Inverse limit of finite sets with surjective maps has surjective projection?

Question

I'm doing some exercises about inverse limits and got stuck in the following:

Exercise

Let $(S_{i},\pi_{ji})$ be an inverse system of finite sets with all $\pi_{ji}$ surjective and let $S=\varprojlim S_{i}$. Use the fact that the inverse limit of an inverse system of nonempty compact Hausdorff spaces is itself nonempty compact Hausdorff to prove that all maps $\pi_{i}\colon S\to S_{i}$ determined by the $\pi_{ji}$ are surjective.

My attempt

First, I quickly note that the notation $\pi_{ji}$ here is for the map $\pi_{ji}\colon S_{j}\to S_{i}$, defined whenever $i\leq j$.

Okay. We start using the given fact to justify that the inverse limit $S$ is nonempty, since equipping each given finite set (assuming nonempty) with the discrete topology gives the compactness and Hausdorff hypotheses.

Now we need to prove that given $s_{i}\in S_{i}$, we can find $s\in S$ such that $\pi_{i}(s)=s_{i}$.

We know that given such $s_{i}$, for every $j\geq i$, we know that there is $s_{j}\in S_{j}$ such that $\pi_{ji}(s_{j})=s_{i}$.

Well it would then suffice to take an element that projects to $s_{j}$ in $S_{j}$ for one of these $j's$.

However I'm not sure why this is supposed to happen.

Also, when he says in the end "... the maps determined by the $\pi_{ij}$ are surjective.", what does determined mean? I believe it is specifying these $\pi_{i}$ because they are (are they?) the only maps that satisfy that commutative diagram property with the $\pi_{ji}$'s.

Any help, tips or comments are greatly appreciated. Thank you very much.

EDIT Just found a similar question on the sidebar, but it is a little different since there the indexing set is the positive integers and here is totally arbitrary. The answer also invokes some set-theorical induction which I know nothing about, so I'm keeping this question open in case there is a possible answer that the other guy missed. Also, I see how "induction" would work, and my argument makes that clear. However, the 'initial' induction step is the trouble here, of course.

Answer 1

First of all: "determined" does exactly mean what you think it means. The $\pi_i$ are the maps that make the respective diagramms for the inverse limit commutative. Now for a proof. We are in the category of sets so we can view the inverse limit as

$$ S=\varprojlim S_i=\left\{(s_i)_i \in \prod_{i\in I}S_i|s_i=\pi_{ji}(s_j) \forall j < i \in I \right\}, $$

where the $\pi_i$ are given by the obvious projections. Equip the $S_i$ with the discrete topology so that they become compact Hausdorff spaces.

We want to prove that there is a $s \in S$ s.t. $\pi_i(s)=x_i$ for a given $x_i\in S_i$. Therefore define for $j>i$:

$$ T_j:=\left\{(s_l)_l \in \prod_{l\in I}S_l| s_i=x_i \text{ and } s_n=\pi_{nj}(s_j) \text{ for all } n <j \right\} \subset \prod_{l\in I}S_l. $$

Now note a few things:

1. Since all the $S_l$ are compact so is $\prod S_l$. This is Tychonoffs's theorem.
2. The $T_j$ are closed in the product topology, since the $S_j$ are Hausdorff.
3. The intersection $\cap_{j>i} T_j$ is a subset of $S$ and every element in this intersection has $i$-coordinate $x_i$.

So it remains to show that the intersection $\cap_{j>i} T_j$ is non empty. Since the $T_j$ are closed and using the characterization of compactness by the finite intersection property it is enough to show that every finite intersection of $T_j$ is nonempty. So let $i_1,\ldots,i_n \subset I$ with $i_j>i$. Now since $I$ is a indexed set there is a $k> i_1,\ldots,i_n$. and therefore

$$T_k\subset T_{i_1}\cap\ldots\cap T_{i_n}. $$ And since all the maps in the inverse system are surjective we have that $T_k$ is non empty. Which precisely says that finite intersections are non empty.