Pass surjection to inverse limits. QuestionLet $I$ be a directed preordered set, and let $(X_i,\alpha_{ij})$, $(Y_i,\beta_{ij})$ be two inductive system of sets indexed by $I$, and denote their inverse limits by $X$ and $Y$ respectively. Let $u_i:X_i\rightarrow Y_i$ be a morphism of inductive systems, and let $u:X\rightarrow Y$ be the morphism induced by $u_i$. Suppose there exists an $i\in I$, such that whenever $k\geq i$, $u_k$ is a surjection. How to prove that $u$ is a surjection?
I can choose an element $y\in Y$, and denote its image in $Y_k$ by $y_k$. When $k\geq i$ I can lift it to an element $x_k\in X_k$, but these $x_k$ may not be compatible with $\alpha_{ij}$s. How can I get a "good" lift that is compatible with $\alpha_{ij}$s? (hence get an element in $X$) Thank you!
 A: This is not possible in general.
Let $I = (\mathbb{N},\leq)$.
For all $i\in I$, let $X_i = \mathbb{N}_{\geq i}$, and when $j\leq i$, let $\alpha_{ij}\colon \mathbb{N}_{\geq i}\to \mathbb{N}_{\geq j}$ be the inclusion.
For all $i\in I$, let $Y_i = \{*\}$, a singleton set, and when $j\leq i$, let $\beta_{ij}\colon \{*\}\to \{*\}$ be the identity.
For all $i\in I$, let $u_i\colon \mathbb{N}_{\geq i}\to \{*\}$ be the unique map. Note that $u_i$ is surjective, and the $u_i$ give a morphism of inductive systems.
Then $X = \emptyset$, and $Y = \{*\}$, so the induced map $u\colon X\to Y$ is not surjective.
On the other hand, if each $X_i$ is finite, it follows from a compactness argument that the induced map $u$ is always surjective. If you're interested in this case and need more details, let me know.

More details, as promised:
Assume each $X_k$ is finite and non-empty. View $X_k$ as a finite (hence compact) topological space with the discrete topology. By Tychonoff's theorem, $\mathbb{X} = \prod_{k\in I} X_i$ is compact.
Let $y\in Y$, and denote by $y_k$ the image of $y$ in $Y_k$. To find $x\in X$ such that $u(x) = y$, it suffices to find a point $(x_k)_{k\in I}\in \mathbb{X}$ such that:

*

*$u_j(x_j) = y_j$ for all $j\in I$.

*$\alpha_{jj'}(x_j) = x_{j'}$ for all $j\geq j'$ in $I$.

For all $j\in I$, let $C_j = \{(x_k)_{k\in I}\in \mathbb{X}\mid u_j(x_j) = y_j\}$, and for all $j\geq j'$ in $I$, let $D_{jj'} = \{(x_k)_{k\in I}\in \mathbb{X}\mid \alpha_{jj'}(x_j) = x_{j'}\}$. Then it suffices to show that $\bigcap_{j\in I} C_j\cap \bigcap_{j\geq j'}D_{jj'}\neq \varnothing$. Since each of the sets $C_j$ and $D_{jj'}$ are closed in $\mathbb{X}$, by compactness it suffices to show that the intersection of any finitely many of them is non-empty.
Given finitely many of the $C_j$ and $D_{jj'}$, pick (by directedness) some $k\in I$ which is above all of the indices $j$ appearing in these sets, and also above the $i\in I$ which ensures that $u_k\colon X_k\to Y_k$ is surjective. Let $x_k\in X_k$ such that $u_k(x_k) = y_k$. For all $j$ such that $k\geq j$, let $x_j = \alpha_{kj}(x_k)$. For all other $j'$, pick an arbitrary $x_{j'}\in X_{j'}$. Then it's straightforward to check that $(x_j)_{j\in I}$ is in the finitely many $C_j$ and $D_{jj'}$ specified.
The point here is that the compactness argument let us reduce the problem from finding an infinite coherent family of lifts down to finding a finite coherent family of lifts, and then (by directedness) down to picking a single lift.
If it seems strange to you to get topology involved here: (1) it's actually not so surprising, since any inverse limit of finite sets carries a natural compact topology, but (2) there are other options - the compactness argument can be just as well phrased using the compactness theorem for propositional logic or (when the indexing order is just $(\mathbb{N},\leq)$) König's Lemma. See my answer here for a variety of different ways to frame the same proof using different compactnesss principles.
