First let me make a definition. Let $M$ be a smooth manifold and 
$S \subset M $ a topological subspace of $M$. We say that $S$ has 
"dimenion" at most $k$ if $S$ is a subset of  

$$ X\_1 \cup X\_2 \ldots \cup X\_n $$ 
such that each of the $X\_i$ is a smooth manifold of dimension less than or 
equal to $k$. Of course $n$ is finite. Note that as per my definition 
if $S$ has dimension at most $4$ then it also has dimension at most $5$. 
Also note that the $X\_i$ need not be closed. 


Suppose $\pi: E \rightarrow M $ be a compact fiber bundle over a compact 
manifold $M$ (everything is smooth). Suppose $S \subset E$ is a topological 
subspace that has dimension at most $k$. Is it true that 
$\pi(S)$ also has dimension at most $k$? 


It seems to me that this should follow from the fact that there can not be 
a smooth surjective map from $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ 
if $n > m$, but I am not sure. 


In case the answer is no, suppose everything was in the complex setting, 
ie $\pi: E \rightarrow M$ is a compact 
complex fiber bundle, $M$ is complex manifold, $\pi$ is a homolomorphic 
and $S$ has complex dimension at most $k$ (there is an obvious definition 
for having complex dimension at most $k$). Is my claim true in that case, 
ie $\pi(S)$ has complex dimsnion at most $k$? 



Is it impossible for the dimension of a topological space to increase under a smooth map?
I guess "smooth manifold" means "smooth submanifold".
In particular a curve which pass through a given point infinitely many times is not a smooth submanifold.


If yes then it is easy to construct an example.
Take a smooth curve $\gamma\colon(0,1)\to \mathbb R^2$ which pass through $(0,0)$ infinitely many times under different angles. Then the curve $t\mapsto (t,\gamma(t))\in\mathbb R^3$ has dimension 1 in your sense while its projection $\gamma(0,1)$ on $\mathbb R^2$ has dimension 2. 


You can do the same construction in the 3-torus, so everything can be made to be compact. 


