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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.

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I can't answer your question (give it a couple of months!) but, regarding your last question, if $\mathcal{C}$ is small then $\{ S(c,c)\, :\, c \in \operatorname{ob}\mathcal{C}\}$ is a set (by replacement), and so by completeness of $\mathcal{X}$ this has a product. So $\prod_{c \in \mathcal{C}}S(c,c)$ exists. [In fact, it exists if $\mathcal{X}$ is small even if $\mathcal{C}$ isn't.] –  Clive Newstead Sep 20 '12 at 13:55
    
I'm asking what happens if C isn't small though –  Paul Slevin Sep 20 '12 at 14:03
    
Even if $\mathcal{C}$ isn't small then $\{ S(c,c)\, :\, c \in \operatorname{ob} \mathcal{C}\}$ is a set since it is a subset of $\operatorname{ob} \mathcal{X}$ (which is a set because $\mathcal{X}$ is small). –  Clive Newstead Sep 20 '12 at 14:05
    
Sometimes it exists, sometimes it doesn't. The usual construction only works when $\mathcal{X}$ has enough limits. –  Zhen Lin Sep 20 '12 at 14:15
    
I haven't been clear... small-complete means has all small limits. forgot the dash –  Paul Slevin Sep 20 '12 at 14:26

2 Answers 2

up vote 2 down vote accepted

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.

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do you need the axiom of universes for that? –  Paul Slevin Sep 20 '12 at 15:30
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That's the easiest way to formalise things. It's somewhat troublesome to define things like $\textrm{Nat}(F, G)$ in NBG or MK class–set theory when the categories involved are large. –  Zhen Lin Sep 20 '12 at 15:32
    
How do we know Nat(F,G) is in Set? –  Paul Slevin Sep 20 '12 at 16:47
    
It isn't always. See the second example. –  Zhen Lin Sep 21 '12 at 1:10
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Cantor's theorem says $2^\kappa > \kappa$, and because $\kappa$ is strongly inaccessible, it must lie outside $V_\kappa$. –  Zhen Lin Sep 22 '12 at 10:20

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..

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This has nothing to do with set theory. If $\mathcal{X}$ is only known to have small limits then you simply can't talk about $\prod_c S(c, c)$ when $c$ varies over a non-small set (i.e. a proper class). –  Zhen Lin Sep 20 '12 at 14:17

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