# Intuition on the Representable Functor

Given a locally small category C, and an object $C$, the functor:

$$\mbox{Hom}_\textbf{C}(C,-):\textbf{C} \longrightarrow \textbf{Sets}$$

that sends objects to hom-sets and arrows $f:A\rightarrow B$ to functions: \begin{align} f_*:\mbox{Hom}_\textbf{C}(C,A) &\longrightarrow \mbox{Hom}_\textbf{C}(C,B) \\ g&\longmapsto f\circ g \end{align}

is called the Representable Functor.

I am looking for some intuition of why is that so, past the fact that it is an obvious functor from C to Sets.

• It is just a name. We also say that $C$ represents the functor $\mathrm{Hom} (C, -)$. – Zhen Lin Jul 23 '15 at 11:24
• You should really read Tom Leinster's "Basic Category Theory", because it answers such motivational questions in great detail! – Martin Brandenburg Jul 23 '15 at 13:04

First of, a functor $F : \mathscr C → \mathrm{Set}$ is called representable (by $C$) if it's isomorphic (not necessarily equal) to $\mathrm{Hom}(C, -)$ for an object $C$ of $\mathscr C$. As for the term, an abstract functor $F$ is represented by the very concrete action of $\mathrm{Hom}(C, -)$.
Take for example the functor $L : \mathrm{Top} → \mathrm{Set}$ sending a topological space $X$ to the set of all loops in $X$. A loop is just a continuous function $S^1 → X$, so we have that $LX ≅ \mathrm{Hom}(S^1, X)$. In fact, it kind of makes sense to say that a loop is an $S^1$-shaped element of $X$, right? This really generalizes the classical elements of $X$, since those correspond to functions $* → X$, where $*$ is the one-point space. Note in particular that every continuous function $f : X → Y$ extends to these generalized elements: if $l : S^1 → X$ is a loop, then it's image in $Y$ is exactly $f∘l = \mathrm{Hom}(S^1, f)(l)$. Now compare: we have no idea how an arbitrary functor $F: \mathrm{Top} → \mathrm{Set}$ might act. But if $F$ is representable and $F ≅ \mathrm{Hom}(C, -)$, then we know that $FX$ are just $C$-shaped elements of $X$, and that $Ff$ simply maps the $C$-elements of $X$ to $C$-elements of $Y$ in the most obvious way. So you represented something possibly completely abstract with a very simple idea.
For contravariant functors, a somewhat higer-level perspective would be that the Yoneda embedding $C ↦ \mathrm{Hom}(-, C)$ embeds $\mathscr C$ into the functor category $[\mathscr C^\mathrm{op}, \mathrm{Set}]$. Looking at it that way, you could say that a representable contravariant functor $F$ literally is (isomorphic to) the element of $\mathscr C$ it's represented by.
• It's $C \mapsto \hom(-,C)$ which embeds $\mathscr C$ into $[\mathscr C^{\rm op},\mathsf{Set}]$. – Pece Jul 23 '15 at 11:47
• Maybe it could be a good idea to mention that when a functor is representable, there are usually many ways for it to be represented (different possible isomorphisms $F \cong \hom(-,a)$ corresponding to elements of $F(a)$)? Like when a manifold is orientable, there are two ways (over the integers anyway) to make it oriented? – Najib Idrissi Jul 23 '15 at 13:07