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Let I be a be a partially ordered set such that for any $i$; $i'$ $\in$ I, there exists $i''$ $\in$ $I$ such that $i'' > i, i'$. Let F be a functor from $I^{op}$ to finite sets. This means that for each $i \in I$, there is a finite set F(i), and for each $i''> i$ there is a map of finite sets $F(i'') \rightarrow F(i)$ such that the relation i > i is sent to the identity map and for i''> i'> i, the composition $F(i'') \rightarrow F(i') \rightarrow F(i)$ equals $F(i'') \rightarrow F(i)$. Given $i'' > i$ and $x_{i''} \in F(i'')$ , let $x_i$ denote the image of $x_{i''}$ under $F(i'') \rightarrow F(i)$. We will use the notation $x_{i''} \rightarrow x_i$.

Let $\varprojlim F \subset \prod_{i \in I} F(i)$ be defined $\varprojlim F = {(x_i)_{i \in I} : x_i \in F(i), \forall{i'} > i, x_{i'} \rightarrow x_i}$

Suppose that for each $i'> i$ the map $F(i') \rightarrow F(i)$ is surjective. Show that the projection map $\pi_{i_0} : \prod_{i \in I} F(i) \rightarrow F(i_0)$ restricts to a surjection $ \varprojlim F \rightarrow F(i_0)$

I think here, I'm just confused. Does it suffice to prove that the inverse limit is a subset of the product?

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No, it does not suffice. You need to show that starting from each element $x$ of $F(i_0)$ you can construct a point in the product which is in the limit and which projectos to $x$. You will need the various surjctivities for this. –  Mariano Suárez-Alvarez Oct 14 '11 at 5:50
    
In fact, that the inverse limit is a subset of the product is given (it's part of the definition), so it's not even something that you need to prove. –  Arturo Magidin Oct 14 '11 at 16:36
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