Indeterminates and forgetful functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:28:20Z http://mathoverflow.net/feeds/question/50410 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50410/indeterminates-and-forgetful-functors Indeterminates and forgetful functors Zhen Lin 2010-12-26T09:46:26Z 2010-12-26T13:01:58Z <p>Forgive me if this is a well-known observation/result, but I'm quite new to graduate-level algebra and I was wondering if there are generalisations of the constructions I describe below.</p> <p>It's straightforward to see that the functor "adjoin an indeterminate to a ring" is the unit of the adjunction $F : \mathbf{Ring} \to \mathbf{Ring}_* : U$, where $U: \mathbf{Ring}_* \to \mathbf{Ring}$ is the forgetful functor from the category of pointed rings to the category of rings. </p> <p>Along similar lines, the functor "disjoint-union an element to a set" is the unit of the adjunction $F : \mathbf{Set} \to \mathbf{Set}_* : U$, and if I'm not mistaken, the functor "coproduct with $\mathbb{Z}$" is the unit of the adjunction $F : \mathbf{Grp} \to \mathbf{Grp}_* : U$. Indeed, in general, it seems that if a category has a notion of "free object on one generator", a notion of "pointed" objects, and binary coproducts, then the forgetful functor from the category of pointed objects has a left adjoint, and the unit of the adjunction is the functor which takes objects to their coproduct with the free object on one generator. </p> <ol> <li>Does this construction generalise when binary coproducts don't exist?</li> <li>What about when the required free object doesn't exist?</li> <li>Is there a (even) more general way to describe this operation of "adjoining an indeterminate", e.g. when there's no notion of pointed objects?</li> </ol> http://mathoverflow.net/questions/50410/indeterminates-and-forgetful-functors/50420#50420 Answer by Todd Trimble for Indeterminates and forgetful functors Todd Trimble 2010-12-26T13:01:58Z 2010-12-26T13:01:58Z <p>In your examples, there is an evident underlying functor $U: C \to Set$ which allows you to talk about points as elements of the underlying set. So a pointed $C$-object would be an object of the comma category $1 \downarrow U$. There is an evident projection $\pi: 1 \downarrow U \to C$. A left adjoint to this functor would be the process of adjoining an indeterminate. </p> <p>A category $C$ equipped with an "underlying" functor $U$ to $Set$ is often called a <i>concrete category</i>, especially when $U$ is faithful (some authors drop this assumption). So concrete categories admit a notion of pointed object, and a notion of adjoining an indeterminate if the associated $\pi$ has a left adjoint $(-)[x]$. A priori this makes sense without the need to mention binary coproducts in $C$ or the free object on one element, although it seems to me instances of that would be somewhat artificial. (For example, if $C$ has an initial object $0$, then $0[x] = 1 \to Uc$ would exhibit $c$ as the free object on one element, precisely because $0[x]$ is initial in $1 \downarrow U$.) </p>