Past exam of the topology course of the University of Oxford / 2014 / 1 / c / ii / Solution

For a nonempty open subset $U$ of the quotient, $p^{-1}(U)$ is nonempty and open in the cofinite space. Thereforeis finite. Surjectivity implies that $(X/\mathcal{R})\setminus U$ is finite as well: it is the image of that finite preimage. Hence every quotient-open set is cofinite or empty, proving the desired inclusion.

Equality is not necessary. Give $X=\mathbb{Z}$ its cofinite topology and use two equivalence classes, the even and odd integers. Neither singleton of the quotient is open, since its preimage and its complement are both infinite. The quotient is the two-point indiscrete topology, whereas the cofinite topology on a finite two-point set is discrete.This is the quotient topology of a cofinite space phenomenon, not a failure of continuity of the projection.