Inverse image of a cover of a compact set by a continuous surjective function Reference to the book "Concepts and Results in Chaotic Dynamics: a Short Course" Proposition 3.9
Given an open cover $\mathcal{U}$ of a compact space $X$ and a continuous surjective function $f:X\to X$. Let $f^{-1}(\mathcal{U})$ denote the set of open sets $\{f^{-1}(U)|U\in\mathcal{U}\}$. For a cover $\mathcal{V}$ of $X$, $H$ is a function such that $H(\mathcal{V})$ is the cardinality of the minimal subset of $\mathcal{V}$ that is still a cover of $X$.
The proposition states that
$$H(\mathcal{U})\geq H(f^{-1}(\mathcal{U}))$$.
There are two points that I cannot understand:


*

*Is $f^{-1}(\mathcal{U})$ still a cover of $X$? Why is the right hand side of the inequality well defined?

*Given the answer of the first question is yes, wouldn't it be $\leq$ rather than $\geq$? Let $\mathcal{V}$ be one of the minimal subsets of $f^{-1}(\mathcal{U})$ such that it is still a cover of $X$. $f(\mathcal{V})$ is evidently still a cover of $X$ with the surjectivity of $f$ and is also a subset of $\mathcal{U}$. So I conclude the opposite direction of the inequality.
 A: *

*Since $f$ is continuous, if $U \in \mathcal{U}$, then $f^{-1}(U)$ is open. Furthermore, since $\mathcal{U}$ covers $X$, one has $\bigcup_{U \in \mathcal{U}} U = X$. It follows that 
$$
\bigcup_{U \in \mathcal{U}} f^{-1}(U) = f^{-1}(\bigcup_{U \in \mathcal{U}}U) = f^{-1}(X) = X.
$$
Thus $f^{-1}(\mathcal{U}) = \{f^{-1}(U) \mid u \in \mathcal{U}\}$ is an open cover of $X$. Thus $H(\mathcal{U})$ is well-defined.

*In your argument, $\mathcal{V}$ is a cover, but is not an open cover since if $U$ is open, $f(U)$ is not necessarily open.
A: The sets $f^{-1}[U]$ are open when $U$ is open. And if $x \in X$, $f(x)$ is covered by some $U \in \mathcal{U}$ (as $\mathcal{U}$ is a cover of $X$ and $f(x) \in X$), and then by definition $x \in f^{-1}[U]$ for that same $U$. So sets of that form do form a cover of $X$ too, which is an open cover by continuity of $f$.
Now if $\{U_1, U_2, U_3, \ldots,U_n\}$ form a finite subcover of $\mathcal{U}$ of size $n$, then $\{f^{-1}[U_1], \ldots, f^{-1}[U_n]\}$ is a finite subcover of $f^{-1}(\mathcal{U})$ of the same size $n$ (see the argument above: $x\in X$ so $f(x) \in U_i$ for some $i$ and then $x \in f^{-1}[U_i]$ for that same $i$).
So if we have $n = H(\mathcal{U})$ (so we have chosen a subcover of $\mathcal{U}$ that realises this minimum), we have at least one subcover of $f^{-1}(\mathcal{U})$ of that size $n$ and the minimal size of a subcover of $f^{-1}(\mathcal{U})$ is thus at most that $n=H(\mathcal{U})$ and hence
$$H(f^{-1}(\mathcal{U})) \le H(\mathcal{U})$$
Note that nowhere I used $f$ being surjective, that's irrelevant for the arguments I have given.
