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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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    In f$a$ct, th$a$t 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.2011-10-14

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