The Gelfand duality for pro-$C^*$-algebras

Question

The Gelfand duality says that $$X\to C(X)$$ is a contravariant equivalence between the category of compact Hausdorff spaces and continuous maps and the category of commutative unital $C^*$-algebras and continuous $*$-homomorphisms. A well known generalization of $C^*$-algebras are pro-$C^*$-algebras. Pro-$C^*$-algebras are topological $*$-algebras that are cofiltered limits of $C^*$-algebras (in the category of topological $*$-algebras). See for instance this paper by Phillips. If $X$ is a weakly Hausdorff compactly generated space, it is not hard to see that $C(X)$, the continuous functions from $X$ to $\mathbb{C}$, is a pro-$C^*$-algebra. Indeed, it is the cofiltered limits of $C(K)$, as $K$ ranges over all compact Hausdorff subspaces of $X$. My question is:

Is the functor $$X\to C(X)$$ a contravariant equivalence between the category of weakly Hausdorff compactly generated spaces and continuous maps and the category of commutative unital pro-$C^*$-algebras and continuous $*$-homomorphisms?

Few remarks:

1. In the paper by Phillips mentioned above, Theorem 2.7, it is shown that the functor above is a contravariant equivalence between the category of completely Hausdorff quasitopological spaces and the category of commutative unital pro-$C^*$-algebras. It is also shown, in Example 2.11, that when restricted to the full subcategory of completely Hausdorff compactly generated spaces, this functor is not essentially surjective. But this still doesn't answer my question, because maybe on weakly Hausdorff compactly generated spaces this functor is essentially surjective.

2. Apparently, the reason Phillips insists on working with completely Hausdorff (quasi)topological spaces is that the way he proves this result is by constructing an inverse equivalence which assigns to any commutative unital pro-$C^*$-algebra its spectrum. The way he defines the spectrum, it can be shown that if $A$ is a commutative unital pro-C*-algebra, the spectrum of $A$ is completely Hausdorff. But, again, this doesn't answer my question, because maybe one can define an inverse equivalence in a different way, or maybe one can just show that this functor is fully-faithful and essentially surjective.

3. The main reason I am asking is that the category of weakly Hausdorff compactly generated spaces is very important in algebraic topology and homotopy theory, and such a result would imply that pro-$C^*$-algebras are exactly the non-commutative version of it.

Remark: I used the terminology of the paper of Phillips that I referred to. Note that the category of pro-$C^*$-algebras is not the pro-category of the category of $C^*$-algebras. In particular, its objects are topological $*$-algebras and not cofiltered diagrams of $C^*$-algebras. Some authors call pro-$C^*$-algebras locally $C^*$-algebras. It can be shown that the pro-category of the category of $C^*$-algebras contains the category of pro-$C^*$-algebras, as a full coreflective subcategory.

Edit: Due to Simon's answer the functor $X\to C(X)$ defined above is clearly not faithful. But the following question remains: Is this functor full and essentially surjective. If so then it would induce a contravariant equivalence of categories between $\overline{CGWH}$ and the category of commutative unital pro-$C^*$-algebras, where $\overline{CGWH}$ is the category of weakly Hausdorff compactly generated spaces and equivalence classes of continuous maps, where two continuous map $X\rightrightarrows Y$ are called equivalent if the induced maps $C(Y)\rightrightarrows C(X)$ are the same.

Answer 1

The answer is No. Rougly, because it is not a good idea to look at continuous $\mathbb{C}$ valued function on a space which is not completely Haussdorff as completely haussdorf is exactly the hypothesis that says "your space can be understood by looing at function over it"... I really don't think you can obtain something different from Philips result by considering the exact same functor.

More precisely: Take any example of a space $X$ which weakly Hausdorff CG, but non completely Hausdorff.

then there is going to be a pair of points $x,y \in X$ such that for any continuous function $f:X \rightarrow \mathbb{C}$. $f(x) = f(y)$

Hence the two functions :$x,y: \{ *\} \rightrightarrows X$ are going to have the same image by your functor which is hence not faithfull, and hence not an equivalence of category.

Edit : This answer the edited version of the question.

If you start from a weakly Haudorf CG space $X$, then you can consider the equivalence relation on $X$ defined by $x \sim y$ if for all continuous $\mathbb{C}$ valued functions $f$ on $X$, $f(x)=f(y)$. Let $Y$ be the quotient of $X$ by this relation. By construction $Y$ is CG (it is the quotient of a CG space), any functions from $X$ to $\mathbb{C}$ is compatible with the equivalence relation hence defines a function from $Y$ to $\mathbb{C}$, hence $Y$ is completely Hausdroff and $C(Y) = C(X)$. So the image of your functor is the same as the image of the functor of Philips when restricted to CG spaces, which as you mentioned in your question is apparently not essentially surjective.

One can also probably prove it is not full by constructing an example where there won't be so many interesting maps from the quotient $Y$ to the initial space $X$ while if the functor was full then the isomorphisms map from $C(X)$ to $C(Y)$ should be represented by a map from $Y$ to $X$...

Answer 2

The answer is yes, provided you change the category on the topological side slightly to: compactly generated functionally Hausdorff topological spaces with a distinguished family of compact sets; with continuous maps that preserve the distinguished family. See Theorem 6 of

Michael Forger, Daniel V. Paulino, Locally $C^*$ Algebras, $C^*$ Bundles and Noncommutative Spaces, arXiv:1307.4458v1