Let $f: A \subset \mathbb{R}^{n + p} \rightarrow \mathbb{R}^p$ class $C^1$ submersion. So $f$ is open and not injective.
(i) As we prove that all submersion is a local diffeomorphism and like all local diffeomorphism it is an open application follows that $f$ is open.
(ii) I don't know how to prove that $f$ is non-injective. Why it doesn't make sense to me if all submerged is not a local diffeomorphism. Does anyone have any ideas?
Write $x=(x_1,..,x_{n+p})\in\mathbb{R}^{n+p}$. Consider $g:\mathbb{R}^{n+p}\rightarrow\mathbb{R}^{n+p}$ defined by $g(x_1,..,x_{n+p})=(f_1(x),..,f_p(x),x_{p+1},..,x_{n+p})$ since $f$ is a submersion, the differential of $g$ is inversible, thus we can apply the local inverse theorem to conclude that $g$ is open, since the projection $p:\mathbb{R}^{n+p}\rightarrow\mathbb{R}^p$ is open, we conclude that $f=p\circ g$ is open.
Let $y=(y_1,..,y_p)=f(x_1,...,x_{n+p})$ there exists neighbourhood, $U$ of $x$ and $V$ of $(y_1,..,y_p,0,..,0)$ such that the restriction of $g$ to $U$ is a diffeomorphism onto $V$, there exists $c\neq 0$ such that $(y_1,...,y_p,c,0,..o)\in V$, this implies that there exists $z\in U$ with $g(u)=(y_1,...,y_p,c,0,..,0$, we deduce that $f(z)=p\circ g(z)=f(x)$ and $f$ is injective. In fact this is the idea behind the implicite function theorem.