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Let $A = \{(\frac{4}{2n+1},0) \mid n=1, 2, 3, \dotsc\}$. Show that there is a surjective group homomorphism from $\pi_{1}(\mathbb{R}^{2}\setminus A)$ to $\mathbb{Z}^{\infty}$ (the product of infinitely many copies $\mathbb{Z}$). In particular, $\pi_{1}(\mathbb{R}^{2} \setminus A)$ contains uncountable many elements.

It can be easily prove that $\mathbb{R}^{2} \setminus A$ deformation retracts to the Hawaiian earring. But my question is, why does the elementary group of $\mathbb{R}^{2} \setminus A$ can correspond to uncountable many elements, given that $A$ is countable? I think we somewhat need Van-Kampen Theorem here, but how? Many thanks for your suggestions or answers!

Jendrik Stelzner
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    $A$ is a countably infinite set. For any subset of it, there is a loop that encloses just the points in that subset. Count such elements. – Arkady Dec 07 '22 at 08:59
  • Since you know that $\mathbb R^2 \setminus A$ deformation retracts to the Hawaiian earring, your question is answered by https://math.stackexchange.com/q/3925195. – Paul Frost Dec 07 '22 at 12:06

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