Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In How to get a group from a semigroup Arturo Madigin writes:

So if we look at the composition of the forgetful functors $\bf Group \to \bf Monoid \to \bf Semigroup$, we obtain a right adjoint by composing the adjoints going the other way, $\bf Semigroup \to \bf Monoid$ (adjoin a 1), and $\bf Monoid \to \bf Group$ (enveloping group). So: first adjoin a 1, then construct the enveloping group.

What are the properties of the composed functor: $F: \bf Group \to \bf Semigroup$? In particular is it full and/or faithful? Or are there multiple inclusion functors from which a particular one needs to be specified?

Faithful functor: $\forall (f,g: A \to B), Ff = Fg \implies f = g$

Full: $\forall (h: FA\to FB), \exists (f: A \to B):Ff = h$

As an FYI, as magma cites from Adamek & Herrlich's Abstract and Contrete Categories, a fully faithful functor reflects isos, sections, and retracts. (I assume it also preserves these properties). That may make it easier to think about it.

The context for this question is to compare and contrast the embedding of groups into semigroups versus metric spaces in topological spaces.

share|cite|improve this question
Every functor preserves isomorphisms, sections, and retractions because all three of these things are defined entirely in terms of composition of morphisms. – Qiaochu Yuan Jan 21 '13 at 20:58
there is some strange problem with this page: I am trying to upvote the question, but I cannot do it. Looks like the title interferes with the up-arrow. Also the link to Arturo's answer does not work properly even if it is pointing correctly. Anybody else has this problem? – magma Jan 21 '13 at 23:14
@magma, hm, that's unusual. Have you tried a different browser? – alancalvitti Jan 22 '13 at 22:23
yes It is browser dependent. In opera it works ok. In Firefox it is as if the title lived on a large transparent button covering arturo's link and the upvote arrow. Maybe there is some problem with the mathml in the title. anyway I managed to upvote the question in Opera. – magma Jan 23 '13 at 15:31
up vote 8 down vote accepted

The forgetful functor from groups to semigroups is fully faithful. This is equivalent to the statement that a group homomorphism $f : G \to H$ between two groups is the same thing as a semigroup homomorphism. To see this, first suppose that $f$ is a semigroup homomorphism. Then $f(e) = f(e^2) = f(e)^2$ (where $e$ is the identity in $G$), from which it follows by cancellation that $f(e)$ is the identity in $H$. Second, $f(e) = f(g^{-1} g) = f(g^{-1}) f(g)$, from which it follows by the uniqueness of inverses that $f(g^{-1}) = f(g)^{-1}$.

(Note that it is not true that the forgetful functor from monoids to semigroups is full. The problem is that the identity of a monoid need not be sent to the identity under a semigroup homomorphism, but just to an idempotent.)

The properties of the forgetful functor from metric spaces to topological spaces depends on what category of metric spaces you're using. A common choice is to take metric spaces and continuous functions, but this is the "wrong" notion of morphism for metric spaces (it's the correct notion of morphism for metrizable spaces); you can't recover the metric from an isomorphism class in this category. If you want to recover metrics from isomorphism classes, the "correct" notion of morphism between metric spaces is a weak contraction.

share|cite|improve this answer
Thanks for the derivations. Interesting that the embedding factors through a non-full functor. – alancalvitti Jan 23 '13 at 1:04
Re proper morphisms in metric spaces, it depends on application area. In data analysis, metric spaces are finite, and I argue here… that the interesting arrows are (neighborhood) graph preserving, to enable testing sensitivity of results on the choice of metric. – alancalvitti Jan 23 '13 at 1:07
If we take weak contractions as our morphisms between metric spaces, how do we recover the metric? I've read that we cannot recover the group operation from an entity of the concrete category $\mathrm{Grp}$, because we cannot distinguish a group from its opposite. – goblin Oct 11 '13 at 8:06
@user: by "recover the metric" I mean "recover the isometry class" and that's clear because in the weak contraction category all isomorphisms are isometries. We absolutely can recover the group operation from an object in $\text{Grp}$; see and for some indication of how to do this. – Qiaochu Yuan Oct 20 '13 at 5:21
@user: to recover the actual metric, consider weak contractions $f : M \to \mathbb{R}$ and set $d'(x, y) = \sup_f |f(x) - f(y)|$. This recovers the metric, since on the one hand $d'(x, y) \le d(x, y)$ and on the other hand we can take $f(x) = d(x, y)$. (This is some kind of Yoneda lemma for metric spaces.) – Qiaochu Yuan Oct 20 '13 at 5:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.