I am working on the construction of a group free over a monoid. Maybe you know where I can find something about the (right) adjointness of forgetful functor $U:\mathbf{Grp}\rightarrow\mathbf{Mon}$. I called it forgetful functor here, but it is clear to me that you could also look at it as an inclusion functor.

It is (right) adjoint isn't it? If not then I am working in vain. I do not exclude the possibility that this question is a duplicate.


Adjoint is ambiguous. Left or right adjoint? In the case of $U : \mathsf{Grp} \to \mathsf{Mon}$, we have a left and a right adjoint (this is quite rare for forgetful functors). The right adjoint maps a monoid to its group of units. The left adjoint "adjoins" inverses to a monoid in order to obtain a group. It is (at least in the commutative case) known as the Grothendieck construction. If $M$ is given as a monoid by generators $X$ and relations $R$, then the Grothendieck group $G(M)$ is given by the same generators $X$ and relations $R$, but now as a group. For example, $\mathbb{N}$ is the free monoid on one generator, hence $G(\mathbb{N})$ is the free group on one generator, i.e. $G(\mathbb{N})=\mathbb{Z}$.

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    $\begingroup$ Do you happen to know of there exists a similar construction for categories (gneeralized monoids) and groupoids (generalized groups)? $\endgroup$ Dec 13 '13 at 12:35
  • $\begingroup$ I was talking about right-adjoint and will make an edit to make that clear. Is there a place where this construction is exposed @StefanHamcke? However do not feel obliged to search for it. $\endgroup$
    – drhab
    Dec 13 '13 at 12:42
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    $\begingroup$ Yes, $\mathsf{Gpd} \to \mathsf{Cat}$ has a right adjoint, it maps a category to its core: This is the subcategory containing all the isomorphisms of the given category. It also has a left adjoint, the localization at all morphisms. In order not to run into set-theoretical troubles, one has to work with small categories and groupoids here. $\endgroup$ Dec 13 '13 at 12:46
  • $\begingroup$ You write: 'The left adjoint "adjoins" inverses to a monoid in order to obtain a group.' Sounds pretty cool, but how can this be possible if the monoid doesn't satisfy the (two-sided) cancellation law? $\endgroup$ Apr 5 '14 at 12:08
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    $\begingroup$ It is always possible, by abstract reasons. If you have no cancellation, the unit (the "inclusion" to the Grothendieck group) just won't be monic. $\endgroup$ Apr 6 '14 at 22:11

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