Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 75

A semigroup is a set $S$ together with a binary operation that is associative. Examples of semigroups are the set of finite strings over a fixed alphabet (under concatenation) and the positive integers (under addition, maximum, or minimum). A monoid is a semigroup with a neutral element. Of course, any group is also a monoid/semigroup.

24 votes

When $X \times Y \cong X \times Z$ implies $Y \cong Z$ (in the category of finite topologica...

The monoid $\mathcal M_{\rm fin}$ is in fact cancellative. To prove this, we start with a Lemma. Let us write $h(T,Y)$ for the cardinality of the set of continuous maps $T\to Y$ and $i(T,Y)$ for t …
Eric Wofsey's user avatar
  • 31.2k
17 votes
Accepted

Haar Measure on Locally Compact Semigroups

Not every locally compact semigroup admits a (locally finite) left-invariant measure. In fact, this has nothing to do with any sort of analytic technicalities and already fails for finite semigroups. …
Eric Wofsey's user avatar
  • 31.2k
7 votes
Accepted

Is the monoid of taking iterated images and inverse images freely generated by the image and...

Yes, it is. The idea is that given an element of $\mathcal{M}$, you can detect whether the last step of it was $L$ or $U$, and then undo the steps one by one to recover a unique expression for it. F …
Eric Wofsey's user avatar
  • 31.2k
6 votes
Accepted

"Exactness" of groupify functor

Let $A=\mathbb{N}\cup\{\infty\}$, considered as a monoid under addition. Let $M=\mathbb{N}$, $N=A\oplus A$, $i_1(n)=(n,0)$ and $i_2(n)=(0,n)$. Then the equalizer of $i_1$ and $i_2$ is $0\to\mathbb{N …
Eric Wofsey's user avatar
  • 31.2k
6 votes

Homological algebra for commutative monoids?

One interpretation of the "right" category for doing homological algebra of commutative monoids would be the category of simplicial commutative monoids. Now up to homotopy, a space is a commutative m …
Eric Wofsey's user avatar
  • 31.2k