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)

  • $\begingroup$ 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

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 $(\Bbb R,\Bbb R^+,1)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.