I think that my question is easily answerable. The question is: What is a nice subcategory of topological spaces where the subobjects are subspaces. I would like the category of compactly generated Haussdorf spaces to be such a category, since this category is convienent in many other ways.


Some backround definitions:


1. A subspace (of X) of a topoogical space is a subset of X with the induced topology.
2. The notion of a subobject (of X) is a categorical notion. It is an equivalence class of monomorphisms of the form, $m:Y\rightarrow X$. This $m$ is equivalent to another subobject, $m':Y'\rightarrow X$ if their exists an isomorphism, $i:Y\rightarrow Y'$ making the evident triangle commute.



A subcategory of top where subspaces and subobjects coincide?
Well, I'd say that compact Hausdorff spaces are at least a *nice* category of spaces where subspaces are equivalent to subobjects. One reason it is nice is that it is a category of algebras for a monad on $Set$, with all the nice properties that entails, for example Barr exactness. It is of course *not* "convenient" in the technical sense of Steenrod and (earlier) Ronnie Brown: it isn't cartesian closed. 


Whatever the candidate is, you absolutely have to worry about whether the subcategory allows the same set to have comparable topologies living as spaces in the same subcategory. For example, in the classically convenient category of compactly generated spaces, you can have for example the real numbers with the standard topology as one object, and the real numbers with the discrete topology as another, and a continuous function $id: \mathbb{R}\_{disc} \to \mathbb{R}$. This is certainly a monomomorphism, but certainly not a subspace. So this convenient category doesn't work. (On the other hand, there cannot be two comparable compact Hausdorff topologies on the same set.) 


Off-hand I'm having trouble thinking of an example of a full subcategory of $Top$ which is both cartesian closed (and complete and cocomplete) and has this property you are looking for, but perhaps I haven't thought hard enough. 


