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Given a map of topological spaces $f:X\rightarrow Y$. Assume, that $X$ has finite Lebesgue dimension. I am wondering, what dim$(f(X))$ might be. Of course, if $f$ is a homeomorphism onto its image, then it's just dim$(X)$. On the other hand there are the space filling curves, that show, that the dimension might increase. So I am wondering, whether there is any nice condition for $f$ (such as open, closed, proper etc), that guarantees, that dim$(f(X))\le$dim$(X)$.

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Might not the same question be asked for Hausdorff dimension? – Anweshi Jul 21 '10 at 11:36
corrected spelling Lebesgue. Lebesgue dimension is also called covering dimension – Gerald Edgar Jul 21 '10 at 11:39
@Anweshi: yes, but that would be a different question. – Gerald Edgar Jul 21 '10 at 11:40
up vote 6 down vote accepted

Some results from Engelking's dimension theory book:

If $f: X \mapsto Y$ is a closed, continuous and surjective function between normal spaces $X$ and $Y$, and $\forall y \in Y: | f^{-1}[{y}] | \le k$ for some integer $k \ge 1$, then $\dim(Y) \le \dim(X) + (k-1)$.

If $f: X \mapsto Y$ is an open, continuous and surjective function between a weakly paracompact space $X$ and a normal space $Y$ with finite fibres, then $\dim(X) = \dim(Y)$.

If $f: X \mapsto Y$ is an open, continuous and surjective function between a locally compact normal space $X$ and a weakly paracompact normal space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.

If $f: X \mapsto Y$ is an open-and-closed, continuous and surjective function between a locally compact normal space $X$ and a paracompact space $Y$ such that all fibres are at most countable, then $\dim(Y) \le \dim(X)$.

In the other direction: If $f: X \mapsto Y$ is a closed, continuous function between a normal space $X$ and a paracompact space $Y$ such that the dimension of the fibres is at most 0 (this includes the empty fibres of dimension -1), then $\dim(X) \le \dim(Y)$.

weakly paracompact includes Hausdorff, and is called metacompact or point-paracompact by other authors: every open cover has a point-finite refinement.

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One result of this kind: if $X$ is zero-dimensional and $f$ is at most $n$ to one, then $\dim f(X) \le n-1$. This is if and only if... Given $Y$ of dimension $m$, there is a zero-dimensional $X$ and surjection $f \colon X \to Y$ that is at most $m+1$ to one.

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Maybe there is a condition like $f$ is open missing. For any space $Y$, one can consider $id_Y:Y\rightarrow Y$, where the first $Y$ is equipped with the discrete topology. Then it is bijective, the first space is zero dimensional and $Y$ could be of any dimension. Am I missing something ? – HenrikRüping Jul 21 '10 at 11:56
Indeed Henrik. We need f to be closed and surjective, and X,Y to be normal (which is pretty mild). The theorem then states that $\dim(Y) \le \dim(X) + (k-1)$ whenever all fibres have size at most $k$. – Henno Brandsma Jul 21 '10 at 15:48
I guess what I was remembering has $X$ and $Y$ compact. – Gerald Edgar Jul 21 '10 at 18:26

I am not sure if this works, but I believe that if $X$ is compact, one can prove that if $f^{-1}(y)$ is compact connected and of diameter bigger or equal to $\delta>0$ for every $y\in f(X)$ then $dim(f(X))< dim(X)$.

The proof goes by induction in the dimension of $f(X)$: If it is zero, it is clear (since every compact connected set of diameter bigger than $0$ has dimension $\geq 1$.

Assuming it holds for $k < n$ consider $f(X)$ of dimension $n$ and consider countably many disjoint subsets of dimension $n-1$, so, the preimage has dimension at least $n$. Assuming that $X$ has dimension $\leq n$ we get countably many disjoint open sets contradicting the compacity of $X$.

EDIT: I have recently found this paper were the notion of cell-like mapping is discussed. There it is shown that this result does not hold unless more hypothesis are added. There is a nice discusion on the possibility of a cell-like mapping increasing dimension. See section 3.

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