Given $X$ the set of all points $[x_0,x_1,x_2] \in {{\mathbb P}_{\mathbb C} ^2}$ st $x_0x_1=0$:
a) find a compact, path-connected topological space which is homotopically equivalent to $X$;
b)compute the 1st homotopy group of $X$.
Here is my proof:
$X=A \cup B$, where $A$ is the set of points in $X$ with $x_0=0$, and similarly $B$ has $x_1=0$. Now, if it is true that both $A$ and $B$ are omeomorphic to ${\mathbb P}_{\mathbb C} ^1$ and so to the sphere $S^2$, and given that $A\cap B$ consists of a single point, I could say that the space $X$ is homeomorphic to two tangent spheres, which are contractible to the intersection point, which is obviously both compact and path-connected, and has the trivial group consisting only of the identity as its fundamental group.
I am not sure of the correctness of the proof because of the assumption that $A$ and $B$ are homeomorphic to two tangent spheres; moreover I have little general knowledge about complex projective spaces, so I fear I could have made some other errors.