Let $A$ be a directed set, let $X_{\alpha}$ be topological spaces for $\alpha \in A$, let $f_{\beta \alpha}: X_{\beta} \to X_{\alpha}$ be continuous, and suppose that if $\alpha \leq \beta \leq \gamma$, we have $f_{\gamma \alpha} = f_{\beta \alpha} \circ f_{\gamma \beta}$. Assume $f_{\alpha \alpha}$ is the identity $X_{\alpha} \to X_{\alpha}$ for each $\alpha$. Let $X: = \{x \in \prod X_{\gamma} | x_{\alpha} = f_{\beta \alpha} (x_{\beta}) \text{ for all } \alpha \leq \beta \}$ denote the inverse limit, which we now equip with subspace topology.
Part of exercise 29C in Willard states that the projection map $\pi_{\alpha}$ restricted to $X$ surjects onto $X_{\alpha}$ for all $\alpha$. I'm having trouble seeing why this is true; any hints are appreciated.
It's not clear to me this is true; I present an example. Take $A= \mathbb N \setminus \{0\}$, take $X_n = [- \frac{1}{n}, \frac{1}{n}]$, and let $f_{mn}: X_m \hookrightarrow X_n$ be the inclusion map for $m \geq n$. Then, the inverse limit is the space $X=\{(0, 0, \ldots )\}$; clearly the projection $\pi_n$ restricted to $X$ does not surject onto $X_n$. Have I made a mistake?
Best Answer
The question whether the maps $p_\alpha = \pi_\alpha \mid_X : X \to X_\alpha$ are surjective is not related the the topologies of the spaces $X_\alpha$, it is an issue of sets.
In fact, you can consider an inverse system $\mathbf X = (X_\alpha, f_{\beta \alpha})$ consisting of sets $X_\alpha$ and functions $f_{\beta \alpha} : X_\beta \to X_\alpha$. Then the same definition $X = \{ (x_\gamma) \in \prod X_{\gamma} | x_{\alpha} = f_{\beta \alpha} (x_{\beta}) \text{ for all } \alpha \leq \beta \}$ yields an inverse limit in the category of sets.
We have $p_\alpha = f_{\beta \alpha} \circ p_\beta$, thus for $p_\alpha$ to be surjective it is necessary that $f_{\beta \alpha}$ is surjective.
Is the surjectivity of the $f_{\beta \alpha}$ sufficient? If $A$ contains a countable cofinal $C$ (which means that each $\alpha' \in A$ admits $\gamma \in C$ such that $\alpha' \le \gamma$), then the answer is ''yes''.
We write $C = \{ \gamma_n \mid n \in \mathbb{N} \}$.
Let $\xi_\alpha \in X_\alpha$. Recursively we can construct a strictly increasing sequence of integers $n_k$ and elements $x_{\gamma_{n_k}} \in X_{ \gamma_{n_k}}$, $k \in \mathbb{N}$, such that for all $k$
(1) $\gamma_k,\gamma_{n_{k-1}} \le \gamma_{n_k}$ (where we set formally $\gamma_{n_0} = \alpha$)
(2) $f_{\gamma_{n_k} \gamma_{n_{k-1}}}(x_{\gamma_{n_k}}) = x_{\gamma_{n_{k-1}}}$ (where we set formally $x_{\gamma_{n_0}} = \xi_\alpha$)
Note that this involves a variant of the axiom of choice (the axiom of dependent choice).
By construction we have $\alpha = \gamma_{n_0} \le \gamma_{n_1} \le \gamma_{n_2} \le \dots$. The set $C' = \{ \gamma_{n_k} \mid k \in \mathbb{N} \}$ is again cofinal in $A$: For $\alpha' \in A$ there exists $k$ such that $\alpha' \le \gamma_k$, hence $\alpha' \le \gamma_{n_k}$.
For each $\gamma \in A$ we now define $x_\gamma = f_{\gamma_{n_k} \gamma} (x_{\gamma_{n_k}})$ with any $k$ such that $\gamma \le \gamma_{n_k}$. It is easy to see that this does not depend on the choice of $k$ and that $x_\alpha = \xi_\alpha$ and $(x_\gamma) \in X$. Hence $p_\alpha((x_\gamma)) = \xi_\alpha$.