Inverse limit of arc-like spaces is arc-like

Question

I came across this in Nadler's book, "Introduction to continuum theory".

First, for a collection of spaces $P$, a compact metric space $X$ is said to be $P$-like if for every $\epsilon > 0$ there's some member $Y_{\epsilon} \in P$ such that there's an onto continuous map $f : X \rightarrow Y_\epsilon$ such that for any $y \in Y_\epsilon$ the diameter of $f^{-1}(y)$ is less than $\epsilon$.

Now what confuses me is the proof the book provides of how if $X = \lim_{\leftarrow} \{X_i,f_i\}$ is an inverse limit in which the maps $f_i$ are onto and each $X_i$ is $P$-like then $X$ is $P$-like. The proof takes the projection $\pi_i:X \rightarrow X_i$ which is onto and a $2^-i$ map and it states that because of this, $X$ is arc like, but I don't see why this is the case. I was thinking of taking a map $g_i: X_i \rightarrow Y_\epsilon$ as specified above, that exists because $X_i$ is $P$-like and take the composition $g_i \circ \pi_i$ but I don't know how to prove that this last map is as required. Any help would be appreciated.

Answer 1

Let $X=\varprojlim\left\{X_i,f_i\right\}_{i=1}^\infty$ be an inverse limit, where for every $i$ we have that $X_i$ is an arc-like continuum. We want to show that $X$ is arc-like, that is for every $\varepsilon>0$ we want to find an $\varepsilon$-map $f_\varepsilon\colon X\to [0,1]$. Fix $\varepsilon>0$ and let $i$ be big enough so that $2^{-i}<\varepsilon$.

We know that $\pi_i\colon X\to X_i$ is a $2^{-i}$ map, and by Exercise 2.33 in Nadler's book we can find $\delta>0$ so that $\mathrm{diam}(\pi_i^{-1}(Z))<2^{-i}$ whenever $Z\subseteq X_i$ is such that $\mathrm{diam}(Z)<\delta$. Since $X_i$ is arc-like there exists a $\delta$-map $g\colon X_i\to [0,1]$. It only remains to verify that $f_\varepsilon=g\circ\pi_i$ is an $\varepsilon$-map $X\to[0,1]$, so let $x\in[0,1]$ be arbitrary. Since $g$ is an $\varepsilon$-map we have that $\mathrm{diam}(g^{-1}(x))<\delta$, which means that $\mathrm{diam}(\pi_i^{-1}(g^{-1}(x)))<2^{-i}$ by our choice of $\delta$, which shows that $g\circ\pi_i$ is an $\varepsilon$-map, since $2^{-i}<\varepsilon$ by construction.