evil properties, higher category theory and well-chosen tensor products - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:05:35Z http://mathoverflow.net/feeds/question/38717 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38717/evil-properties-higher-category-theory-and-well-chosen-tensor-products evil properties, higher category theory and well-chosen tensor products Martin Brandenburg 2010-09-14T17:22:20Z 2010-09-14T21:03:24Z <p>Let's start with the following random example: If $F$ is a presheaf, then for every chain of open subsets $U \subseteq V \subseteq W$, the morphisms $F(W) \to F(V) \to F(U)$ and $F(W) \to F(U)$ coindice. But this may be an evil (<a href="http://ncatlab.org/nlab/show/evil" rel="nofollow">nlab link</a>) definition, especially when the values of $F$ are categories (which occurs in my current research). So we should just impose that these morphisms are equivalent. But then we arrive at compatibility conditions between these equivalences, which are, again, equalities, and may be evil. So what is the consequence of this: Should every mathematical theory take place in a $\infty$-category? Or is 'real' mathematics basically evil? I know this question is quite imprecise, but currently I just don't see where and why this process of increasing the "depth" of category theory should end. Anyway, I have to admit that my knowledge of higher category theory is very, very rudimental.</p> <p>Here is a evil question, which is part of my confusion: Let $Ring$ denote the category of rings, ring homomorphisms and $AbCat$ denote the category of abelian categories, functors. Then $Ring \to AbCat, A \to Mod_A$ should be a functor, right? If $A \to B$ is a ring homomorphism, just tensor with $B$ over $A$. But then functoriality is only satisfied up to equivalence ($2$-isomorphism) in $AbCat$. To get a honest functor, we might mod out these equivalences to get a $1$-category $\tilde{AbCat}$. But I don't think that this is the most natural way to handle this. Or we may define quasi-functors (cf. the presheaf example). Anyway, may we <em>think</em> of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M \otimes_A B$ carefully? I think the latter is interesting, although I know that nobody really cares about such evil equalities like $(M \otimes_A B) \otimes_B C = M \otimes_A C$.</p> http://mathoverflow.net/questions/38717/evil-properties-higher-category-theory-and-well-chosen-tensor-products/38734#38734 Answer by Mike Shulman for evil properties, higher category theory and well-chosen tensor products Mike Shulman 2010-09-14T21:03:24Z 2010-09-14T21:03:24Z <p>There are a lot of questions here, but I'll try to answer them all.</p> <blockquote> <p>Should every mathematical theory take place in a ∞-category? Or is 'real' mathematics basically evil?</p> </blockquote> <p>I would say that all mathematics should take place in its natural context. Sometimes you have things that are sets where equality makes sense, like an ordinary presheaf, and then you work in a 1-category. Sometimes you have things where only isomorphism makes sense, like a presheaf of categories, and then you work in a 2-category. Etc.</p> <p>It is true that any n-category for finite n can be considered a special case of an &infin;-category with only identity cells above n, so in this degenerate sense all n-categories are &infin;-categories, and thus one might say that "all mathematics takes place in an &infin;-category" &mdash; at least if one believes that all mathematics takes place in an n-category for some n! But even that is not clear, e.g. some mathematics naturally takes place in other categorical structures, such as a double category or a proarrow equipment. Some mathematics uses no category theory at all (at least as far as anyone has noticed so far), and so it would be a stretch to say that it takes place in any sort of category.</p> <blockquote> <p>Anyway, may we <em>think</em> of it as a usual functor, without turning into troubles? Or is it important, in practice, to have this higher category theoretic point of view? Or is it possible to turn this functor into a honest functor, by choosing the tensor products $M\otimes_A B$ carefully?</p> </blockquote> <p>I would say qualified yes, yes, and yes, respectively. You can think of it as a usual functor as long as doing so doesn't cause you to think that it behaves in any way that a pseudofunctor doesn't! Which is sort of a vacuous statement, but the point is that pseudofunctors really shouldn't be a very scary concept (as opposed to a technical definition, which might be a bit complicated, though cf. Harry's comment) &mdash; they really are just like ordinary functors, except that you're dealing with things (e.g. categories) for which it doesn't really make sense to ask morphisms to be equal, only isomorphic.</p> <p>On the other hand, the "higher category theoretic" fact that pseudofunctors are <em>not</em> all strict functors <em>is</em> very important. I believe that Benabou, the inventor of bicategories, once said that the important thing about bicategories is not that they themselves are "weak," but that the <em>morphisms</em> between them are weak. In particular, although every bicategory is equivalent to a strict 2-category, not every pseudofunctor between bicategories is equivalent to a strict functor.</p> <p>But on the third hard, it <em>is</em> true that any pseudofunctor with values in the 2-category Cat <em>is</em> equivalent to a strict functor. In the language of fibrations, this says that any fibration is equivalent to a split one. Tyler mentioned one construction of an equivalent strict functor in the case of modules and tensor products. There is also a general construction which, applied to the case of modules, will replace $Mod_A$ by a category whose objects are pairs (M,&phi;) where M is an R-module and &phi;:R&rarr;A is a ring homomorphism. We regard such a pair as a formal representative of $M\otimes_R A$ and define morphisms between them accordingly, to get a category eequivalent to $Mod_A$. Now the extension-of-scalars functor $\psi_!:Mod_A \to Mod_B$ is represented by the functor taking a pair (M,&phi;) to (M,&psi;&phi;), which is strictly functorial since composition of ring homomorphisms is so.</p>