Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 4F (Metric and Topological Spaces)

The equation implies $|z|\ge 1$. Its intersections with $\{z>0\}$ and $\{z<0\}$ are disjoint, nonempty subsets open in the subspace topology, and together cover the entire surface. They therefore form a separation, proving the surface is disconnected. More precisely its two sheets are $z=\pm \sqrt{1+x^2+y^2}$; each is homeomorphic to the connected plane by projection, so these are exactly its two connected components.

Factorization gives $y=x$ or $y=-x$. Each point can be joined to the origin by a straight segment lying on its own line. Two such segments join any pair of points, so the union is path-connected and hence connected. Alternatively it is the union of two connected lines with a common point.

The map $(x,y)\mapsto (x,y,x^2+y^2)$ is a continuous surjection from the path-connected plane onto this surface. It is a homeomorphism because coordinate projection is its continuous inverse. Thus the paraboloid is path-connected and connected. Explicitly, lifting any straight planar segment by this map produces a path on the surface.