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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

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It isn't important at all if your question is homework or not: the tag only serves for purposes of how deep a hint is given or not. People will still try to help you. –  DonAntonio Dec 30 '12 at 13:28
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But $\pi_{1} (S^{2})=0$,so how are you getting $\mathbb Z$? –  Shraddha Srivastava Dec 30 '12 at 13:35
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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. –  JSchlather Dec 30 '12 at 13:37
    
@ShraddhaSrivastava, I thought we were working with copies of $\,S^1\,$ instead... –  DonAntonio Dec 30 '12 at 13:39
    
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. –  Andrew Dec 30 '12 at 13:42

2 Answers 2

Unfortunately there is a flaw in this solution. It is in the statement "To see that $F$ is also continuous when $t=1$, it suffices to prove that for any neighborhood $U$ of the origin $O$ in $X$, $f^{−1}(U)$ contains all but finitely many intervals of $(a_i,b_i)$."

The problem is that if the homotopies are chosen as described above, but otherwise arbitrary, continuity can fail. Example: suppose that the $(a_i,b_i)$ cluster at some point $s_0$. Suppose the initial loop is trivial. Suppose that the homotopy "shrinking" the small loop on $[a_i,b_i]$ sweeps over the sphere $S_1$ before collapsing back down to $O$. (Perverse, but not excluded by the description. N.B.: always the same sphere.) Then for $t$ arbitrarily close to 1 and points $s$ arbitrarily close to $s_0$, we have $F(s,t)$ always at least some fixed distance (say 1/2) from $O$.

To fix this, we need to exert some control on the homotopies. If the small loop on $(a_i,b_i)$ does not have the ''opposite pole'' (the point furthest from $O$) in its range, then shrink the loop by sliding along the meridians from the opposite pole to $O$. This creates a distance decreasing property for these homotopies. If the opposite pole is in the range, then any homotopy will work.

Now we use the continuity of the initial loop. It follows, as in the previous poster's argument, that the $F(\cdot,t_k)$ is continuous for $t_k=1-(1/k)$. Also for $t_k$ sufficiently close to 1, $F(\cdot,t_k)$ is as close as you like to $O$. We now have a dichotomy for loops yet to be shrunk. Either the support of the loop is in a ''small'' sphere, or else the opposite pole is not in the range. Either way, the homotopies will keep us close to $O$.

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The only problem is to prove that any loop $f:I\rightarrow X$ that passes through the origin for infinitely many times is nullhomotopic. Here is the idea:

Let $X=\bigvee_n S_{1/n}^2$ and $O$ be the origin, then the subset $E:=f^{-1}(X\backslash\{O\})$ should be open in $I=[0,1]$. Therefore $E$ can be written as disjoint union of infinitely many open intervals , i.e. $E=\bigcup_{i=1}^{\infty}(a_i,b_i)$. Then we can define a sequence of homotopies $F_i:I\times [1-\frac{1}{i},1-\frac{1}{i+1}]\rightarrow X$, such that $F_i(\cdot,1-\frac{1}{i})=f_i$ and $F_i(\cdot,1-\frac{1}{i+1})=f_{i+1}$, where $f_i$ is the loop we get after having shrunk the first $i-1$ small loops $f((a_k,b_k))$ to $O$. Note that $f_1=f$. Finally we can construct a homotopy $F:I\times I\rightarrow X$ by gluing $F_i$ together, i.e. $F(\cdot,t)=F_i(\cdot,t)$ if $t\in[1-\frac{1}{i},1-\frac{1}{i+1}]$ and $F(\cdot,t)=O$ if $t=1$. By pasting lemma, $F$ is continuous whenever $t<1$. To see that $F$ is also continuous when $t=1$, it suffices to prove that for any neighborhood $U$ of the origin $O$ in $X$, $f^{-1}(U)$ contains all but finitely many intervals of $(a_i,b_i)$. This is true because all the intervals $(a_i,b_i)$ form a disjoint open cover of $f^{-1}(X\backslash U)$, which is compact in $I$.

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