Let $N, M$ be two smooth manifolds and $f: N \rightarrow M$ a smooth immersion.
(i)-If $f$ is a homeomorphism on its image the $f(N)$. Then $f(N)$ is a submanifold.
(ii)- If the dimension of $N, M$ are equal, then $f$ is an open mapping.
for (ii) we have :
It suffices to show that $f$ maps "small" enough open sets to open sets. For that, given $p \in M$, let $U$ be a chart around $p$ such that $f$ is locally the identity. That is, $U$ and $V$ are such that the following diagram commutes.
$\require{AMScd}$ \begin{CD} U @>f>> V\\ @V \phi V V @VV \psi V\\ U'\subset \mathbb{R}^n @>>Id> V' \subset \mathbb{R}^n. \end{CD}
This is a consequence of the local form of immersions (since $\dim M=\dim N$! Otherwise, $Id$ should be replaced by a inclusion). Now, $f(U)=\psi^{-1} \circ Id \circ \phi$, and each one is an open map ($\phi$ and $\psi^{-1}$ by assumption about charts, and $Id$ since it is $Id$).