If the underlying set of a Banach space, treated as an object in a closed category, is the closed unit ball, then how to do we describe this as this set with extra structure. We can do this very concretely!

]]>linking to absolute convergence

Anonymous

]]>just to highlight that the above comment refers to an edit in this paragraph.

]]>The image of linear operators between Banach spaces need not have closed image, and so the usual definition of cokernels need not yield a Banach space, as quotients by non-closed subspaces of a Banach space can be non-complete. However, the category of Banach spaces still admits coequalizers by taking the quotient by the closure of the image of $f-g$ instead.

Javier Villar

]]>Added a reference

- Jiří Rosický,
*Are Banach spaces monadic?*, (arXiv:2011.07543)

It’s the old issue of whether the category of foos has foos as objects or morphisms. I’m a partisan of the former position (which is the same as making the distinction between a group and its delooping), but the latter is obviously inviting sometimes!

]]>Toby: In #16 I originally wrote “sets-and-injections” in describing $\mathcal{M}$, then corrected it. I guess for some reason that is an easy mistake to make!

]]>I corrected the description of the $\mathcal{M}$-category $\mathcal{M}$

I don’t know why I could define $\mathcal{M}$ correctly as a category but not as an $\mathcal{M}$-category! Thanks.

]]>we should remember that the “underlying ordinary category” in the usual sense of enriched category theory actually has $T$ as its hom-sets.

And this only proves that the category theorists were right about $Ban$ all along! (^_^)

]]>Oh, you’re right Mike! It’s because the monoidal unit is $(1, 1)$, and $hom((1, 1), (T, L)) \cong T$.

]]>Thanks, Toby – I hadn’t created it yet because I wasn’t sure about what to call it. But $\mathcal{M}$-category is certainly fine. The page looks great! I corrected the description of the $\mathcal{M}$-category $\mathcal{M}$, and modified the Idea section so it doesn’t suggest that $\mathcal{F}$-categories are an “incomplete” categorification of $\mathcal{M}$-categories (which I don’t think they are – there are just also other possible categorifications, which might have other applications).

Re 24: Yes – but just so we don’t get confused, we should remember that the “underlying ordinary category” in the usual sense of enriched category theory actually has $T$ as its hom-sets.

]]>Rod, if $x, y$ are objects of an $M$-category, then $hom(x, y)$ is by definition a pair of sets $(T, L)$ where $T\subset L$. An element of the first component $T$ is called a *tight* morphism from $x$ to $y$. We can think of the hom-set of the underlying ordinary category as just $L$, so every morphism of the ordinary category is *loose* by default, but some morphisms, the ones that land in $T$, are also tight.

ummm, M-category never says what “tight” and “loose” mean.

]]>Thanks, Toby. Always nice when an nForum discussion leads to a useful nLab entry. :-)

]]>I created M-category.

]]>I really think the $\mathcal{M}$-category viewpoint has a lot to recommend it. Any $\mathcal{M}$-category has two kinds of isomorphism, which we might in general call “tight” and “loose”, as Steve and I did for $\mathcal{F}$-categories. But in lots of examples these two already have specific names.

In a $\dagger$-category, like Hilbert spaces, the tight isomorphisms are called

*unitary*.In the $\mathcal{M}$-category of material-sets, the tight isomorphisms are called

*equality*and the loose ones*bijections*.In the orbit category of a group $G$, whose objects are the $G$-sets $G/H$, whose loose morphisms are $G$-maps, and whose tight morphisms are those that commute with the quotient maps from $G$, a loose isomorphism $G/H \cong G/K$ means that $H$ and $K$ are

*conjugate*, whereas a tight one means they are*equal*(as subgroups of $G$).

So in the $\mathcal{M}$-category of Banach spaces, there’s no problem with saying that the tight isomorphisms are called *isometries* while the loose isomorphisms are just *isomorphisms*.

