Remark 3

Let I be an ordered set and let $(E_\alpha, f_{\alpha\beta})$ be an inverse system of sets relative to I. For each finite subset J of I, let $F_J$ be the inverse limit of the (finite) inverse system obtained from $(E_\alpha, f_{\alpha\beta})$ by restricting the index set to J. If J and K are any two finite subsets of I such that $J \subset K$, let $g_{JK}$ denote the canonical mapping (3) of $F_K$ into $F_J$. Then the relation (4) shows that $(F_J, g_{JK})$ is an inverse system of sets relative to the directed set (with respect to the relation $\subset$) $\mathfrak{F}(I)$ of finite subsets of I. Next, for each $J \in \mathfrak{F}(I)$ let $h_J : E \to F_J$ be the canonical mapping (3). By virtue of (4) and with the abuse of language mentioned in Remark (2), $(h_J)$ is an inverse system of mappings. Put $h = \varprojlim h_J : E \to F = \varprojlim F_J$, and let us show that $h$ is a bijection (called canonical). Indeed, let $y = (y_J) \in F$. By definition we have $y_J = (x_{\alpha, J})_{\alpha \in J}$, where $x_{\alpha, J} \in E_\alpha$ for all $\alpha \in J$. If $J \subset K$, then by definition of the mapping $g_{JK}$ and because $y_J = g_{JK}(y_K)$, we have $x_{\alpha, J} = x_{\alpha, K}$ for all $\alpha \in J$. Hence, given $\alpha \in I$, there is a unique element $x_\alpha \in E_\alpha$ such that $x_\alpha = x_{\alpha, J}$ for all finite subsets J of I which contain $\alpha$. If $\alpha \leqslant \beta$, there is a finite subset J of I which contains both $\alpha$ and $\beta$; hence $x_\alpha = f_{\alpha\beta}(x_\beta)$ by definition. Consequently $x = (x_\alpha)$ is the unique element of E such that $h(x) = y$.