Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathscr{J}$ be a small category and $\mathscr{C}$ the category of sets. Suppose we have a functor $F:\mathscr{J} \to \mathscr{C}$ (so that there is an $A_i \in \mathscr{C}$ for each $i \in \mathscr{J}$).

The claim is that the (inverse) limit is given by $$\{ (a_i)_{i \in \mathscr{J}} \in \prod_i A_i:F(m)(a_j)=a_k \text{ for all } m \in \text{Hom}_\mathscr{J}(j,k)\},$$ along with the projection maps to each $A_i$ (which I denote $f_i$).

First of all I don't think it is to hard to see that $f_k = F(m) f_j$ so I am mainly concerned with showing that the universal property holds.

So assume I have another set $W$ with maps $g_i:W \to A_i$ commutating with the $F(m)$. I need to construct a map $g:W \to \varprojlim_{\mathscr{J}} A_i$

The only thing I can really think of is to appeal to the universal property of products. We have obvious maps $\pi_j:\prod_j A_j \to A_j$ and by the universal property of products these (and the $g_j$) give the existence of a map $W \to \prod_i A_i$. Can I just compose this with the natural map $\prod_i A_i \to \varprojlim_{\mathscr{J}} A_i$? (I mean this really is just the product modulo some relations)

share|improve this question
3  
You should check that the map $W \to \prod_i A_i$ takes its value in the candidate of the inverse limit (which is a subset of the product, not a quotient). –  t.b. Jan 8 '12 at 7:39
    
@t.b. Thanks. You mean that the map $W \to \prod_i A_i$ really ends up in the inverse limit? You can post this as an answer if you like and I will accept! –  Juan S Jan 17 '12 at 3:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.