(June ’04 # A2) Let \(X\) be the unit sphere in \(\mathbb R^3\) and define an equivalence relation on \(X\) by \[ (x,y,z)\sim(x',y',z')\Longleftrightarrow z=z'.\] Let \(Z=X/\sim\) be the quotient space under this equivalence relation, with the quotient topology. Show that \(Z\) is homeomorphic to the interval \([-1,1]\).
Solution
Let \(q:X\to Z=X/\!\sim\) be the quotient map and define \[ h:X\to[-1,1],\qquad h(x,y,z)=z. \] The map \(h\) is continuous and constant exactly on the equivalence classes, so by the universal property of the quotient there is a unique continuous map \[ \bar h:Z\to[-1,1] \] with \(h=\bar h\circ q\).
The map \(\bar h\) is bijective: every \(t\in[-1,1]\) occurs as the height of a point of the sphere, and two points of \(X\) have the same image under \(h\) exactly when they are equivalent.
The sphere \(X=S^2\) is compact, so its quotient \(Z\) is compact. The interval \([-1,1]\) is Hausdorff. Therefore the continuous bijection \[ \bar h:Z\longrightarrow[-1,1] \] from a compact space to a Hausdorff space is a homeomorphism.