Hurewicz theorem on mappings that lower dimension

Question

A form of Hurewicz theorem on mappings that lower dimension states that: Let $X$ and $Y$ be compact metric spaces and $f:X\to Y$ a continuous map. Suppose that there is some $n$ so that for every $y\in Y$, ${\rm dim}(f^{-1}y)\le n$. Then ${\rm dim} X ≤ {\rm dim} Y + n$.

A proof is refered to the book of R. Engelking where it used the induction on ${\rm Ind} Y={\rm dim} Y$. I would like to know whether there is anohter proof without using the induction on ${\rm Ind} Y$. Could we prove it by using covering dimension ${\rm dim}$ directly? Thanks.

Answer 1

In Engelking's Theory of Dimensions, Finite and Infinite, Thm 3.3.10 (p. 200) proves the more general result

If $f: X \to Y$ is a closed mapping from a normal space $X$ to a weakly paracompact normal space $Y$ and there exists an integer $k \ge 0$ such that $\dim(f^{-1}(y)) \le k$ for every $y \in Y$, then $\dim(X) \le \dim(Y)+ k$

(Of course if $X,Y$ are compact metric, the conditions on $f,X,Y$ are easily satisfied).

The proof takes almost the whole section 3.3 of which this is a part and is quite technical, but purely uses $\dim$ and no $\operatorname{Ind}$ or the fact that these coincide on compact metric spaces. It does use the compactification $\beta X$ as a tool, to reduce to compact spaces. It's illustrative to check out the historical notes in Engelking's book as well, which is a veritable encyclopedia on dimension theory.

So yes, you can do in with only dim. But it’s quite involved. You could study Engelking’s proof and see where you have shortcuts (unnecessary steps, e.g.) in your restricted compact metric case.