Given a closed irreducible set C, x is a generic point for C if the closure of x yields C.
If every closed irreducible set has a generic point, the space is generic.
Map S into T by taking any x to its closure, which is an irreducible closed set. Thus points of S map to points of T.
The preimage of a closed set in T is the collection of points whose closures all live in a closed set C in S. Now x satisfies this criterion iff x lives in C, hence the preimage is closed, and the map is continuous.
Now start with a closed set C in S. The closure of every point in C remains in C, hence the image in T is closed, and the map is bicontinuous. Note, to be technically correct: the map from S onto its image in T, using the subspace topology inherited from T, is bicontinuous. After all, there may be points of T that don't map back to S.
Assume x and y have the same closure, and map to the same point in T. Every open set containing x contains y, and every open set containing y contains x. Turn this around and express it this way. If S is half hausdorff, or T0, then the map from S into its generic space T is an embedding. In fact it is a bicontinuous embedding, which means S is a subspace of T. Expanding S into its generic space sort of fills in the gaps. The topology of S, within T, has not changed.
If S is hausdorff, each point is already closed and irreducible, and these are the only irreducible sets, hence S and T are homeomorphic.
We already know closed sets map forward to closed sets; we only need show the map is onto to complete the correspondence. Yet every closed set D in T comes from a closed set C in S by construction. Points in C have closures that lie in D, and the image of S faithfully captures every closed set in T. The closed sets of S and T correspond one to one.
Since containment is preserved in both directions, chains of closed sets map to chains of closed sets. Therefore S is noetherian iff T is noetherian.
Let C be an irreducible closed set in S. We already know the image of C is closed in T; is it irreducible? If not, then two proper closed sets in T cover the image, and these pull back to proper closed sets in S that cover C, which is a contradiction. In general, the continuous image of an irreducible closed set is an irreducible closed set.
Since the map is bicontinuous, this runs in both directions. A closed set is irreducible iff its image or preimage is irreducible. Apply this to chains, and the dimension of S equals the dimension of T.
Irreducible correspondence seems odd, since the irreducible closed sets of S also correspond to the points of T. How can they correspond to points of T and irreducible closed sets in T at the same time?
To illustrate, let C be an irreducible closed set in S, with three other irreducible closed sets inside it. Over in T, C corresponds to a point y, and the image of C, under our bicontinuous map, is a set of four points with y at the top.
Each point of T acts as a group leader for an irreducible closed set in T, which corresponds to all the irreducible closed sets in the preimage of y.
Continuing the above, the closure of any other point in D would produce a proper subset of D. Thus y is the generic point for D.
Given two points y and z in T, they will have nothing to do with each other, or one will be the group leader for the other. In either case, one point can be placed in an irreducible closed set that omits the other. Take the complement, and one point is in an open set that omits the other; hence T is T0.
Suppose we transform T into U. Each irreducible closed set D in T has a generic point y whose closure gives D. The map from U back to T can be reversed, hence the map from T to U is onto. Since T is T0, the map is also an embedding. therefore T and U are homeomorphic spaces.
Transforming a generic T0 space changes nothing, hence there is no need to transform S twice. This may remind you of the completion of a metric space, which is complete. There is no need to complete a metric space twice.
Let C be the category of topological spaces and continuous maps, and let G be the category of generic spaces and continuous maps. The transformation, as described above, is a functor between the two categories.
Clearly it takes objects to objects, so let's look at morphisms. Let f be a continuous map from S1 into S2. We already said f caries irreducible closed sets to irreducible closed sets, hence it induces a well defined map from T1 into T2. The preimage of a closed set in T2 is associated with a closed set in S1, and a closed set in T1, hence the induced function downstairs is continuous, and is a valid morphism in the category of generic spaces.
Finally, f(S1) into S2, combined with g(S2) into S3, defines h(S1) into S3, and the same is true of the induced functions downstairs. The diagram commutes, and we have a valid functor.