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

4 votes

functors reflecting "isomorphism relations"?

One paper about this is Elliott's Towards a theory of classification, where such functors are called classification functors, and there is a nice collection of examples. (Though note that a category …
Tobias Fritz's user avatar
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2 votes

Spans as binary relations: reflexivity, transitivity, and completeness?

This is not really an answer to your question, but I'd like to point out that your "extension" definition probably isn't quite what you want yet. Of course this may well depend on your intended applic …
Tobias Fritz's user avatar
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4 votes

Quotient-free monoidal categories

Here are some additional thoughts and examples. 1) The argument presented in the OP applies whenever one has a monoidal category $(\mathsf{C},\otimes,1)$ in which the unit object $1$ is a separator a …
Tobias Fritz's user avatar
  • 6,406
10 votes
Accepted

Why is subcoalgebra structure unique?

Let us say that a monomorphism $U\subseteq A$ is nontrivial if $U$ is nonempty. Since every nontrivial monomorphism in $Set$ is split, every $F : Set\to Set$ takes nontrivial monomorphisms to monomorp …
Tobias Fritz's user avatar
  • 6,406
7 votes

Slicing up monads on categories with pullbacks: what are these mysterious "zerosumfree" monads"

Here are some remarks and pointers to the literature which are too long for a comment. The map $$ \mathsf{Dist}(X_1) \times_{\mathsf{Dist}(X)} \mathsf{Dist}(X_2) \longrightarrow \mathsf{Dist}(X_1 \t …
Tobias Fritz's user avatar
  • 6,406
8 votes
Accepted

Strongly connected components as adjoint functor?

The existence of such an adjoint triple in particular requires the strongly connected components functor to be a left adjoint. But in fact this fact is not a left adjoint, since it doesn't preserve co …
Tobias Fritz's user avatar
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4 votes

In which category is a measure on a measurable space a morphism?

A very natural enlargement of the category of measurable spaces is $\mathsf{Stoch}$, the category of measurable spaces and Markov kernels, with composition given by the Chapman-Kolmogorov equation. Th …
Tobias Fritz's user avatar
  • 6,406
8 votes

Why Kleisli Markov categories and not the Eilenberg-Moore categories of the associated monads

What can you do with the Kleisli category (of a probability monad) that you can't do with the Eilenberg-Moore category? The first paragraph of kirk sturtz's answer provides a good high-level summary …
Tobias Fritz's user avatar
  • 6,406
8 votes

Are there noncartesian monoidal categories with $A \otimes B = A \times B$?

Note: As pointed out by Peter LeFanu Lumsdaine in the comments, the example constructed in this answer does not work in its current form, since naturality of the associator fails. (This only affects t …
Tobias Fritz's user avatar
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4 votes
Accepted

Algebras for probability monad

The algebras of this monad and closely related ones have been introduced by Pumplün and Röhrl. For example the introduction of a paper by Börger and Kemper provides a good summary, Pumplün and Röhrl …
Tobias Fritz's user avatar
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3 votes
Accepted

Symetrical simplex category

You haven't specified what the morphisms in your "symmetric simplex category" are supposed to be, and there are two natural choices: Functors, or Natural isomorphism classes of functors. In the fi …
Tobias Fritz's user avatar
  • 6,406
10 votes
1 answer
813 views

Tannaka duality for C*-algebras?

Tannaka-Krein duality shows how to recover a group $G$ from its category $\mathbf{Rep}(G)$ of finite-dimensional complex representations and the forgetful functor $F:\mathbf{Rep}(G)\to \mathbf{Vect}_{ …
Tobias Fritz's user avatar
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23 votes
Accepted

Reconstructing base manifold from its category of smooth vector bundles

Yes, it is possible to recover the manifold through the following steps: Smooth Serre-Swan theorem: $\mathbf{Vect}(X)$ is equivalent to the category of finitely generated projective modules over $C^\ …
Tobias Fritz's user avatar
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3 votes
Accepted

Elements that commute with $1$ in the pushout of a $C^{\ast}$-algebra

The answer is yes: $1 \ast c = c \ast 1$ implies $c \in B$. To see this, let me first translate the statement into fully categorical language. If we restrict to $\|c\| \le 1$ wlog, then the elements …
Tobias Fritz's user avatar
  • 6,406
13 votes

Another notion of exactness: how to refine it, and where does it fit?

I'd like to argue that the current definition is too minimal to allow for much theory development, since the only substantial axiom is the pasting condition. In particular, it would be possible to tak …
Tobias Fritz's user avatar
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