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The category of inverse semigroups is isomorphic to the category of etale groupoids whose unit and arrow spaces are Alexandrov spaces and the poset associated to the unit space is required to be a mee …

Do you mean take the monoid with the following presentation? Take the arrows of C as generators and add the relations that f.g = fg if f and g are composable and that each identity of C be equivalent …

Profunctor composition is not strictly associative even after putting on the equivalence relation. One is in a bicatgeory setting here. The equivalence relation is among other things to make identit …

No, this is false. Let $C$ be the monoid $\lbrace 1,\ldots, 2^n\rbrace$ with $\max$ as the operation and let $D$ be the power set of $\lbrace 1,\ldots, n\rbrace$ with $\cup$ as the operation. These …

The universal abelian group for a finite poset is the free abelian group on the edges of the Hasse diagram. Consider the posets $P=\lbrace 0,a,b,1\rbrace$ with $0\lneq a,b\lneq 1$ and $a,b$ incompar …

Probably you want to look at pushouts of categories along a common set of objects. For example, the free product of monoids is the pushout along the inclusion of the identity. Such things and their wo …

Let $\mathbf C$ be a category with finite limits. Then a left exact functor $F\colon \mathbf C\to \mathbf{Set}$ is pro-representable and hence extends to the pro-completion $\mathbf C$. My question …

No. There is a monoid with trivial group image whose classifying space is a sphere. See Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to …

This is an extended comment to Tyler's answer. The Kurosh theorem implies the equalizer is a free product of a free group with conjugates of subgroups of the free factors. But each non-trivial subgrou …

I believe the following is an example of a category whose leanification is discrete but which is not a groupoid. It should be possible to simplify it. There may be details to work out. Let $B$ be t …

Conceptual answer. There can be no such functor. Let $C$ be any concrete category of finite sets and mappings such that the only automorphisms in $C$ are trivial. I claim there is no underlying set …

What you are looking at is called a heap or groud. The category of heaps is equivalent to the categor of torsors but is the universal algebra viewpoint from the sixties. The wiki article I linked to g …

Look at the paper of Cockett and Hofstra on Turing categories. (Here is another link and also DOI: 10.1016/j.apal.2008.04.005.)

There are two natural definitions of a profinite category. You can look at inverse limits of finite categories or you can look at topological categories whose underlying spaces are profinite (call th …

I am a complete novice in the art of categorification, so this may not be a great question. Background. The groupoid $\mathbf {FSet}$ of finite sets and bijections categorifies the natural numbers an …