Exercise 2, § 7

This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Let $I$ be a directed ordered set and $(E_\alpha, f_{\alpha\beta})$ an inverse system of sets relative to $I$ such that each transition map $f_{\alpha\beta} : E_\beta \to E_\alpha$ (for $\alpha \leq \beta$) is injective. Let $E = \varprojlim E_\alpha$ be the inverse limit and, for each $\alpha \in I$, let $f_\alpha : E \to E_\alpha$ be the canonical mapping.

We must show that $f_\alpha$ is injective for every $\alpha \in I$.

Fix $\alpha \in I$ and take $x, y \in E$ with $f_\alpha(x) = f_\alpha(y)$. Since $E$ is a subset of the product $\prod_{\beta \in I} E_\beta$, the elements $x$ and $y$ are families $(x_\beta)_{\beta \in I}$ and $(y_\beta)_{\beta \in I}$ satisfying the compatibility condition $x_\alpha = f_{\alpha\beta}(x_\beta)$ and $y_\alpha = f_{\alpha\beta}(y_\beta)$ whenever $\alpha \leq \beta$; this is the definition of the inverse limit, and the canonical mappings are the projections $f_\alpha(x) = x_\alpha$. In particular, for any $\beta \in I$ with $\beta \geq \alpha$ we have $f_\alpha(x) = f_{\alpha\beta}(f_\beta(x))$ and $f_\alpha(y) = f_{\alpha\beta}(f_\beta(y))$ (formula (2) of §7 no. 1).

Now let $\beta \in I$ be arbitrary. Because $I$ is directed, there exists an index $\gamma \in I$ such that $\gamma \geq \alpha$ and $\gamma \geq \beta$. Applying the compatibility relations at $\gamma$, we obtain

Since $\alpha$ was arbitrary, all canonical mappings $f_\alpha$ are injective.