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.


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.

  • $\begingroup$ Thank you very much. Is it necessary to prove that in both cases the intersection of the open covering sets is path connected? $\endgroup$ – W. Rether Jul 30 '16 at 11:54
  • $\begingroup$ Yes, good remark. You are right: in the computation of $\pi_1(Y)$ and $\pi_1(Z)$ in order to use Van Kampen's Theorem (even in the simplified version I'm using) you should check that the intersections are connected. $\endgroup$ – Antonio Alfieri Jul 30 '16 at 15:49
  • $\begingroup$ Actually, I did a mistake when I described the neighbourhood of infinity in the computation of $\pi_1(Y)$. I did an edit, and corrected the mistake. $\endgroup$ – Antonio Alfieri Jul 30 '16 at 15:51

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