?2012-12-30
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    The only thing you need to worry about is a loop that is on infinitely many of the spheres. But it seems to me you ought to be able to define a homotopy by being clever enough.2012-12-30
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    @ShraddhaSrivastava, I thought we were working with copies of $\,S^1\,$ instead...2012-12-30
  • 0
    After some homotopy, we can assume that the loop you are looking at avoid at least one point on each sphere. Deleting one point from each sphere makes the space contractible.2012-12-30
  • 0
    There's still some nuance, because a wedge of contractible spaces need not be contractible.2012-12-30
  • 0
    yes,but I think if the spaces are locally contractible as well ,then the wedge of contractible spaces will be contractible.2012-12-30
  • 0
    Well,It seems to me that if the loop doest stay in a single sphere then the parts of the loop in any sphere must be a loop(because of continuity).Then each part of this loop is homotopic to the constant map of the basepoint ,so the whole loop will be homotopic to the constant map at the basepoint.2012-12-30
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    @JacobSchlather: I think this the hint you were giving to me(please read my previous comment).2012-12-30
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    But still it doest seem to me very precisely :( .2012-12-30

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