In the category of Banach spaces, where the objects are Banach spaces and the morphisms are continuous linear maps, what are there coproducts? Are they the typical direct sum of Banach spaces? If so, does it accept finite and infinite coproducts?
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1This is a poorly behaved category of Banach spaces. A much better behaved category is the category of Banach spaces and maps of norm at most $1$; isomorphism in this category is isometric isomorphism, so all universal constructions in this category really give you a Banach space (as in a space with a norm) and not a Lipschitz equivalence class of Banach spaces. This category is about as nice as it gets, and in particular is complete, cocomplete, and closed symmetric monoidal; see http://qchu.wordpress.com/2012/06/23/banach-spaces-and-lawvere-metrics-and-closed-categories/ for details. – Qiaochu Yuan Jun 27 '14 at 15:58
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@Qiaochu: But it is not additive ... which is quite important for doing a lot of things! – Martin Brandenburg Jun 27 '14 at 18:56
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@Martin: it is additive if it's regarded as being enriched over itself. The subtlety is that there are now two ways to regard this enriched category as an ordinary category, one using the usual forgetful functor and one using the unit ball forgetful functor $\text{Hom}(1, -)$, which disagree. Even if you don't use this enrichment, it's enriched over totally convex spaces, and in particular you can make sense of, say, $\frac{f - g}{2}$ for parallel morphisms $f, g$, which is enough to replace (co)equalizers with (co)kernels. – Qiaochu Yuan Jun 27 '14 at 20:02
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Thank you both, great insight, I'll look at your blog entry to get up to speed. – Richard Jennings Jun 27 '14 at 23:52
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Yes, the coproduct of two Banach spaces $(X,\|~\|_X)$ and $(Y,\| ~ \|_Y)$ is $(X \oplus Y,\|~\|_{X\oplus Y})$ with $\|(x,y)\|_{X \oplus Y} = \|x\|_X + \|y\|_Y$. This is just an example for a suitable norm, equivalent norms such as $\sqrt{\|x\|_X^2 + \|y\|_Y^2}$ give isomorphic Banach spaces. The universal property is easily verified. It follows that finite coproducts exist. But infinite coproducts don't exist. I've proven this here recently for Hilbert spaces, cannot find the question right now (someone else?). The same argument applies here.
Martin Sleziak
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Martin Brandenburg
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I think the relevant link is here, where Martin shows that infinite coproducts of 1-dimensional HIlbert spaces don't exist. – tcamps Sep 06 '15 at 22:13
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Doesn't that also follow from the fact that we have a preadditive category (i.e. ab-enrichment + zero object) and an obvious product? Because if I'm not mistaken, in preadditive category, products give rise to coproducts in a canonical way and vice versa, yielding the concept of a “biproduct”. – Lukas Juhrich Jul 05 '21 at 15:40
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