Let $f: X\to Y$ be a continuous surjective map between compact metric spaces. Suppose the fibre $f^{-1}(y)$ has zero topological dimension for each $y\in Y$. Then by Hurewicz dimension lowering theorem, one has $${\rm dim}(X)\le {\rm dim}(Y)+\sup\_{y\in Y} {\rm dim}(f^{-1}(y))={\rm dim}(Y).$$ I would like whether there is a direct/simple proof of ${\rm dim}(X)\le {\rm dim}(Y)$ without using Hurewicz dimension lowering theorem. Thanks.



A continuous map with zero-dimensional fibres
Yes, there is such a proof. A reference is Theorem 6.4.11 from Pears *Dimension Theory of General Spaces*, which is a great reference for dimension theory in general even though it's not as well known as other books on the topic.


The other good news is that, assuming a small lemma which might even be a definition for you if you are only working with metrizable spaces, the proof is short and clear enough to be completely written here, let's start with said lemma (which is Corollary 3.4.4 in the aforementioned book and I will not prove it here).


**Lemma:** Let $X$ be a normal space with $\dim X\leq n$ and let $\{G\_i\}\_{i\in I}$ be a locally finite open cover of $X$. Then there exist and open cover $\{H\_i\}\_{i\in I}$ of $X$ such that the order of $\{H\_i\}\_{i\in I}$ is at most $n$ and $H\_i\subseteq G\_i$ for all $i$.


With this lemma available we can prove the result you want (I'm copying the statement and the proof from the book by Pears in full generality but you can ignore most point-set technicalities since you are interested in compact metric spaces).


**Theorem:** Let $X$ be a normal space, let $Y$ be a paracompact Hausdorff space and let $f\colon X\to Y$ be a continuous closed surjection such that $\dim f^{-1}(y)=0$ if $y\in Y$. Then $\dim X\leq\dim Y$.


**Proof:** We shall show that if $\dim Y\leq n$ then $\dim X\leq n$. Let $$\{U\_1,\ldots U\_k\}$$ be an open covering of $X$. If $y\in Y$, then since $Y$ is a $T\_1$-space, $f^{-1}(y)$ is a closed subspace of $X$ and thus is a normal space. Since $$\dim f^{-1}(y)=0,$$ there exist a disjoint closed covering $\{F\_{1y},\ldots, F\_{ky}\}$ of $f^{-1}(y)$ such that $F\_{iy}\subseteq U\_i$ for each $i$. Since $X$ is a normal space and the sets $F\_{iy}$ are closed in $X$, there exist disjoint open sets $G\_{iy}$ such that
$$F\_{iy}\subseteq G\_{iy}\subseteq U\_i,\text{ for } i=1,\ldots,k.$$
Since $f^{-1}(y)\subseteq \bigcup\_{i=1}^k G\_{iy}$ and $f$ is a closed mapping, there exist an open neighbourhood $W\_y$ of $y$ in $Y$ such that $f^{-1}(W\_y)\subseteq\bigcup\_{i=1}^k G\_{iy}$. Since $Y$ is a paracompact normal space such that $\dim Y\leq n$, it follows from Corollary 3.4.4. that there exists an open covering $\{V\_y\}\_{y\in Y}$ of $Y$ of order not exceeding $n$ such that $V\_y\subseteq W\_y$ for each $y$. If $y\in Y$ and $i=1,\ldots,k$ let $H\_{iy}=G\_{iy}\cap f^{-1}(V\_y)$. Then $\{H\_{iy}\}\_{y\in Y,i=1,\ldots k}$is an open covering of $X$ of order not exceeding $n$ which is a refinement of $\{U\_1,\ldots, U\_k\}$.Thus $\dim X\leq n$. $\square$


