Proving $\varprojlim S_i \ \cong \ \varprojlim S_j $ where $J \subseteq I$ cofinal.

Question

Let $I$ be a directed poset and $J \subseteq I$ cofinal. Let $(S_i)_{i \in I}$ be a collection of sets, with a collection of maps $$\{ f_{ij} \ : \ S_i \rightarrow S_j \ | \ i,j \in I \text{ and } i \geq j \}$$ that satisfies the usual commutation relations. Prove that there is a canonical bijection: $$ \varprojlim_{i \in I} S_i \ \cong \ \varprojlim_{j \in J} S_j $$

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My own efforts

In what follows I will drop certain subscripts to maintain a readable notation.

The limit $\varprojlim S_i$ is a subset of $\prod_{i \in I}S_i$, and the limit $\varprojlim S_j$ is a subset of $\prod_{j \in J}S_j$. This means that we already have a projection map $$ \pi \ : \ \prod S_i \ \longrightarrow \ \prod S_j \ : \ (x_i)_{i \in I} \ \longmapsto \ (x_j)_{j \in J} $$ that we can restrict to $\varprojlim S_i$ . It is easy to show that $\pi \left( \varprojlim S_i \right) \subseteq \varprojlim S_j$, hence we obtain a well-defined restriction $\rho$ of $\pi$: $$ \rho \ : \ \varprojlim S_i \ \longrightarrow \ \varprojlim S_j $$ All we need to show is to find an inverse $\rho^{-1}$. We need to "extend" elements $(x_j)_{j \in J}$ to elements $(x_i)_{i \in I}$ by choosing $x_i$ for $i \in J \setminus I$.

Let $i \in I$. By cofinality $\exists j \in J$ such that $j \geq i$, so we can choose $x_i:= f_{ji}(x_j)$. However, if we want the map to be well-defined, we need the property $$ \#\{f_{ji}(x_j) \ | \ j \in J \text{ and } j \geq i \} \ = \ 1, \quad \text{ for all } i \in I. $$ This seems to hold when $J$ is totally ordered, but if $J$ is just a poset I don't know how it could be true. I still attempted to make use of directedness of $J$ but I failed.

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I can spell out the attemps of use of directedness of $J$ if you like, but I think it wouldn't be useful. Could you please give me a hint?

Answer 1

You have defined $\rho\colon \varprojlim_{i\in I}S_i \rightarrow \varprojlim_{j\in J}S_j$ and want to define the inverse. Let $(x_j)_{j\in J}\in \varprojlim_{j\in J}S_j$. For $i\in I\setminus J$ there is some $j\in J$ with $j\ge i$ by cofinality of $J$. Thus, you define $x_i:= f_{ji}(x_i)$. What still has to be shown is that this definition is independent of the choice of $j$.

So let $j'\in J$ be another element with $j'\ge i$. Since $J$ is directed as a cofinal subset of a directed set, you find $k\in J$ with $k\ge j$ and $k\ge j'$. This gives $f_{kj}(x_k) = x_j$ and $f_{kj'}(x_k) = x_{j'}$ and thus $$ f_{ji}(x_j) = f_{ji}(f_{kj}(x_k)) = f_{ki}(x_k) = f_{j'i}( f_{kj'}(x_k) = f_{j'i}(x_{j'}), $$ which shows that the definition of $x_i$ indeed does not depend on $j$.

Hence we have found $(x_i)_{i\in I}\in \prod_{i\in I}S_i$ with $\pi((x_i)_{i\in I}) = (x_j)_{j\in J}\in \varprojlim_{j\in J}S_j$. What is left to show, is that $(x_i)_{i\in I}\in \varprojlim_{i\in I}S_i$.
For this, let $i,i'\in I$ with $i\ge i'$. By cofinality, we find $j\in J$ with $j\ge i$, and hence also $j\ge i'$ by transitivity. Because of $f_{ji'}(x_j) = x_{i'}$ and $f_{ji}(x_j) = x_i$ by the above discussion, it follows that $$ x_{i'} = f_{ji'}(x_j) = f_{ii'}(f_{ji}(x_j)) = f_{ii'}(x_i). $$ This shows $(x_i)_{i\in I}\in \varprojlim_{i\in I}S_i$ and we are done.