The existence of ends of functors.

Question

Let $\mathcal {C}$ be a small category. In MacLane's book we have a theorem:

If $\mathcal X$ is small complete and $\mathcal C$ is small, then every functor $S \colon \mathcal C^{\text op} \times \mathcal C \to \mathcal X$ has an end in $\mathcal X$.

Can we drop the "$\mathcal C$ is small" from this? I am skeptical. I am reading a paper just now which has this statement without mentioning smallness and it is also mentioned on wikpedia (http://en.wikipedia.org/wiki/End_(category_theory)). I don't see why from the assumption that $\mathcal X$ is complete we can assume that $\prod_{c \in \mathcal C} S(c,c)$ exists.

Thanks for any help.

Answer 1

Here is an example of an end that exists even though the indexing category is big. Let $\mathcal{S} = \textbf{Set}$, and let $H = \mathcal{S}(-, -) : \mathcal{S}^\textrm{op} \times \mathcal{S} \to \mathcal{S}$ be the hom functor. I claim $$1 \cong \int_{X : \mathcal{S}} H(X, X)$$ Indeed, by the end form of the Yoneda lemma (or direct verification), it is known that $$\textrm{Nat}(F, G) \cong \int_{c : \mathcal{C}} \mathcal{D}(F c, G c)$$ for all functors $F, G : \mathcal{C} \to \mathcal{D}$; but the ordinary Yoneda lemma says $$\textrm{Nat}(\textrm{id}_{\mathcal{S}}, \textrm{id}_{\mathcal{S}}) = \{ \textrm{id}_{\textrm{id}} \}$$ so the claim follows. I think $\mathcal{S}$ can be replaced with any well-pointed topos in this argument.

Of course, there are also ends that don't exist. Let $\mathcal{C}$ be the disjoint union of $\kappa$ copies of a non-trivial group $G$, where $\kappa$ is the size of the universe, and let $K = \mathcal{C}(-, -) : \mathcal{C}^\textrm{op} \times \mathcal{C} \to \textbf{Set}$ be the hom functor again. This makes sense because $\mathcal{C}$ is locally small by construction. Since $\mathcal{C}$ is very nearly discrete, it is easy to compute $\textrm{Nat}(\textrm{id}_\mathcal{C}, \textrm{id}_\mathcal{C})$: it is just the $\kappa$-fold power of $G$. In particular, it is not a small set, so the end of $K$ does not exist.

Answer 2

As I interpret, $\mathcal X$ needs not be small, but $\mathcal C$ yes, because we cannot index a product by a proper class.. - at least I doubt if the definition of completeness would apply to 'proper class' diagrams. As the product can be given as a limit of a diagram (of set size), by completeness your product $\prod_c S(c,c)$ must exist.

Anyway, this is the situation only for the ZF[C] set theory, but there could be some other base set theories to work in (like some New Foundation theories -- http://en.wikipedia.org/wiki/New_Foundations), where this kind of questions don't cause any problem at all..