It just needs to be constant on the fibers , right? Suppose that is constant on the fibers and define by taking where is some choice function (in simpler terms, if is constant on the fibers of then define a map from by taking each element to the unique value of taken on . This evidently satisfies and is thus continuous by the universal property of quotient maps. The converse is equally as easy since evidently if ( must be continuous by the universal property again, there is no need to assume it is) then for any one has that so that is constant on the fibers.