Given $S\subset \mathbb{R}^{m+n}$ a smooth submanifold, i.e. for all $x\in S$ there exists an open $A\ni x$, open $B\ni 0$ in $\mathbb{R}^{m+n}$ and smooth diffeomorphism $f:A\rightarrow B$ such that $f(A\cap S)=(\mathbb{R}^d\times\{0\})\cap B$ for $d$ is the dimension of $S$, and also satisfying $x\in S\Rightarrow x_{m+1}\ldots=x_{m+n}=0$, so $S$ really lies on the $x_1\ldots x_m$ plane, is the projection of $S$ onto $\mathbb{R}^m$ a submanifold?
My intuition says yes but so far I can't even prove that it's a manifold.
Say the dimension of $S$ is $k$. Consider a point $s$ on $S$. We can take $k$ coordinates $x_{i_1}$, $\ldots$, $x_{i_k}$ so that around $s$ $S$ will be the graph a smooth function $\phi: \mathbb{R}^k \to \mathbb{R}^{m+n-k}$. Since the last $n$ coordinates of any point in $S$ are $0$ we must have $\{i_1, \ldots, i_k\} \subset \{1, \ldots, m\}$. The $m-k$ components of $\phi$ that are of $\mathbb{R}^m$ will produce a smooth function $\psi$ so that $S$ is the graph of $\psi \colon \mathbb{R}^k \to \mathbb{R}^{m-k}$.
Hence $S$ is a submanifold of $\mathbb{R}^m$.
In general: if $S \subset M$ and $M$ is a submanifold of $M'$, then $S$ is a submanifold of $M'$ if and only if $S$ is a ubmanifold of $M$. It can be reduced the the particular case above. The main ideas: a submanifold is locally the graph of a smooth function, and, for a function $\phi = ( \psi, 0)$, $\phi$ is smooth if and only if $\psi$ is smooth.