I tried to solve an exercise of algebraic topology. Let $X$ be $(\mathbb{R}^3-\{(0,0,0)\})/\sim$, where $x\sim y$ means $x=y$ or $x=-y$. The question is finding homology groups of (Alexandroff) one-point compactification of $X$.
I think X is homotopy equivalence to the real projective plane $RP^2$. But $RP^2$ is compact, so I think one-point compactification of X is also $RP^2$. Then one-point compactification is meaningless. Am I wrong something?
EDIT 1 : One-point compactification $Z$ of $X$ is strictly same to the Thom space of the trivial line bundle over $\mathbb{R}P^2$. So we have: \begin{equation} H_i(Z;\mathbb{Z})\cong H_{i-1}(\mathbb{R}P^2;\mathbb{Z})\ (i \neq0), H_0(Z;\mathbb{Z})\cong \mathbb{Z}. \end{equation}
EDIT 2 : We can calculate the homology group by using an elementary Mayer-Vietoris argument without using any Thom isomorphism as follows. Let $M$ be a compact space and $X:=M\times\mathbb{R}$ (in the case of your question, $M=\mathbb{R}P^2$) . Now we want to calculate the homology group of one-point compactification $Z=X\cup\{\infty\}$. We have an open covering $\{U+,U-\}$ where $U_{+}:=M\times(-1,\infty)\cup\{\infty\}$ and $U_{-}:=M\times(-\infty,1)\cup\{\infty\}$. Note that $U_{\pm}$ are contractible and $U_+\cap U_-$ is homotopy equivalent to $M\sqcup\{\infty\}$. Using Mayer-Vietoris exact sequence, we have $\tilde H_i(Z;\mathbb{Z})\cong\tilde H_{i-1}(M\sqcup\{\infty\};\mathbb{Z})$.