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Enriched categories, topoi, abelian categories, monoidal categories, homological algebra.

8 votes
2 answers
469 views

Reference request: (co)limits in Eilenberg--Moore (V-)categories

The following result seems to be well known: If T is a (V-)monad on a (V-)category C, then the forgetful functor $U^T \colon C^T \to C$ creates any limits that exist in C, and any colimits that exis …
15 votes
Accepted

Any example of a non-strong monad?

Here is a class of examples different to Tom's: if your underlying monoidal category C is closed, then a strong monad on C is the same as a C-enriched monad, i.e. one that respects the enrichment of C …
Martin Sleziak's user avatar
27 votes

morita equivalence for categories

That their Cauchy completions are equivalent.
misseuler's user avatar
4 votes
Accepted

Category enriched over a monoidal 2-category

Yes, here: R. Garner, M. Shulman, Enriched categories as a free cocompletion. arxiv.
Finn Lawler's user avatar
  • 3,622
4 votes

Kan extensions of pseudofunctors

I haven't seen this written down anywhere, but I've worked it out myself in a paper I'm working on. The paper isn't quite ready, so I'll just sketch the idea here. As Zhen says, you need to generali …
Finn Lawler's user avatar
  • 3,622
2 votes
Accepted

Functor generalisation

Here is one way to look at it: if V is a monoidal category and $\mathbf{B} V$ is the corresponding one-object bicategory, then a V-category in the usual sense is the same thing as a lax functor $\math …
Finn Lawler's user avatar
  • 3,622
2 votes

Basic category theory: Universality of adjunction unit is justified by Yoneda Proposition in...

Our adjunction isomorphism $\phi$ does not necessarily involve the assignment above. Yes, it does, by Yoneda's lemma. The bijection $\phi$ is natural, hence its value $\eta_x$ at the identity de …
Finn Lawler's user avatar
  • 3,622
8 votes
Accepted

A question on the Grothendieck construction

The bicategory of elements of a Cat-valued functor is defined in e.g. Street's Fibrations in bicategories; it's the same as the usual one, with 2-cells as described here. Its property of classifying …
Finn Lawler's user avatar
  • 3,622
5 votes

Ends and coends as Kan extensions (without using the subdivision category of Mac Lane)?

I don't know of any definition involving Kan extensions, but (co)ends can be expressed as (co)limits weighted by a hom functor (see e.g. here), so that for $F \colon C^{op} \times C \to D$ the end $\i …
Ricardo Andrade's user avatar
5 votes

Categorification of coends and ends

I don't think I have seen anything like this published before, but I have written up a similar definition here (see here too). One thing in your definition I would take issue with is that your 2-coen …
Finn Lawler's user avatar
  • 3,622
3 votes

Codomain fibration.

I assume that by 'cartesian' you mean 'having cartesian products'. In that case the answer is no. You've more or less said this yourself: for $C^{\mathbf{3}} \to C^{\mathbf{2}}$ or the composite $C^ …
Finn Lawler's user avatar
  • 3,622
5 votes

Monadicity and sheaves.

The answer (to the first question) is yes: reflections are always monadic, and the associated monad is idempotent.
Finn Lawler's user avatar
  • 3,622
21 votes

The main theorems of category theory and their applications

Monadicity theorems like Beck's give sufficient conditions on a functor $U \colon B \to A$ with a left adjoint F to be equivalent (in the slice category Cat/A) to the forgetful functor out of the cate …
Finn Lawler's user avatar
  • 3,622
11 votes

The main theorems of category theory and their applications

There was some discussion here about special cases of Yoneda's lemma, including the usual examples of Cayley's theorem for groups and Dedekind's embedding theorem for posets. It also seems that a goo …
Community's user avatar
  • 1
14 votes

Is there a nice application of category theory to functional/complex/harmonic analysis?

First, a disclaimer: I am not even close to being an analyst. Second, I don't know of any applications of category theory to the areas of analysis that you mention. I don't think we have got to that …
Finn Lawler's user avatar
  • 3,622

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