Of course one doesn’t have to spend one’s whole life in a single category, but for me, having a category (or at least a groupoid, or most generally an $\omega$-groupoid) in mind is necessary to fully define what a Banach space (or whatever concept you’re thinking about) *is*. This is why people are inspired to say ‘Banachable [topological vector] space’ for things that are considered the same when a mere linear homomorphism exists between them; when you see the definition of Banach space as an abstract structure, you wouldn’t naturally think that two such things are the same unless there was a surjective linear isometry between them: a bijection which (and whose inverse which) preserves *all* of the structure in that definition.

I keep thinking of group theory, and how sometimes group theorists are interested in arbitrary functions between groups. They even have a term for an isomorphism class in the category of groups and functions (‘order’). Yet nobody has any doubt about what ‘the category of groups’ means or when two groups should be considered the same (although they may be equivalent for certain purposes without being the same, as having the same order, being Morita-equivalent, etc).

Of course, we should not steal the term ‘isomorphism’. The functional analysts have decided what that means in the context of Banach spaces, and we shouldn’t denigrate it just because the same term was adopted in category theory for something else.

]]>Mike wrote:

There's a paper by John Power, I think, where he makes the same observation in this context that we did for F-categories: they can be regarded as categories enriched over the category of injections.

It's *Premonoidal categories as categories with algebraic structure*, Theoretical Comp. Sci. 278, 2002.

Re Toby #7: It’s worth keeping in mind that, as with many (most?) other uses of “isomorphism” in mathematics, analysts’ usual usage of the term predates the invention of category theory.

Re Toby #9: I’m not convinced that there’s any utility to having a single category which encompasses those two different notions, but I’m open to persuasion.

]]>having a subcategory of “special” morphisms… Is there such a thing?

Indeed! It doesn’t have a standard name yet, but I’ve been starting to see it all over the place. I first noticed it in the 2-categorical version, where Steve Lack and I called it an F-category. Here are some other important examples:

The Kleisli category of a monad, where the “special” morphisms are those arising from morphisms in the base category. This matters in computer science where the monad builds in “side effects” and the “special morphisms” are those with no side effects. There’s a paper by John Power, I think, where he makes the same observation in this context that we did for F-categories: they can be regarded as categories enriched over the category of injections.

The category of sets in a material set theory, where the “special” morphisms are literal subset inclusions (

*not*arbitrary injections). A generalized version of this called a “directed structural system of inclusions” plays an important role in this paperMore generally, any category of subobjects (or quotients) of some given object, with morphisms that don’t necessarily respect the inclusions, where the “special” morphisms are those that do. (The previous example is where the given object is the class of all sets.) For instance, Peter May’s Galois theory example of an “evil” category on the categories mailing list: subfields of an extension field, but with field homomorphisms that don’t necessarily respect the inclusion.

I’ve thought of calling these “$\mathcal{M}$-categories” since they can be identified with categories enriched over the category of monomorphisms in Set (a category which one might call $\mathcal{M}$), but if anyone has a better name, I’m all ears.

]]>With regard to Paul Siegel’s comment at http://mathoverflow.net/questions/80567/what-is-an-isomorphism-of-banach-spaces, that seems like a common enough concept: of having a subcategory of “special” morphisms. Something like a dagger category but not as strong. Is there such a thing?

]]>In the interests of completeness, and to see if we can get more people involved, I just asked about this on MO:

http://mathoverflow.net/questions/80567/what-is-an-isomorphism-of-banach-spaces

My motivation is just so that we don’t fix up this article only to find someone else complaining that we’ve left out *their* favourite type of morphism!

@13 Yemon, I, like you, can think of several good reasons for doing homological algebra with banach algebras!!!

]]>We are probably all in agreement that both categories are needed, right? Doing operator theory in Bang looks to me a bit Procrustean. Maybe Bant should be viewed as a category with some kind of enrichment that detects the extra structure in the “Hom-sets”?

As it happens, if one tries to do homological algebra with Banach(able) algebras and Banach(able) modules – I know, why would you want to do that? – then one sometimes wants to perform “standard constructions with norm control”. So I really do want to say that certain $n$-cocycles cobound with norm at most $3^n$, and it has never been clear to me whether the arguments would more naturally take place in Bang or in Bant.

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