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Can want to define a category $\mathcal{D}$ as follows:

the objets are quadrupels $(G,H,X,r)$, where

1) G is a preordered abelian group with order unit,

2) H is an abelian group,

3) X is a compact Hausdorff space

4) $r:X\to S(G)$ is a continuous map ( $S(G)$ is a compact Hausdorff space)

With morphisms $(f,g,h):(G,H,X,r)\to (G',H',X',r')$, where

1) $f:G\to G'$ is a morphism of preordered groups preserving the order unit

2)$g:H\to H'$ is a group homomorphism

3)$h:X'\to X$ is a continuous map which is compatible with $r$ in the following way, that the following diagram commutes:

$$\require{AMScd}\begin{CD} X' @>h>> X \\ @VV r' V @VV r V \\ S(G')@>S(f)>> S(G) \\ \end{CD}$$ where $S(f)(\chi)=\chi\circ f$.

The composition of morphisms is the canonical one. I checked if this satisfies the definition of a category and I don't see any problems. But I'm not sure of I overlook something: is there everything ok or is there something wrong with this definition? I'm asking because I'm not sure if I have to add $S(G)$ and $S(\alpha)$ somehow to the data..

And my main question is (if everything is correct): Is this catgeory isomorphic to a certain product of categories of well-known ones, for example something similar to (category of preordered abelian groups with order unit)$\times$(compact hausdorff space)$^{op}$? (It can't be isomorphic to (category of preordered groups with order unit)$\times$(compact hausdorff space)$^{op}$, since we have this pariring map $r$ as well)

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  • $\begingroup$ What is $S(G)$ exactly? From your definition of $S(f)$, it looks like it should be a subset of the set of morphism $G\to Y$ for some object $Y$ in some category... $\endgroup$ – Arnaud D. Aug 5 '17 at 20:02
  • $\begingroup$ oops, I forgot..Yes, it is. $S(G)$ is the set of homomorphisms $f:(G,G^+,u)\to (\mathbb{R},\mathbb{R}^+,1)$ of preordered groups, endowed with the weak*-topology. $\endgroup$ – Sabrina G. Aug 5 '17 at 20:15
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I don't know how to express this as a product of categories, but your construction looks a lot like a comma category (so it is actually a fibered product of categories!). In fact, it is (up to the order of your terms) the category $\mathbf{Ab}\times (Id_{\mathbf{CHaus}^{op}}\downarrow S)$, where $\mathbf{Ab}$ denotes the category of abelian groups, $\mathbf{CHaus}$ denotes the category of compact Hausdorff spaces, and $S=Hom(\_ ,(\Bbb R,\Bbb R^+,1))^{op}:\mathbf{PrAb}\to \mathbf{CHaus}^{op}$ is (the dual of) the contravariant functor represented by $(\Bbb R,\Bbb R^+,1)$.

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  • $\begingroup$ thank you! Your answer is very satisfying, I was looking for exactly something like this. $\endgroup$ – Sabrina G. Aug 5 '17 at 21:06
  • $\begingroup$ one question: the functor $Id_{\mathbf{CHaus}^{op}}$ is not contravariant here, or am I wrong? $\endgroup$ – Sabrina G. Sep 8 '17 at 12:04
  • $\begingroup$ @toto No, it's a covariant functor $\mathbf{CHaus}^{op}\to \mathbf{CHaus}^{op}$. Alternatively, you can see it as $Id_{\mathbf{CHaus}}^{op}$, i.e. the dual of the identity functor on $\mathbf{CHaus}$. $\endgroup$ – Arnaud D. Sep 8 '17 at 12:47
  • $\begingroup$ ok, thx. but $h$ is contravariant and I guess, $Id_{\mathbf{CHaus}^{op}}$ should be the functor $h$, or does this make no difference? Or shall I replace $Id_{\mathbf{CHaus}^{op}}$ with the "identity" $\mathbf{CHaus}\to \mathbf{CHaus}^{op}$ instead? $\endgroup$ – Sabrina G. Sep 8 '17 at 14:08

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