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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 3-cells to single point namely origin but then I realize the space I will get is wedge sum of spheres 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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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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