# Properties of Group $\to$ Monoid $\to$ Semigroup.

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.

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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

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 $f(g^{-1}) = f(g)^{-1}$.
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 qchu.wordpress.com/2011/01/21/structures-on-hom-sets and qchu.wordpress.com/2013/06/09/operations-and-lawvere-theories 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