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 $m>0$?



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


