I'm trying to calculate the homology of $\mathbb{R}^3$ removing $3$ coordinate axes.
Let this space be $X$. I was able to show that $\pi_1(X) = \mathbb{Z} \ast \mathbb{Z}$ (via reducing the problem to calculating $\pi_1$ of $\mathbb{S}^2 - \{3 \text{ points }\}$. But how about its higher homology groups?
Instinct told me that they are all zero but I'm having a hard time justifying it.
In particular, I tried to use Mayer-Vietoris sequences on the $3$ complement of the hyperplanes of $\mathbb{R}^3$ so that their intersection gives the desired space but I haven't seen Mayer-Vietoris sequences applied on triple of subspaces.
Any comments will be appreciated!
The space deformation retracts onto sphere with 6 points removed. It's a non-compact 2-manifold, therefore $H_n(X) = 0$ for $n \geq 3$, and also $H_2(X) = 0$, because sphere with 6 points removed is not compact.