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I have three questions.

1:

Does the groupification of a semigroup always exist? I believe this should be yes because for every $x$ in the semigroup one could just define an element $x'$ that should work as its inverse. But what would then happen to the product $x'y$ for $x,y$ elements of the semigroup? It feels like we get choices (or maybe not) here that messes things up.

2:

When defining the groupification, $G$, of a semigroup $S$ one require it to come with a morphism (of semigroups) $S \rightarrow G$ such that any other morphism (of semigroups) from $S$ to another group $G'$ factorizes through the previous map. Exactly which type of objects can be groupified? I guess one cannot groupify a topological space.

3:

This is a broad question but is there some sense of -ification? In the example one could replace "group" by "topological space" and talk about topologyfication. Now, no such word seem to exist so I guess one could not "topologyfy".

We can (i think) consider the groupification functor from the category of semigroups to the category of groups and it should be adjoint to the forgetful functor from the category of groups to the category of semigroups. This would suggest that we need some sense of a forgetful functor in the first place to talk about a -ification.

Apologies for this bad question, sometimes asking the right question is just as hard as answering it.

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    $\begingroup$ Your last statement is the point here. We want to take a structure and "add a little bit more structure" to make something else. In the case of semi-groups to groups, we are adding inverses. In the case of sheaves and the sheafification functor, we take a presheaf and require the germs be compatible. I think that the example of sheafification is perhaps the most clear of this sort of construction. This idea of looking at the right adjoint to the forgetful functor gives us a universal construction, which is precisely what we need for an -ification. Universality says your map is well-defined. $\endgroup$
    – BBischof
    Aug 16, 2010 at 18:39
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    $\begingroup$ Also, You might want to look at Freyd's Adjoint Functor Theorem. The conditions in this theorem already get you thinking about universality, and it very nearly looks like -ification already. When you consider one map to be forgetful, things look very transparent. $\endgroup$
    – BBischof
    Aug 16, 2010 at 18:40
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    $\begingroup$ I am sorry about all these comments, but in response to 2). If you just consider the forgetful functor from topological groups to topological spaces and you construct that functors adjoint we can groupify a top space. $\endgroup$
    – BBischof
    Aug 16, 2010 at 18:42
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    $\begingroup$ Oh man, another comment... Anyways, I just wanted to say that I do NOT think this is a bad question. In fact, I think that your question is very well posed. Both the level and structure of your question. The only complaint at most would be to divide it into two questions, but that would look bad. In short: +1 :) $\endgroup$
    – BBischof
    Aug 16, 2010 at 21:12

4 Answers 4

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I take the liberty of answering only the parts of your question, to which I think I can give a precise answer.

1: Yes, it is called the Grothendieck group $G(N)$ of the given semigroup $S$. This construction is functorial.

2: There is always a canonical semigroup homomorphism $S\to G(S)$, but this need not be injective in general. For example, the Grothendieck group corresponding to $\mathbb N\cup\{\infty\}$ with the obvious addition $(n+\infty=\infty)$ is trivial!

3: Concerning -ification in general, at the moment I have nothing to add to BBischof's great comments above.

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As BBischof says, the standard notion of -ification is to take the left adjoint of a forgetful functor. This includes the following as special cases:

  • The groupification of a monoid or semigroup,
  • The free group on a set, the free vector space on a set, etc.
  • The abelianization of a group,
  • The group ring of a group (the forgetful functor here sends a ring to its group of units),

and many other examples. I do not think one can reasonably talk about universal -ification without a specific choice of forgetful functor; if no good choice exists, you won't get a good notion of -ification, and on the other hand there may be more than one choice.

(Notably, I remember reading that in at least one example the natural construction is to take the right adjoint, but I don't remember what this example was.)

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    $\begingroup$ Right adjoints to forgetful functors are more often thought of as “cores”; e.g. a monoid has its “group core” of invertible elements, a category has its “groupoid core” — these contrast nicely with the group(oid)ification. (“Core” is also a handy mnemonic for the abstract-nonsense term for it: “co-reflection”.) $\endgroup$ Nov 21, 2010 at 9:32
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    $\begingroup$ The torsion subgroup of an abelian group is another nice example of a coreflection. $\endgroup$ Jan 16, 2011 at 11:00
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The name of the really really general construction (which is indeed in many cases the left adjoint of the forgetful functor) is called the completion of an object. The basic idea is to find the 'most natural' or 'smallest' object having the original as a subobject, which means we can extend the idea to, say, the closure of a subset of a topological space.

Here's the n-lab page if you're interested...

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  • $\begingroup$ a-ha! I forgot about that! I was totally excited to learn this a couple months ago. So easily we forget. $\endgroup$
    – BBischof
    Aug 16, 2010 at 20:50
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To add to Rasmus's answer to 1. One can't avoid elements like $x^{-1}y$ in the Grothendieck group. Consider say the natural numbers. It's Grothendieck group is the integers. As it's additive we write say four plus the inverse of 6 as $4-6$. Things aren't too bad in the commutative case: (in additive notation) all elements of $G(S)$ have the form $a-b$ with $a$, $b\in S$. Moreover $a-b=c-d$ iff there is $s\in S$ with $a+d+s=b+c+s$ (and we can forget about $s$ if $S$ is cancellative).

But things are nastier in the noncommutative world. Returning to multiplicative notation, you now get elements in $G(s)$ like $ab^{-1}cd^{-1}e$ which perhaps don't simplify to shorter ones. And it's not so easy to give a criterion for when two elements are the same (or equivalently when one element equals the identity). C'est la vie!

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