# A continuous map with zero-dimensional fibres

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.

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 $$i$$.
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 $$\dim X\leq n$$. $$\square$$