In Algebraic Geometry one has a very useful formula for the dimension of fibers. Specifically I am thinking about a statement of the following form:

Let $C$ be a curve over $\mathbb{C}$, and let $S$ be an irreducible constructible subset $S \subset C^n$. Here we take the Zariski topology on $C^n$ and constructible means relatively closed $S \subset_{cl} U \subset_{op} C^n$, irreducible means there are no constructibles $S_1, S_2 \subset_{cl} S$ such that $S = S_1 \cup S_2$ and $S_1 \not\subset S_2$, $S_2 \not\subset S_1$. Let $\pi: C^n \rightarrow C^m$ be a projection. Then $$\dim S = \dim \pi(S) + \min_{a \in \pi(S)} \dim(\pi^{-1}(a) \cap S).$$

I wonder, is there a similar statement for topological dimensions?

In particular, consider a compact Hausdorff space $X$ (or maybe seperable metric or similar) and a projection $\pi : X^n \rightarrow X^m$. For which subsets $S \subset X^n$ do we have $$\dim S = \dim \pi(S) + \min_{a \in \pi(S)} \dim(\pi^{-1}(a) \cap S)?$$

I don't know a lot about topological dimension theory, but some very basic first examples with the unit cube $I^n$ seem to work out for constructible subsets, i.e. locally compact subsets here.

Also note that if $Y \subset X^n$ has $\dim Y = 0$, then $\dim \pi(Y) = 0$ because $\pi$ is continuous open-and-closed.

Any ideas, help, or pointers to literature greatly welcome!


1 Answer 1


Your question can be viewed as a general dimension theory question, and it is answered by the standard textbooks on the topological dimension theory.

Indeed, let $\ S\subseteq X^k\ $ be embedded in $\ X^k.\ $ Let $\ f:S\rightarrow X^m\ $ be an arbitrary map. Let $ n:=k+m,\ $ so that $\ X^n\ =\ X^k\times X^m.\ $ Define

$$ S'\ := \{(x\ \ f(x)): x\in S\} $$

Of course $\ S'\ $ is canonically homeomorphic to $\ S.\ $ Consider your projection $\ \pi:X^n\rightarrow X^m.\ $ Then the restriction $\ \pi|S'\ $ doesn't differ topologically from f. Thus there is no gain or difference from considering a projection as opposed to a general continuous map.

Let's remember that finite d-dimensional metric compact spaces can be embedded into $\ \mathbb I^{2\cdot d+1}\ $ (hence into $\ \mathbb R^{2\cdot d+1}$).

Two kinds of examples to keep in mind:

A.Kolmogorov provided the following (or something better?): glue together the end-points of the removed open intervals of the standard Cantor set. Then your space (the Cantor set) has dimension $\ 0,\ $ and the "fibers" (call them "fibers" in the context of your Question) have dimension $\ 0\ $ (they are single points or 2-point sets), while the "base" i.e. the image of the given surjection, has dimension $\ 1,\ $ hence $\ 0 < 0+1$.

L.Pontryagin provided a pair of 2-dim metric compacta for which their product has dimension $\ 3,\ $ and V. Boltyansky has obtained even sharper examples where that pair of such compacta were homeomorphic.

  • $\begingroup$ Thank you very much for your reply! Although I am not sure your example of Kolmogorov would contradict the fiber dimension formula. Let $S = \mathcal{C} \subset [0,1] = X$ be the Cantor set, and let $f: S \rightarrow X$ be the map you describe by glueing endpoints of the removed open intervals. Then the corresponding projection as you mention in the first part would be $\pi : \{ (x,f(x)) \; : \; x \in S \} \rightarrow X$ where $\pi((x, f(x))) = f(x)$. Hence $\dim \pi(S') = \dim \pi(S \times f(S)) = \dim f(S) = 1$ and equality holds. Or do I miss something? $\endgroup$
    – Niki
    Mar 27, 2017 at 12:43
  • $\begingroup$ I am a bit confused, something cannot be right here because if $\dim S' = 0$ then $\dim \pi(S') = 0$ as mentioned above because $\pi$ is continuous open and closed. $\endgroup$
    – Niki
    Mar 27, 2017 at 13:39

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