A topological category which is a subcategory of Set

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In category theory it is possible to freely chose what are objects and what are morphisms as long as the definitions fulfills the axioms for a category.

Now, I'm trying to construct a natural category of topological spaces, equivalent with Top, but being a subcategory of Set (which Top isn't). My approach is letting the objects be the Kuratowski closure operators, usually denoted $\text{cl}$.

I have some problems with the morphisms and the approach however (see Does this category explain continuity?).

My question is simply:

would a function $\hat f:\text{cl}\to\text{cl}'$ defined by $\hat f(S,T)=(f(S),f(T))$ make sense for continuous functions $f:X\to X'$?

Are there more to check except that $\text{Im}\hat f\subseteq \text{cl}'$, for these functions to be adequate morphisms?

It's understand that $(S,T)\in \text{cl}\subseteq \mathcal P(X)\times \mathcal P(X)$.

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Since you consider $cl\subset \mathcal{P}(X)\times\mathcal{P}(X)$, you cannot plug any pair $(S,T)$ into $\hat f:cl\to cl'$, but only those $(S,T)\in cl$, and these are exactly the pairs $(S,\overline{S})$. Hence the map $$\hat f:cl\to cl':(S,\overline{S})\mapsto (f(S),f(\bar{S}))$$ is well defined iff $(f(S),f(\bar{S}))\in cl'$ for all $S\in X$. I.e. $(f(S),f(\overline{S}))=(f(S),\overline{f(S)})$, i.e. $f(\overline{S})=\overline{f(S)}$ for all $S\subset X$. And this is not true for general continuous $f$.


I didn't check it, but I think you'll want to consider a contravariant functor, since generally pre-images are better behaved wrt the topology then images.