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 $k$?