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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – freakish
    Dec 15, 2022 at 14:14
  • $\begingroup$ @freakish, The first homology group seems to be $\mathbb{Z}^7$ $\endgroup$ Dec 15, 2022 at 16:41
  • 1
    $\begingroup$ @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. $\endgroup$
    – Frank
    Dec 15, 2022 at 18:23
  • $\begingroup$ @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. $\endgroup$
    – freakish
    Dec 15, 2022 at 19:45
  • $\begingroup$ @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 $\endgroup$ Dec 15, 2022 at 19:46

1 Answer 1


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.

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

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