Let f be a function from S onto Q, taking each point to its equivalence class. By construction, f is continuous. But it need not be bicontinuous. Let S consist of 3 points a b and c, where {a} is an open set. Let r equate a and b, and leave c alone. Thus Q consists of two points, ab and c. Only ∅ and Q are open in Q. The open set {a} maps to the set ab in Q, which is not open.
A quotient space often compresses a portion of S down to a point. If T is a subspace of S, let r declare all the points of T as equivalent, whence they are represented by a single point p. Every point outside of T remains in its own equivalence class. Let's look at an example.
Let S be the unit disk in 2 dimensions and let T be its boundary. Thus T is the unit circle. Taking advantage of the third dimension, pull the unit circle out of plane, turning the disk into a bowl. Then draw the rim of the bowl together into a single point p. The result is a sphere. In general, the quotient of an n dimensional ball by its boundary gives an n dimensional sphere.
The above is pretty intuitive; here is a more technical argument. Note that S-T is the open ball, and T is mapped onto the point p, separate from the open ball. An open set O containing T has a closed complement C that misses T. In fact C, closed and bounded, is compact. Thus O is the complement of a compact set within our open ball. Therefore the quotient space is the compactification of the open ball, where p is the point at infinity. The result is a sphere of the same dimension.
Conversely, let f(S) be a continuous function that is constant across equivalence classes. This induces a well defined function through g. The preimage of an open set O, under f, yields an open set U in S. Pull O back through g to get a set V in Q. Since U includes all the points in the equivalence classes of V, V is open by definition. Therefore g is continuous.