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Let $n>1$. I need to show that the space $X=\mathbb{R}^2-\{x_1,x_2,\dots,x_n\}$ does not have the structure of topological group.

This is an exercise about the Van Kampen theorem. Certainly, we should prove it by contradiction, but I do not know how to get this contradiction.

HSN
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noname1014
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  • Any connected topological group has abelian fundamental group, while $X$ does not. Indeed, the fundamental group of $X$ is the (non-abelian) free group of rank $n > 1$. – Crostul Mar 25 '16 at 10:43
  • Related: http://math.stackexchange.com/questions/5309/topological-group-multiplying-two-loops-is-homotopic-to-linking-these-paths – Najib Idrissi Mar 25 '16 at 14:08

1 Answers1

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Fundamental group of a topological group is abelian always. And it's easy to show the given space has non-abelian fundamental group.

Proof of the statement in bold:

Let $a$ and $b$ be two loops in a topological group $(G,\bullet )$ starting at the identity element $e$. We need to show $ a\ast b \simeq b\ast a$, where "$\ast$" is the fundamental group operation.

Now for each $t,s\in [0,1]$, define

$F_t(s)=a(st)\ast(a(t)\bullet b(s))\ast \bar a(st)$

Clearly {$F_t$} gives the homotopy between $b$ and $a\ast b \ast \bar a$.

( We can assume the topological group path connected )

Paladin
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  • thnaks, but how to prove given space has non-abelian fundamental group?no matter what topology structure it has. – noname1014 Mar 25 '16 at 10:54
  • Show that the space has deformation retract to wedge of n-circles.... and by vancampen you can compute the fundamental group of wedge of circles which is free group on n generators and for n>1 the group is non-abelian – Paladin Mar 25 '16 at 10:56
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    @noname1014 Note that the topological structure is given, in your problem; what's not fixed at the outset is the algebraic structure. – egreg Mar 25 '16 at 10:58
  • @noname1014 egreg makes an important point. If we were not inheriting the topological structure of $\mathbb R^2$, then the set in question can simply be bijected to $\mathbb R^2$ (with no points removed), or to any topological group of continuum cardinality, and then we can pull back a topological group structure along such a bijection. But here we implicitly inherit the topology from $\mathbb R^2$. – zibadawa timmy Mar 25 '16 at 14:03