# Left adjoint for forgetful functor

I am aware that forgetful functors usually have a left adjoint, but I'm more interested in developing my technique.

Let $$\mbox{Cat}$$ be the category of small categories and $$\mbox{Set}$$ the category of sets.

Show that the forgetful functor $$U:\mbox{Cat}\to \mbox{Set}$$ has a left adjoint.

We need a functor $$F:\mbox{Set} \to\mbox{Cat}$$ such that we have an adjunction with $$F$$ as left and $$U$$ as right adjoint.

An adjunction is determined by finding an assignment $$F_0 (X) \in \mbox{Cat}_0, X\in\mbox{Set}_0,$$ and for every $$X\in \mbox{Set}_0$$ a universal morphism $$\eta _A : A\to U(F_0(X))$$ from object $$X$$ to functor $$U$$. More precisely, this universal morphism is an (the) initial object of the category $$(X\downarrow U)$$, where $$(X\downarrow U) _0 = \left\lbrace (h,\mathcal C) \mid \mathcal C \in \mbox{Cat}_0 \ \&\ h : X \to \mathcal C_0 \right\rbrace$$ and a morphism $$(h,\mathcal C) \to (k,\mathcal D)$$ is a functor $$f :\mathcal C\to \mathcal D$$ such that $$(U(f)h)(x) = k(x),\quad x\in X.$$

For a fixed set $$X$$ the initial object is surely something that involves the identity map on $$X$$. But there's a problem. We must have a (small) category whose set of objects is $$X$$. Denote this as $$\mathcal C^X$$. The universal morphism would be $$\mbox{id}_X : X\to \mathcal C_0^X = X$$.

• Question. How do we pick category $$\mathcal C^X$$?

Once we have that sorted we can define $$F :\mbox{Set} \to \mbox{Cat}$$ s.t $$F(X) = F_0(X)$$ and for a map $$v : X\to Y$$ the morphism (functor) $$F(v) : \mathcal C_X\to\mathcal C_Y$$ would have to satisfy $$U(F(v)) \eta _X = \eta _Y v \quad \mbox{i.e in this case}\quad (UF)(v) = v$$

While I am aware of something like constructing a vector space from a (non-empty) set (freeness), I don't know what this would mean for small categories. How to make progress?

• Re your first sentence: Not all forgetful functors have left adjoints, see mathoverflow.net/questions/6376/… Commented May 18, 2019 at 11:07
• @HagenvonEitzen thanks for the remark. Commented May 18, 2019 at 11:07

## 1 Answer

The left adjoint to $$U$$ is the "discrete category functor" $$F=D$$, which sends a set $$X$$ to the discrete category $$D(X)=D_X$$ with the set $$X$$ as the set of objects, and a mapping $$f\colon X\to Y$$ to the functor $$D(f)=D_f\colon D_X\to D_Y$$ between discrete categories, such that $$f$$ is the mapping on objects of $$D_f$$. It is obvious that $$D\colon\mathbf{Set}\to\mathbf{Cat}$$ is indeed a functor. Now the unit mapping of adjunction $$(D,U)$$ is simply an identity mapping $$id(X)\colon X\to U(D_X)$$.

• Is there a way to cook up this left adjoint without knowing a priori what it is? The question might be stupid, though, I could similarly ask how we can give (nontrivial) morphisms between arbitrary groups or what not. Commented May 18, 2019 at 13:28
• @AlvinLepik There are some ways to construct free objects in a very general situation (e.g. via proof of the Freyd's theorem), but I guess they are too difficult seeing our aim is to construct an adjoint to some reasonable functor as above. Commented May 18, 2019 at 13:53
• @AlvinLepik Another way is to look at the isomorphism $\hom(F(X),C)\cong hom(X,U(C))$ and recall that functors from discrete categories correspond to mappings between sets of objects. Commented May 18, 2019 at 13:54
• @AlvinLepik But personally I think that constructing a left adjoint (in such simple cases) is like a game, where you should find the easiest (and the most natural) way to construct objects of a one category by objects of another. Commented May 18, 2019 at 13:54