Homology of $\mathbb{R}^2$ under the equivalence $x \sim 2x$ I was computing some examples of homologies of quotient spaces and I thought of the following.  Does anyone know how to compute the homology groups of $\mathbb R^2/\sim$, where $\sim$ is the equivalence $x \sim 2x$.
Thank you
 A: Unless I've made a mistake, I think all the homology groups are trivial.  In fact, I think the space is contractible.  More generally, it seems as though the proof shows that if $Y$ is any topological space and $Z = Y\cup\{p\}$ where the only open set around $p$ is $Z$ itself, then $Z$ is contractible.
I will use $X$ to denote the quotient.
As Stefan H. says in the comments, $X$ is homeomorphic to $T^2\cup\{p\}$ where $T^2$ has the usual topology and the only open set containing $p$ is the whole space.  I'll freely identity $X$ with $T^2\cup\{p\}$.
Consider the map $F:X\times I\rightarrow X$ with $$F(x,t) = \begin{cases} x & t < 1\\ p & t = 1\end{cases}$$ so $F$, if continuous, is a homotopy between the identity map (at time $0$) and the constant map (at time $1$).
I claim that $F$ is continuous.  If $U\subseteq X$ contains $p$, then $U = X$ and clearly $F^{-1}(U) = X\times [0,1]$, so is open.  Hence, we may assume $U$ doesn't contain $p$, so $U$ is just a regular open set on $T^2$.  I claim that $$F^{-1}(U) = U\times [0,1)$$ so is open in $X\times [0,1].$
Proof of $\subseteq$: To see that $F^{-1}(U)\subseteq U\times [0,1)$, pick any $(x,t)\in F^{-1}(U)$.  This means that $F(x,t)\in U\subseteq T^2$.  This implies $t < 1$.  But when $t < 1$, $F(x,t) = x$, we must have $x\in U$.  Thus, $(x,t)\in U\times [0,1)$.
Proof of $\supseteq$: To see that $U\times [0,1)\subseteq F^{-1}(U)$, pick any $(x,t)\in U\times [0,1)$.  Since $t<1$, $F(x,t) = x\in U$, so $(x,t)\in F^{-1}(U)$.
