Homology of Union of Planes-Point

I was trying to compute the singular homology of the following space

Consider $$X= \{(x,y,z) \in \mathbb{R}^3: xyz=0 \}$$ and consider let $$A=X \setminus (0,0,0)$$.

What are the singular homology groups of $$A$$. How do I compute it?

I thought about using Mayer-Vietoris Sequence but cannot find convenient open covers. Or I am not sure what this space is homotopy equivalent to. It seems to me that this is homotopy equivalent to wedge of $$6$$ circles. But I am not sure.

Bonus:What is the fundamental group of this space?

• It is homotopy equivalent to $\{(x,y,z)\in S^2\ |\ xyz=0\}$ through normalization and linear homotopy. And this space consists of 3 circles, perpendicular to each other on $S^2$. This has $8$ holes, and can be shown to be homotopy equivalent to wedge sum of $8$ circles, right? Thus $H_1$ is $\mathbb{Z}^8$ (and vanishes in higher dimensions) while $\pi_1$ is the free group $F(a_1,\ldots,a_8)$, and also vanishes in higher dimensions. Commented Dec 15, 2022 at 14:14
• @freakish, The first homology group seems to be $\mathbb{Z}^7$ Commented Dec 15, 2022 at 16:41
• @freakish's idea is correct, but the set $\{(x, y, z) \in S^2 \mid xyz = 0\}$ isn't homotopy equivalent to $8$ circles. But it is homotopy equivalent to $S^2$ minus $8$ points, which is homotopy equivalent to $\mathbb{R}^2$ minus $7$ points, which is then homotopy equivalent to the wedge of $7$ circles. So that gives you $\mathbb{Z}^7$ for both. Commented Dec 15, 2022 at 18:23
• @Frank you mean "homeomorphic to $\mathbb{R}^2$ minus 7 points" (homotopy equivalence does not have to preserve point removal). Yes, other than that you are correct. My mistake. Commented Dec 15, 2022 at 19:45
• @Frank, I still couldn't get it. By normalisation I understood that its homotopy equivalent to union of circles in $xy$-plane, $yz$-plane, $zx$-plane. But after that I cannot understand Commented Dec 15, 2022 at 19:46

So first of all your $$A$$ deformation retracts onto $$B=\{(x,y,z)\in S^2\ |\ xyz=0\}$$, where $$S^2$$ denotes the $$2$$-dimensional sphere. The deformation retraction is as follows:

$$(v,t)\mapsto (1-t)\cdot v+t\cdot \frac{v}{\lVert v\rVert}$$

which is just normalization connected linearly to the identity. And so $$A$$ is homotopy equivalent to $$B$$.

So what is $$B$$ exactly? These are $$3$$ circles on $$S^2$$, perpendicular to each other. And those circles make $$8$$ "holes" (empty spaces) on the sphere. Choose $$8$$ points inside those holes, say $$\{v_1,\ldots, v_8\}$$. Then $$C=S^2\backslash\{v_1,\ldots,v_8\}$$ deformation retracts onto $$B$$. That's because a disk without point deformation retracts onto its boundary, and every hole is homeomorphic to a disc. We just need to glue $$8$$ such deformations together. And so $$B$$ is homotopy equivalent to $$C$$.

Next apply the stereographic projection to $$C$$, based on say $$v_1$$. This gives us that $$C$$ is homeomorphic to $$D=\mathbb{R}^2\backslash\{w_1,\ldots,w_7\}$$ for some vectors $$w_1,\ldots, w_7\in\mathbb{R}^2$$. Note that we have $$7$$ vectors here: we lose one due to stereographic projection.

And finally $$D$$ is homotopy equivalent to the wedge sum of $$7$$ circles, see this: $R^2$ with $n$ points removed is a bouquet of $n$ circles

Putting it all together (and applying some well known results on fundamental group and homology group) we get:

$$\pi_1(A)=F(a_1,\ldots,a_7)$$ $$H_1(A)=\mathbb{Z}^7$$

where $$F(a_1,\ldots,a_7)$$ denotes the free group on $$7$$ letters. Both homotopy and homology groups vanish in higher dimensions.

• thanks a lot but I had understood it after going through the comments. Commented Dec 16, 2022 at 12:12