TITLE: Calculate the fundamental group of a subspace of $\mathbb R^3$, $S^3$ and $\mathbb P^3(\mathbb R)$
QUESTION [4 upvotes]: Consider the topological space $$X=\{(x,y,z)\mid (x^2+y^2)(x^2+z^2)(y^2+z^2)(x^2+y^2-1)=0\}$$ in $\mathbb R^3$. Let $Y$ be its closure in $S^3$ (seen as the Alexandrov compactification of $\mathbb R^3$) and $Z$ the closure of $X$ in $\mathbb P^3(\mathbb R)$ (obtainable from $Y\subseteq S^3$). Calculate the fundamental group of $X,Y,Z$ and of any of the connected components of their complements.

$X$ is a the union of cilinder with the coordinate planes. $\pi_1(X)$ is easily found by retractions. One can retract $X$ to a "letter X inscribed in a circle" which has fundamental group $\mathbb Z* \mathbb Z* \mathbb Z* \mathbb Z$. I tried to apply Van Kampen's theorem to find $\pi_1(Y)$, but I have some problems in describing the intersection of the two open sets, wich should be $X$ itself and an open neighbour of the additive "Alexandrov" point.
Also I think that $\pi_1(Z)$ should be derived from $\pi_1(Y)$ in some way.
Also general techniques and references are very welcome.

REPLY [3 votes]: First of all notice that $\pi_1(X)=1$. In fact (as you said) $X$ is a union of an "infinite" cylinder with the coordinate planes. This space retracts on the plane $z=0$ via the retraction 
$$R_t(x,y,z)= ( x,  y , tz) \ \ \ \ \ \ \ t \in [0,1] \ . $$
Since $\pi_1(X)=1$ we also have that $\pi_1(Y)=1$ provided that a neighbourhood of infinity is simply connected (actually this is a simplified formulation of Van Kampen's Theorem). On the other hand, a neighbourhood of $\infty \in Y$ is homeomorphic to the union of the three  coordinate planes in $\mathbb{R}^3$ with the $z$-axis collapsed to one point, that is contractible. Thus $\pi_1(Y)=1$.
Finally, the projective closure of $X$ in $\mathbb{P}^3$ is given by the projective hypersurface
$$Z : \ \ xyz(x^2+y^2-t^2)=0 \ .$$
We decompose it as the union $Z=A \cup B$, where $A$ denotes the part of $Z$ lying in the chart $t\not=0$ (this is actually $X$) and $B= Z- [0,0,0,1]$. Now $\pi_1(A)=\pi_1(X)=1$, and $\pi_1(B)=1$ since $B$ retracts on $p=[0,0,1,0]$ via the retraction 
$$R_\epsilon[x,y,z,t]= [\epsilon x, \epsilon y , z, \epsilon t] \ \ \ \ \ \ \ \epsilon \in [0,1] \ . $$
Thus because of Van Kampen's Theorem (again) we can conclude that also $\pi_1(Z)$ vanish.
{"set_name": "stack_exchange", "score": 4, "question_id": 1874844}