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$(X)$.



Lebesgue dimension of images
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.


