Let $X$ be the space obtained from an annulus by identifying each point $(x,y)$ on the inner circle with the point $(-2x,-2y)$ on the outer circle. Compute the homology groups $H_\bullet(X)$.
The question was taken from here: https://math.illinois.edu/system/files/inline-files/comp525_may19.pdf
Here is my attempt:
Consider the following CW complex $\emptyset = Y^{-1} \subset Y^0 \subset Y^1 \subset Y^2$:
We obtain $X$ by identifying $\alpha \sim -\beta$ and naturally $x\sim y$. The chain complex is then: $$ \begin{align*} \dots \to 0 \to \mathbb{Z} \to \mathbb{Z}^2 \to \mathbb{Z} \to 0 \end{align*} $$ Since $\partial_2 A = \beta -\gamma -\beta + \gamma = 0 $, $H_2(X) = \ker \partial_2 \cong \mathbb{Z}$.
Next, $\partial_1 \equiv 0$ and since $\text{im}\partial_2=0$, $H_1(X) \cong \mathbb{Z}^2$.
Finally $H_0(X) = \mathbb{Z}$ is trivial.
Is this computation right? I'm a bit confused because at first I didn't think we had $\alpha \sim -\beta$ and this gave me a different result.