Given a map of topological spaces $f:X\rightarrow Y$. Assume, that $X$ has finite Lebesque 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$(X)$.

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$(X)$.