# A question from Hatcher's Algebraic Topology

I'm working on the following exercise from Hatcher's algebraic topology:

Let $$\phi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$$ be the linear transformation $$\phi(x,y) = (2x,y/2)$$. This generates an action of $$\mathbb{Z}$$ on $$X = \mathbb{R}^2\setminus \big\{0\big\}$$. Show that this action is a covering space action and compute $$\pi_1(X/\mathbb{Z})$$. Show that the orbit space $$X/\mathbb{Z}$$ is non-Hausdorff, and describe how it is a union of four subspaces homeomorphic to $$S^1\times \mathbb{R}$$, coming from the complementary components of the $$x$$-axis and the $$y$$-axis.

I have been able to prove everything but the fundamental group. Here is what I have so far on this:

I let $$p$$ be the covering map $$p:X \rightarrow X/\mathbb{Z}$$. $$X$$ is path connected and locally path connected. Given this, we know from Proposition 1.40 in Hatcher that $$\mathbb{Z}\cong \pi_1(X/\mathbb{Z})/p_*(\pi_1(X))$$. Since $$p$$ is a covering map, $$p_*$$ is injective and, therefore, $$p_*(\pi_1(X)) \simeq \pi_1(X) \simeq \mathbb{Z}$$. Here we used the fact that $$X \cong S^1$$. Hence, we conclude that $$\pi_1(X/\mathbb{Z}) \simeq \mathbb{Z} \rtimes \mathbb{Z}$$. This semi-direct product is isomorphic to $$\mathbb{Z} \times \mathbb{Z}$$ or a non-abelian group. We claim that $$\pi_1(X/\mathbb{Z}) \cong \mathbb{Z} \times \mathbb{Z}$$. It is sufficient to show that $$\pi_1(X/\mathbb{Z})$$ is abelian, which can be shown via its generators. The generators of $$\pi_1(X/\mathbb{Z})$$ are given by paths in $$X/\mathbb{Z}$$ as follows: one is the image of the loop $$\gamma$$ in $$X$$ which generates $$\pi_1(X,x_0)$$, and the other is the image of a path $$\alpha$$ in $$X$$ joining the basepoint $$x_0$$ to $$\phi(x_0)$$.

From this point it seems like we need to show that $$\phi(\gamma) \alpha \overline{\gamma}\overline{\alpha}$$ is homotopic to the constant map. This, along with the injectivity of $$p_*$$, would give us the abelian property we're after. However, I'm not sure how to do this. Would it be sufficient to say that $$\phi(\gamma) \alpha \overline{\gamma}\overline{\alpha}$$ is not homotopic to $$\gamma$$ since $$\gamma$$ is the only non-trivial path up to homotopy?

I have one more idea. Is it true that $$X/\mathbb{Z}$$ is a topological group? This would tell us that $$\pi_1(X/\mathbb{Z})$$ is abelian.

• The covering group acts by trivial automorphisms on $\pi_1(X)$, hence, the fundamental group of the quotient is the direct product. Commented Feb 17, 2022 at 19:08
• @MoisheKohan I'm not familiar with this type of argument. Could you please expand? Commented Feb 18, 2022 at 0:01

Consider a generator $$\gamma\in \pi_1(X)$$ as shown in blue:

Here $$\gamma$$ is represented by the unit circle, taking $$(1,0)$$ as base point.

We will next pick $$\alpha\in \pi_1(X/\mathbb{Z})$$ which maps to $$\phi$$ the generator of $$\mathbb{Z}$$, in the short exact sequence:$$1\to \pi_1(X)\to\pi_1(X/\mathbb{Z})\to \mathbb{Z}(=\langle\phi\rangle)\to 1$$

The lift of $$\alpha$$ to $$X$$ will be a path from the basepoint $$(1,0)$$ to $$\phi(1,0)=(2,0)$$. We may thus represent some such $$\alpha$$ by the following path in $$X$$, indicated in red:

Now in $$\pi_1(X/\mathbb{Z})$$ we have the composition: $$\alpha\gamma\alpha^{-1}$$. The lift of this to $$X$$ will be the path $$\alpha$$ followed by $$\phi$$ applied to the loop $$\gamma$$ followed by the path $$\alpha$$ in reverse:

By considering the winding number about $$(0,0)$$ we see that: $$\alpha\gamma\alpha^{-1}=\gamma,$$ as required.

• This is a nice explanation. Instead of appealing to the winding number, could we use the fact that the loop you have pictured in your third diagram is homotopic to $\gamma$? Commented Feb 18, 2022 at 0:55
• Yes absolutely.
– tkf
Commented Feb 18, 2022 at 0:56
• Oh nice! The result would follow from this fact as well. Thank you! Commented Feb 18, 2022 at 0:57