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Let us denote the set of all objects of a small complete category by $C^{\bullet}$. My question is concerned with the limit of the diagram $$C^{\bullet} \longrightarrow C$$ which sends every morphism of $C^{\bullet}$ which they all happen to be identities, to the identities. What kind of object is the limit (or colimit for that matter) of the above diagram. For example the category of finite sets doesn't have the product of all of its objects. Perhaps I must look for more peculiar categories than FinSet for meeting such a beast.

Thanks.

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What is the category of finite products? Maybe you wanted to say the category of finite sets? –  Oskar May 1 '13 at 19:26
    
My bad. corrected and thanks. –  Hooman May 1 '13 at 19:29
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I'm not sure what you mean. Often the limit just doesn't exist. When it does, it's the product of all of the objects. –  Qiaochu Yuan May 1 '13 at 19:34
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For example, if $C$ is the category given by a Poset, then such an object would have to be a unique minimal element of the category. So posets with no minimal element or multiple minimal elements would not yield a category with such a product. –  Thomas Andrews May 1 '13 at 19:41
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$\text{FinSet}$ is not small complete. –  Qiaochu Yuan May 1 '13 at 21:59

3 Answers 3

up vote 6 down vote accepted

Here is a relevant result showing that such categories are probably rare. Freyd showed that if a small category has all small limits, then it must be a preorder. So we can more or less reduce to the case that $C$ is a poset, in which case the product of all of the objects is a smallest element (if it exists).

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Having the product of all objects is not the same as having all limits. FinSet has the product of all objects; the empty set. (It doesn't on the other hand have the product of all non empty sets). –  aws May 1 '13 at 20:13
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I thought that at first, but actually it's not the case in general. It's not true for $\mathbf{FinSet}^{op}$! –  aws May 1 '13 at 20:19
    
@aws: oh, double hmm. I'll have to think about this some more. –  Qiaochu Yuan May 1 '13 at 20:20
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For Freyd's theorem to go through one just needs to be able to take very large powers of every object. –  Zhen Lin May 1 '13 at 21:01

I would dispute the above answers that we can (mostly) restrict our attention to preorders; for example, the category $\mathsf{FinSet}$ does have a product of all objects, and it is simply the empty set with empty projections. Similarly, $\mathsf{Set}$ and $\mathsf{Top}$ contain products of all objects, and many more similar examples can be given.

Edit: Ah, aws posted this a minute before me; I did not see the comment in time.

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This is not an answer. If you have criticism of another answer, you should comment on that answer. Of course, you need 50 reputation to comment on someone else's post. –  robjohn May 1 '13 at 20:58
    
On the other hand, the product of all non-empty finite sets does not exist in $\mathsf{FinSet}$. –  Martin Brandenburg May 3 '13 at 16:05

A simple example: let $X$ be a set, $P_X$ be the preorder of its subsets, ordered by inclusion. Then the product of all objects in $P_X$ will be $\varnothing$(empty subset of $X$), and coproduct of all objects in $P_X$(which is colimit of your diagram) will be $X$(which is also subset of $X$).

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