Topological dimension of the image of continuous surjective functions

Question

Consider two topological spaces $(X,\tau)$ and $(Y,\omega)$ and a continuous surjective function $f\colon X\to Y$.

Let $\mathrm{dim}(X)$ and $\mathrm{dim}(Y)$ denote the Lebesgue covering dimension of $X$ and $Y$ respectively.

It is clear that the covering dimension is a topological invariant, in other words if $f$ is also injective with continuous inverse, then $\mathrm{dim}(X) = \mathrm{dim}(Y)$.

However, I am interested in the case when $f$ is only surjective but not injective. Is there some other property of $f$ from which I can conclude that the topological dimension is at least not increased, i.e. $\mathrm{dim}(Y) \leq \mathrm{dim}(X)$. It will be necessary to exclude cases like space-filling curves. I was thinking about something like openess of $f$ but was not able to find such a result anywhere.

Any hints on necessary or sufficient conditions on $f$ would be welcome. Also, if it helps $X$ and $Y$ can be assumed as ''nice'' as necessary (e.g. metrizable, separable, normal, or even as subspaces of Euclidean spaces).

Answer 1

I did some more reading on the topic and found out a few results in Dimension Theory, R. Engelking, Ch. 1, § 12, that I want to share in case anyone else is interested. They apply for separable metrizable spaces (in which case the covering dimension is equal to the small and large inductive dimension).

If $f\colon X\to Y$ is an open, continuous, surjective function and $X$ and $Y$ are separable metrizable spaces, then the following statements hold:

(i) If $f^{-1}(y)$ has an isolated point for any $y\in Y$, then $\mathrm{dim}(Y)\leq \mathrm{dim}(X)$.

(ii) If $f^{-1}(y)$ is a discrete subspace of $X$ for any $y\in Y$, then $\mathrm{dim}(Y)= \mathrm{dim}(X)$.

(iii) If $X$ is locally compact and $f^{-1}(y)$ is at most countable for any $y\in Y$, then $\mathrm{dim}(Y)= \mathrm{dim}(X)$.

Answer 2

A map $f:X \to Y$ is ring-like if for every point $x \in X$ and every pair $U$ and $V$ of open neighborhoods of $x$ and $f(x)$ respectively, there is an open $W$ such that $f(x) \in W \subseteq V$ and $f^{−1}(\partial W) \subseteq U$. Ring-like maps were introduced by V. V. Fedorchuk in Bicompacta with noncoinciding dimensionalities, Dokl. Akad. Nauk SSSR 182 (1968), 275-277. Note that if $Y$ has a base of clopen subsets then any $f:X \to Y$ is ring-like.

A ring-like map between compact spaces does not increase dimension (see Proposition 1.8 of this article).

Answer 3

A proper map is a map such that the preimage of each compact subset is compact. A cell-like map $f: X \to Y$ is a proper surjection between metrizable spaces such that $f^{-1}(y)$ is a cell-like set (i.e., for finite-dimensional $X$, $f^{-1}(y)$ admits a cellular embedding in a Euclidean space.)

In 1968, Kozlowski proved the following result.

Theorem If $X \to Y$ is a cell-like map defined on a subset of a $3$-manifold, then $\dim(Y)\leq 3$.

Succinctly speaking, the idea is cell-like images of topological $n$-manifolds are always $\mathbb{Z}$-homology $n$-manifolds, and for these cohomological and covering dimensions agree if $n\leq 3$.

Other than some special cases, for $n\geq 5$, it's known that the direct generalizations of the theorem are not true. The 4-dimensional analog is a long-standing open problem. For more details regarding this "cell-like dimension raising mapping problem", see this survey.