I was trying to understand what is an example of a topological space that has an uncountable fundamental group. I was reading this answer but I don't understand what $L_q \equiv 0$ and $L_q \equiv 1$ mean.
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2$\Bbb R^2\setminus\Bbb Q^2$ is a simple example of such a space, the Hawaiian earring is another one. $L_q$ in the linked answer is a square with the given four vertices – Alessandro Codenotti Jun 23 '20 at 17:59
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Yes I understand this part, but what does "$\equiv 1$" mean? – roi_saumon Jun 23 '20 at 20:35
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1$\pi_1$ of the punctured plane is $\Bbb Z$, the smaller loop is homotopic to the trivial one in the punctured plane, so it corresponds to $0$ in the fundamental group of the punctured plane, the other one turns around the puncture once, so it corresponds to $1$ in the fundamental group of the punctured plane – Alessandro Codenotti Jun 23 '20 at 21:01
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@AlessandroCodenotti, thank you! – roi_saumon Jun 24 '20 at 23:03
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$L_q \equiv 0$ means that the loop $L_p$ is null-homotopic in $\mathbb{R}^2-(r,r)$. $L_q \equiv 1$ is certainly an infelicitous wording. However, it means that the loop is the canonical generator of $\pi_1(\mathbb{R}^2-(r,r)) \approx \mathbb Z$.
If the loops $L_{q_1}$ and $L_{q_2}$ would be homotopic in $\mathbb{R}^2- \mathbb Q^2$, then also in the bigger space $\mathbb{R}^2-(r,r)$. This is not true in the given situation.
Paul Frost
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Oh, I see. But if $\forall q_1 < q_2$, $L_{q_1}\neq L_{q_2}$ how do you deduce that the set of non-equivalent $L_q$'s is uncountable? Isn't there then $\mid \mathbb{Q} \mid$ of these $L_q$'s? – roi_saumon Jun 24 '20 at 23:16
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