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?