Definition
A $k$-manifold with $k>0$ in $\Bbb R^n$ of class $C^r$ is a subspace $M$ of $\Bbb R^n$ such that for any $y\in M$ there exists an open set $V$ of $M$ containing $y$, and open set $U$ in either $\Bbb R^k$ or $H^k$ and a continuous map $\alpha:U\rightarrow V$ carrying $U$ onto $V$ in one-to-one fashion such that
- $\alpha$ is of class $C^r$;
- $\alpha^{-1}:V\rightarrow U$ is continuous;
- $D\alpha(x)$ has rank $k$ for each $x\in U$.
The map $\alpha$ is called a coordinate patch on $M$ about $y$.
Well I ask to prove that If $f$ is a function defined in a open set $A$ of $\Bbb R^k$ with values in $\Bbb R^n$ and with derivative $Df(x)$ of rank $k$ at each point $x$ of $A$ then the set $M:=f[A]$ is a $k$-manifold.
So by this relevant result we know that the map $f$ is locally injective and so any element $x$ of $A$ has a neighborhood $U_x$ where $f$ is injective so that we have only to prove that the set $f[U_x]$ is open in $M$ and so that the local inverse of $f$ at $U_x$ is continuous but unfortunately I do not able to do it. So could someone help me, please?
MY ATTEMPT
So by the link I posted I know there exist $i_1,...,i_k=1,...,n$ such that the function $$ \phi:=\pi_{i_1,...,i_k}\circ f $$ is a local diffeomorphism where $\pi_{i_1,...,i_k}$ is a function from $\Bbb R^n$ onto $\Bbb R^k$ defined through the equation $$ \big[\pi_{i_1,...,i_k}(x)\big](h):=x(i_h) $$ for any $x\in\Bbb R^n$ and for any $h=1,...,k$. So for any $x\in A$ there exist a open neighborhood $U_x$ where $\phi$ is injective and so $f$ too. Now by the continuity of $\pi_{i_1,...,i_k}$ the set $$ V_{f(x)}:=\pi^{-1}_{i_1,...,i_k}\Big[\phi[U_x]\Big]\cap M $$ is an open neighborhood of $f(x)$ in $M$ so observig that $$ y\in f[U_x]\Rightarrow\pi_{i_1,...,i_k}(y)\in\pi_{i_1,...,i_k}\Big[f[U_x]\Big]=\phi[U_x]\Rightarrow y\in\pi^{-1}_{i_1,...,i_k}\Big[\phi[U_x]\Big]\cap M=V_{f(x)}\Rightarrow f[U_x]\subseteq V_{f(x)}\Rightarrow U_x=f^{-1}\Big[f[U_x]\Big]\subseteq f^{-1}[V_{f(x)}] $$ so that the statement folows showing that $$ V_y:=f[U_x] $$ because in this case the continuous function $\phi^{-1}\circ\pi_{i_1,...,i_k}$ would be the inverse of $f$ at $U_x$. So in this case we observe that $$ y\in f^{-1}[V_{f(x)}]\Rightarrow f(y)\in V_{f(x)}\Rightarrow \phi(y)=\pi_{i_1,...,i_k}\big(f(y)\big)\in\pi_{i_1,...,i_k}[V_y]\subseteq\pi_{i_1,...,i_k}\Biggl[\pi^{-1}_{i_1,...,i_k}\Big[\phi[U_x]\Big]\Biggl]=\phi[U_x]\Rightarrow\phi\Big[f^{-1}[V_{f(x)}]\Big]\subseteq\phi[U_x]\Rightarrow f^{-1}[V_{f(x)}]\subseteq U_x $$ and thus we conclude that $$ f^{-1}[V_{f(x)}]=U_x $$ and so the statement follows. But I am not sure about the correctdness of the proof of the last inclusion. So are my arguments correct?