# Epimorphism going out of an inverse limit into a finite dimensional module

Let $$k$$ be a field and $$A$$ a finite dimensional $$k$$-algebra. Given a sequence of inclusions $$M_1 \subseteq M_2 \subseteq \dots$$ of $$A$$-modules consider the direct limit $$M:= \bigcup_{i=1}^\infty M_i$$. For a finite dimensional module $$X$$ suppose we have $$X\subseteq M$$. Then $$X$$ is generated by finitely many elements $$x_1, x_2, \dots, x_n$$. Hence we find $$m>0$$ such that $$x_i\in M_m$$ for all $$m$$. It follows that $$X\subseteq M_m$$. Now consider the dual setting:

Let $$M_1 \twoheadleftarrow M_2 \twoheadleftarrow \dots$$ be a sequence of $$A$$-modules, let $$M$$ be its inverse limit and suppose we find an epimorphism $$M\twoheadrightarrow X$$, where $$X$$ is a finite dimensional $$A$$-module. Is it true in general, that we obtain an epimorphism $$M_m \twoheadrightarrow X$$ for some $$m>0$$?

The answer is no. First, let me point out that the $$A$$ here plays no role: because each $$M_{i+1}\to M_i$$ is an epimorphism, $$M\to M_i$$ is one too, and so if there is a $$k$$-linear factorization $$M\to M_i\to X$$, then it is automatically $$A$$-linear.

So we can simply focus on $$k$$-vector spaces and worry about the analogous statement. Let $$f: M\to X$$ denote our morphism. ​Let $$K_i\subset M$$ be the kernel of $$M\to M_i$$. $$K_i$$ is a nonincreasing sequence of subspaces of $$M$$, therefore the same is true of $$f(K_i)$$ inside $$X$$.

Because $$X$$ is finite dimensional, this nonincreasing sequence of subspaces stabilizes. $$f$$ factors through $$M_i$$ if and only if $$f(K_i) = 0$$, so the point is to find an example where no such $$i$$ exists.

For instance, take $$k$$ to be a finite field, pick a nonprincipal ultrafilter $$\mathcal U$$ on $$\mathbb N$$ and define $$k^\mathbb N\to k$$ by $$u\mapsto \lim_\mathcal U u$$ (we take the limit along the ultrafilter in a topological sense using the discrete topology on $$k$$, which is compact). More concretely, this partitions $$\mathbb N$$ as $$\coprod_{x\in k} u^{-1}(x)$$ and decides which of these is in $$\mathcal U$$. Write $$k^\mathbb N$$ as $$\lim_n k^n$$. The kernel $$K_n$$ is $$0^n\times \prod_{k> n}k$$ and the image of $$f$$ on that is always $$k$$: take the sequence $$u$$ which is $$0$$ until $$n$$, and $$1$$ afterwards.