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Problem 1.3.25 in Hatcher's "Algebraic Topology" is concerned with the quotient (call it $Y$) of $X=\mathbb{R}^2 \setminus \{0\}$ under the $\mathbb{Z}$-action generated by $(x,y) \mapsto (2x,y/2)$. Some parts of the problem are obvious to me ($Y$ is not Hausdorff, $Y$ is the union of four cylinders $\cong S^1 \times \mathbb{R}$) but I can't find $\pi_1(Y)$, although it fits into the exact sequence: $$ 1\to\mathbb{Z}\to\pi_1(Y)\to\mathbb{Z}\to1 $$ from proposition 1.40(c) ($G\cong\pi_1(Z/G)/p_\ast (\pi_1(Z))$ for a cover $p: Z\to Z/G$ coming from a "covering space action" of $G$ on $Z$). Perhaps if I read the proofs more carefully I could unpackage this isomorphism, but I don't see how to find $\pi_1(Y)$ at the moment.

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I did this problem recently myself. It's tricky, but there are a few ways to proceed.

One way to attack the problem along the lines you have mentioned comes from that exact sequence. How many groups fit there? If the group is abelian, then it turns out there only one can fit there (use the fundamental theorem of finitely generated abelian groups, for example). So you should show it's abelian. How? You might find two generators, and show their order doesn't matter. In fact, the unique lifting property can give you an explicit commutator and show it's trivial.

Alternately, you can just find the two generators. One is in the covering space, and so very obvious. The other, though, isn't contained in the covering space, but comes from going between successive 'nodes,' or equivalent spots on the hyperbola, of the action (the endpoints get mapped to the same point, and so it maps to a loop). You might come up with some homomorphism, show it's injective and surjective, and well-defined (not in that order, I suppose).

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