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for forgetful functors, we can usually find their left adjoint as some "free objects", e.g. the forgetful functor: AbGp -> Set, its left adjoint sends a set to the "free ab. gp gen. by it". This happens even in some non-trivial cases. So my question is, why these happen? i.e. why that a functor forgets some structure (in certain cases) implies that they have a left adjoint? Thanks.

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up vote 32 down vote accepted

Forgetful functors usually have a left adjoint because they usually preserve limits. For example, the underlying set of the direct product of two groups is the direct product of the underlying sets, and similarly for equalizers (that gives you all finite limits).

However, functors that preserve limits don't have to have left adjoints, because once in a while what you want to do to construct a free object results in a proper class. An example is complete lattices. Freyd's Adjoint Functor Theorem gives a necessary and sufficient condition for a limit-preserving functor to have a left adjoint. The proof and related results is discussed in section 1.9 of Toposes, Triples and Theories.

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Forgetful functors usually preserve limits because they are usually representable. – KotelKanim Oct 22 '11 at 16:59
In fact, if the forgetful functor is $\mathsf{Set}$-valued, then it has a left adjoint if and only if it is representable. For base categories other than $\mathsf{Set}$, it's not quite so clear what it means to be representable. – Tim Campion Jan 8 '15 at 17:21

Not every forgetful functor has a left adjoint. If you take comodules over a k-coalgebra A, the forgetful functor to k-modules has a right adjoint, not a left adjoint. So you have cofree comodules, of the form A tensor_k V, but in general no free comodules. So I would argue that forgetful functors will have left adjoints when your structures are defined by maps into your object (like A tensor M into M), but right adjoints when your structures are defined by maps out of your object (like M into A tensor M). This is a universal algebra-style answer, not perhaps so helpful for non-algebraic situations.

But let me point out that the forgetful functor often has both adjoints in non-algebraic situations. For example, the forgetful functor from topological spaces to sets has left adjoint defined by the discrete topology, and right adjoint defined by the indiscrete topology.

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I guess a more opaque way to answer the question along these lines is to say that forgetful functors usually have left adjoints because we (meaning mathematicians in general) usually work with monadic algebraic structures. – S. Carnahan Nov 21 '09 at 20:49

Many standard examples of algebraic "forgetful" functors $U : C \to \mathrm{Set}$ have the following form:

  • $C$ is a presentable category, i.e., there is a small category $I$ and a collection $S$ of cones of $I$ such that $C$ is equivalent to the full subcategory of functors $I \to \mathrm{Set}$ consisting of those functors which send the cones of $S$ to limit diagrams in $\mathrm{Set}$;
  • $U$ is evaluation at an object $u \in I$.

For example, if $C$ is the category of monoids, take $I = \Delta^{\mathrm{op}}$ so that functors $I \to \mathrm{Set}$ are simplicial sets and choose $S$ so that the objects of $C$ are those simplicial sets $X$ such that $X_0 = \ast$ and $X_{i+j} \to X_{i} \times X_{j}$ is an isomorphism (where this map is induced by the inclusions of the first $i+1$ and last $j+1$ elements of an ordered $i+j+1$ element set). The object $u$ is the two-element set $[1]$. (One actually needs only the full subcategory of $\Delta^\mathrm{op}$ on the objects $[0]$, $[1]$, $[2]$, $[3]$, and the cones involving these objects; expanding this gives a possibly more familiar presentation of the notion of monoid.)

In these cases (which include models of any essentially algebraic theory) the existence of a left adjoint is guaranteed by the theory of presentable categories. Indeed, the inclusion of $C$ into $\mathrm{Set}^I$ has a left adjoint which we compose with the constant diagram functor $\mathrm{Set} \to \mathrm{Set}^I$ to obtain a left adjoint to $U$. See Adamek and Rosicky, Locally presentable and accessible categories, for an excellent introduction to the subject.

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The term "forgetful functor" is not perfectly well defined. Depending on context, I've seen it defined as "faithful functor with a left adjoint", because most notions of "Forget" should have a corresponding notion of "Free".

Edit: I should emphasize that there are many notions of "forgetful functor", and it is not a canonically-defined word. J. Baez has thought about what the right notions of "forget" are. For him, a functor "forgets at most properties" if it is full and faithful, "forgets at most structure" if it is faithful, and "forgets at most stuff" if it is (empty list). I think it is not necessarily true that a full and faithful functor has a left adjoint.

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agreed.. so far, I haven't seen any reference that really rigorously defines what is category-therotically meant by a "forgetful functor". – Jose Capco Nov 21 '09 at 21:24
I don't think I've ever seen it defined as such, but it has surely been strongly hinted at. – alekzander Nov 23 '09 at 1:47
I strongly object to the notion that a forgetful functor ought to be faithful, and I also strongly object to the notion that a forgetful functor ought to have a left adjoint... – goblin Apr 28 at 8:52

Many forgetful functors in an algebraic setting are representable; for example, the forgetful functor $\text{Ab} \to \text{Set}$ is represented by $(\mathbb{Z}, 1)$. Functors with left adjoints are also representable, and sometimes the converse holds.

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the answer of reid barton is perfect and general. for categories of $\tau$-algebras, where $\tau$ is a type (thus consisting of function symbols and identities), we also have the following: let $\tau \to \sigma$ be a homomorphism of types, then there is a functor from $\sigma$-algebras to $\tau$-algebras and it has a left adjoint. this can be proved using freyd's adjoint functor theorem. this yields tons of examples:

  • the forgetful functor of $\tau$-algebras to sets has a left adjoint. in particular, free monoids, groups, modules, lie algebras etc. exist.
  • if $R \to S$ is a ring homomorphism, $S-Alg \to R-Alg$ has a left adjoint.
  • every ring has a free unital ring.
  • forgetful functors of algebra may be seen as the functors from $\sigma$-algebras to $\tau$-algebras, where $\tau \subseteq \sigma$ is a subtype; these have a left-adjoint. for example, from groups to monoids, from rings to abelian groups, from R-algebras to R-modules, and so on.

basically, it's all about representing functors, and since subobjects of Set are well-behaved in a certain sense, freyd's adjoint functor theorem tells you that you can do everything you want.

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I think the "left adjoint"/"right adjoint" thing is mostly just terminology, although probably now that I've said that someone's going to point me to a n-Cafe post or a SBS where they explain that it's deeply tied to TQFT or something.

I asked a question a couple weeks back, the answers to which helped me understand why so many functors did have left adjoints.

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In what way do you mean "mostly just terminology"? As Mark Hovey's topological example (an exercise in MacLane, also) points out, one may have both a left and a right adjoint which are clearly distinct. – alekzander Nov 23 '09 at 1:46
Erm, I meant that, as far as I know, the reason we call left adjoints left and right adjoints right is basically entirely historical and has much more to do with arbitrary mathematical notation than any real difference between them. (I mean, yes, they're different, but not that different.) – Harrison Brown Nov 23 '09 at 3:05
If you're just talking about the adjectives left/right, well .. that's terminology, sure, since you're talking about .. terms. There was some usage historically of "adjoint"/"coadjoint", but it's unnatural AFAIAC to say that one is in the "correct" (I guess I have to avoid the word "right" here) direction. However, we do have "unit"/"counit", so it wouldn't be a stretch to associate "adjoint" with "unit", and co-. I think this isn't done mostly because it was historically not used consistently this way (or that's what I gleaned from a comment in MacLane). – alekzander Nov 23 '09 at 9:22

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