Inverse limit of finite sets, questions about surjectivity

Question

Let $X_n, n\geq 1$ be non-empty finite sets and $f^{n+1}_{n}:X_{n+1}\rightarrow X_n$ be a surjective map for each $n\geq 1$. Now the set of projective sequences, $$X_{\infty}:=\{(x_i)_{i\geq 1}:x_n\in X_n, f^{n+1}_{n}(x_{n+1})=x_n~\text{for all}~n\geq 1\},$$ is non-empty and for each $n\geq 1$ the map $$f_n:X_{\infty}\rightarrow X_n, (x_i)_{i\geq 1}\mapsto x_n$$ is surjective with $f^{n+1}_n\circ f_{n+1}=f_n$ for all $n\geq 1$. Now let $Y$ be another set equipped with surjective maps $$g_n:Y\rightarrow X_n$$ such that $f^{n+1}_n\circ g_{n+1}=g_n$ for all $n\geq 1$. So you get the map $$g:Y\rightarrow X_{\infty},~y\mapsto (g_i(y))_{i\geq 1}.$$ I know examples where $g$ is not injective, but I don't know any where $g$ is not surjective. Can you show me one? Or is $g$ always surjective?

Answer 1

Let $X_n$ be the set of all binary sequence of length $n$. So $X_1 = \{0,1\}$, $X_2 = \{00,01,10,11\}$, etc.

Let $f^{n+1}_n$ be the map which deletes the last digit. For example, $f^3_2(011) = 01$.

Then $X_\infty$ can be identified with the set of all infinite binary sequences, where the map $f_n\colon X_\infty\to X_n$ extracts the first $n$ digits.

Now let $Y$ be the set of all infinite binary sequence which are eventually $0$, and let $g_n\colon Y\to X_n$ be the map which extracts the first $n$ digits. The maps $g_n$ are surjective, since any finite binary sequence can be extended to a sequence in $Y$ by appending infinitely many $0$s.

But the induced map $Y\to X_\infty$ is just the inclusion, which is clearly not surjective. In fact, $Y$ is countable, while $X_\infty$ has cardinality continuum.

A topological point of view on this example is that $X$ is the Cantor set and $Y$ is a countable dense subset.