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As I mentioned in my comment above, Gruenberg gave a direct bijection between $\operatorname{Ext}^1_{\mathbb ZG}(I_G,A)$ and group extensions $1\to A\to H\to G\to 1$. The details can be found in eith …

Here is how the proof works for monoids. This doesn't really answer the question so I made this communitywiki, but maybe the OP will find it useful. Let $X$ be a compact Hausdorff space. Call an equ …

The answer to your question is "no". A counterexample is given in this paper by Leroux in Example 1.6.

First note that the definition of equivalence you are using for topological groupoids is too strong and is not the one people generally use. It is really too much to ask for a continuous quasi-invers …

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.)

A Lawvere theory is a small category with finite products such that every object is isomorphic to a finite product of copies of a distinguished object x. A model of the theory in a category with finit …

No. Mark Sapir and Marcel Jackson even showed it is undecidable if the factors embed in an amalgamated free product of finite monoids. See the intro of Jackson, Marcel. "The embeddability of ring and …

The problem of counting semigroups and monoids of order $n$ up to isomorphism and anti-isomorphism (i.e., contravariant equvialence) is a very classical problem whose answer is conjectured but nobody …

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 …

Here is an example of a variety where every object is regular-projective. I am not sure what you mean by a product of varieties, so I don’t know for sure if it fits into that context but I suspect it …

Defining conjugacy for monoids is a dicey subject because many different notions that are equivalent for groups are different for monoids and it is not clear which of these is interesting. The one y …

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 …

George Bergman characterizes representable endofunctors of monoids in 10.6 of https://math.berkeley.edu/~gbergman/245/3.2.pdf by classifying the comonoids in the category of monoids.

I think describing all monoids satisfying these conditions is an essentially impossible task. But for a finite band (monoid where each element is idempotent) this condition is equivalent to being a di …

In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following. …