# What is the difference between a presheaf and a contravariant Hom functor

I'm currently delving into category theory and came across two concepts that appear similar: presheaves and contravariant Hom-functors. Both of these constructs are associated with the mapping from $$\mathbb{C}^{op}$$ to Set, which underlines their apparent similarity.

Both a presheaf and a contravariant Hom-functor $$Hom(\_, A)$$ have the form of $$F: \mathbb{C}^{op} \rightarrow Set$$, and Wikipedia states that a contravariant Hom-functor is a presheaf, meanwhile other pages states that presheaf that is naturally isomorphic to the $$Hom(\_, A)$$.

The situation is similar with copresheaves and covariant Hom-functor $$Hom(A,\_)$$, which both mapping from $$\mathbb{C}$$ to $$Set$$.

What is it that sets a presheaf and a contravariant Hom-functor apart, given that both can be seen as contravariant functors from $$\mathbb{C}^{op}$$ to $$Set$$?

Despite this understanding, I find myself somewhat muddled when trying to pin down the differences between a general presheaf and a contravariant Hom-functor. Also both Wikipedia and nLab mentioned that,

A functor $$F : C → Set$$ that is naturally isomorphic to $$Hom(A, \_)$$ for some $$A$$ in $$C$$ is called a representable functor

Functors of the form $$C^{op}→Set$$ are called presheaves on $$C$$, and functors naturally isomorphic to hom(-,C) are called representable functors or representable presheaves on $$C$$.

These actually made me more confused. Could anyone provide a clearer or more intuitive way to understand this distinction? Is there any functor that takes $$C → Set$$ is not a Hom-functor? Also, is a presheaf/Hom-functor itself a $$Set$$? I really can't imagine that.

• A presheaf is exactly the same as a contravariant functor. A Hom-functor is a special kind of functor. They (and everything isomorphic to any of them) are also called representable functors or representable presheaves. Aug 1, 2023 at 15:47

For an example of a presheaf which is not representable, let $$\mathscr{C} = \mathbf{Grp}$$ be the category of groups and let $$F:\mathscr{C}^{\text{op}} \to \mathbf{Set}$$ be the functor which sends a group to the empty set $$\emptyset$$ and sends a morphism $$f:G \to H$$ in $$\mathbf{Grp}^{\text{op}}$$ to $$\operatorname{id}_{\emptyset}$$. Then for this to be a representable presheaf you would need to be able to find a group $$G$$ such that for any group $$X$$ there is an isomorphism $$\emptyset = F(X) \cong \mathbf{Grp}(X,G)$$ which is natural in $$\mathbf{Grp}^{\text{op}}$$. However, as every pair of groups $$X$$ and $$Y$$ have the trivial map $$X \to \lbrace \ast \rbrace \xrightarrow{\ast \mapsto 1_Y} Y$$ (and so there is a map $$Y \to \lbrace \ast \rbrace \to X$$ in the opposite category) we have that $$\mathbf{Grp}(X,Y) \ne \emptyset$$ for all groups $$X$$ and $$Y$$. Consequently it cannot be the case that $$F$$ is representable.
Addenda: I think what may help you in your confusion is writing down some examples of presheaves and hom functors in order to build your intuition a bit. The trick is that while a presheaf $$F$$ is not a set (it is a functor valued in the category $$\mathbf{Set}$$ of sets), for any object $$X$$ of your base category $$\mathscr{C}$$ the value $$F(X)$$ is a set. However, $$F$$ comes with more information: it also has functions $$F(f):F(X) \to F(Y)$$ whenever $$f:X \to Y$$ is a morphism in $$\mathscr{C}$$. A nice way to think about what a presheaf is comes down to having a tool with which you can use sets to study and understand $$\mathscr{C}$$.
By the way: A handy trick to try and build non-representable presheaves is to use constant presheaves, i.e., presheaves $$F$$ with the property that there is a set $$S$$ for which $$F(X) = S$$ for all objects $$X$$ and $$F(f) = \operatorname{id}_S$$ for all morphisms $$f$$. Then for $$F$$ to be representable you need to have an object $$Y$$ such that $$\mathscr{C}(X,Y) \cong S$$ for all objects $$X$$, which is a big ask!
• Thank you for the clarification. For the exercise, I think since $id_{S}$ is unique, but for any $X \neq Y$, there are more than 1 morphism in $\mathscr{C}(X,Y)$, so $F$ is not presentable. Aug 3, 2023 at 12:30
• @whymgang I'm not quite sure what you mean by the exercise, but if you mean the paragraph I was just suggesting a general strategy for finding non-representable presheaves. Determining if something is representable or not will generically depend heavily on both the category $\mathscr{C}$ you're studying and the set $S$ you feed it. For instance, if you want $S = \lbrace 0 \rbrace$ then the constant presheaf at $S$ is representable if and only if $\mathscr{C}$ has a terminal object. Also remember that you want an isomorphism with $S$ not equality. Aug 3, 2023 at 14:17
• Also what is $\ast$ mean in your first example $X \to \lbrace \ast \rbrace \xrightarrow{\ast \mapsto 1_Y} Y$, can we just say $X \to Y$ always exists? Aug 4, 2023 at 1:29
• So if $\mathscr{C}$is a field or empty set that contains no initial or terminal objects, and $\mathbf{S}$ is a constant , then $[\mathscr{C}^{op},\mathbf{S}]$ is not representable right? Aug 4, 2023 at 1:32