Let $C$ be a category, and $o\in C$ an object. Does $o$ have a universal property? My first reaction: of course it has:
$$E_o(x):\Leftrightarrow \hom(x,-)\cong\hom(o,-).$$
is the universal property of $o$ in the sense that for any object $a\in C$, if $E_o(a)$, then $a\cong o$ (and of course $E_o(o)$). So this property determines the object $o$ up to isomorphism.
But I just noticed that wikipedia uses another definition of universal property: a universal property is given by a functor $F\colon C\to D$ and an object $X\in D$. We say that an object $o\in C$ equipped with an arrow $u\colon X\to F(o)$ satisfies the universal property determined by $(F,X)$ if for any object $o'\in C$ and any morphism $f\colon X\to F(o')$ in $D$ there is a unique morphism $h\colon o\to o'$ such that $F(h)\circ u=f$. Of course, the universal property given by $(F,X)$ pins down $o$ up to isomorphism.
Let's ask the question above again, but this time for this new definition: given any object $o\in C$ is there an arrow $u\colon X\to F(o)$, a functor $F\colon C\to D$, and an object $X\in D$ such that the universal property corresponding to $(F,X)$ is satisfied by $(o,u)$? (Bonus question: is it possible to find such a pair $(F,X)$ for any pair $(o,u)$, i.e., if $u$ is given additionally to $o$?)
Best Answer
Yes, any object $X \in \mathcal{C}$ represents the functor $\mathrm{Hom}(X,-) : \mathcal{C} \to \mathbf{Set}$ and therefore has a (trivial) universal property.