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Oct 5, 2021 at 12:09 comment added Z. M (cont'd) and that is why I was trying to think of compact Hausdorff spaces, since these are another instances which give rise to (natural) bornological sets. In addition, to clarify, unit balls no longer appear, and I think of the category of Banach spaces with morphisms being continuous linear maps, not just of norm $\le1$.
Oct 5, 2021 at 12:08 comment added Z. M Maybe I was mistaken about the compact Hausdorff spaces. For sets $S$, I was just rephrasing your last equivalence (not enriched, but just the underlying sets): the left hand side $\ell^\infty$ is precisely the collection of bounded maps $S\to E$, i.e., the Hom-set in the category of bornological spaces. So $\ell^1$ is at least a "partial left adjoint" to the forgetful functor from Banach spaces to bornological sets, defined on those coming from sets $S$. I suspect the the left adjoint exists, and thus it coincides with $\ell^1$ when restricting to these $S$ coming from sets...
Oct 5, 2021 at 11:34 comment added Yemon Choi I still don't understand what precisely you are claiming about the space of Radon measures. So now you are working with the category of compact Hausdorff spaces and continuous maps, and you want to make M(S) into a functor from this category to Bang which is a left adjoint? But the unit ball of M(S) is not compact in the norm topology, which is the only topology that can be seen in Bang. This is why I assumed that you were actually talking about a functor from CompHaus to Wael
Oct 5, 2021 at 11:32 comment added Yemon Choi @Z.M I am not fluent in all this condensed stuff but I have been thinking categorically about Banach spaces for 10-15 years, so perhaps we are just talking about the same thing in different language. So what you are saying is that we should consider a category BornSet, and then $\ell^1$ is the left adjoint to the "unit ball functor" from Bang to BornSet?
Oct 5, 2021 at 11:26 comment added Z. M "Initial among all maps from $S$ to a Banach space" means to be the initial object in the slice category of Banach spaces $E$ along with a bounded map $S\to E$, which is just rephrasing the adjunction. And yes, $\mathcal M(S)$ is naturally a Smith space at least when $S$ is profinite, but I tend to believe that the norm topology is its "Banachification".
Oct 5, 2021 at 11:19 comment added Yemon Choi @Z.M If you are seeking to use $M(S)$ as some kind of left adjoint, then my instinct is that you need to be using its structure as a dual space (or if you like a certain POV, Waelbroeck space / Smith space) and not just its normed structure. This is why I am not sure what you mean precisely by "initial among all maps"
Oct 5, 2021 at 11:17 comment added Yemon Choi @Z.M Initial in what sense? i.e. what is the category that you are comparing with Ban? I would not be surprised if there is a generalization, but I wrote my answer to answer the particular question that was asked, not to attempt some kind of encyclopaedia of similar results.
Oct 5, 2021 at 11:13 comment added Z. M I suspect that there is a further generalization. Note that $\ell^1(S)$ is the space of signed measures on $S$. We know another analogue: when $S$ is a compact Hausdorff space, the map $S\to\mathcal M(S)$ to the Banach space of Radon measures with the canonical norm is initial among all maps to a Banach space, if I am not mistaken.
Oct 5, 2021 at 11:00 comment added Z. M We could the last paragraph as follows: consider the category of bornological sets with bounded maps as morphisms, and view $S$ as a bornological space with the bornology being all subsets. Then $S\to\ell^1(S)$ seems to be the initial map from the bornological set $S$ to a Banach space.
Jan 6, 2021 at 21:09 comment added Yemon Choi While I was writing this rambling account, I think Qiaochu said it more concisely and coherently
Jan 6, 2021 at 21:08 history answered Yemon Choi CC BY-SA 4.0