# Surjective linear operator onto finite-dimensional vector space

Let $(X,\|\|_X),(Y,\|\|_Y)$ be two normed vector spaces ovef $K$ s.t. $dim_k(Y)<\infty$. Let $A:X\to Y$ be a surjective linear operator. I want to prove that $A$ is open. My attempt goes like this:

Since $Y$ is finite-dimensional, let $dim_K(Y)=n$ and let $\beta=\{e_i\}_{i=1}^n$ be a normalized base for $Y$ (i.e. $\|e_i\|=1\;\;\forall i\in\{1,2,...,n\}$).

So, we get that $\;\forall y\in Y\;\;\exists\;a_1,...a_n\in K$ s.t. $y=\sum_{i=1}^na_ie_i$.

Then, since $A$ is surjective, we get that $\;\;\forall e_i\in\beta\;\;\exists x_i\in X$ s.t. $\;e_i=A(x_i)\;\;\forall i\in\{1,2,,...,n\}$ and lets define $\;\varphi:Y\to X$ s.t. $\varphi(y)=\sum_{i=1}^n a_ix_i\;\;\forall y=\sum_{i=1}^n a_ie_i\in Y$.

I've already proved that $\varphi$ is linear, bounded and thus continuous. And also that $\;Id_Y=A\circ\varphi$

Then, let $U\subseteq X$ be open so to finish the prove I need to show that $A(U)$ is open, but I haven't managed to link $\varphi$ with $A$ in way that gives me the openess of A( I started to prove if $\varphi$ is open, but haven't got anything). Any help or ideas would be appreciated.

The following fact may help: A linear map between normed vector spaces is open iff the image of the unit ball contains a ball centred at $0$. To show this, observe that linear maps commute with translation and scaling.
• Ok, so I need to prove that: $$\exists\; r_0>0\text{ s.t. }B_{r_0}(0)\subseteq A(B_{1}(0))$$ So, lets define $\|\;\|_2:Y\to K$ s.t. $\|y\|_2=\max_{i\in\{1,2,...,n\}}(|\alpha_i|)\;\;\forall y=\sum_{i=1}^n \alpha_i e_i$ with $\alpha_i\in K\;\forall i\in\{1,2,...,n\}$. Clearly, $\|\;\|_2$ is a norm in $Y$ and, since $Y$ is finite-dimensional, $\|\;\|_Y$ and $\|\;\|_2$ are equivalent, thus there $\;\exists c>0$ s.t. $c\|y\|_2\le\|y\|_Y\;\;\forall y\in Y$ So, taking $$r_0=\frac{c}{\sum_{i=1}^n\|x_i\|_X}\;\text{ where }x_i=A(e_i)\;\forall i\in\{1,2,...,n\}$$ – Arnulf Mar 11 '16 at 23:51
• then, if $y\in B_{r_0}(0)$ we take $x=\sum_{i=1}^n\alpha_ix_i$, where $y=\sum_{i=1}^n\alpha_ie_i$, clearly $A(x)=y$ and $$\|x\|_X\le\sum_{i=1}^n|\alpha_i|\|x_i\|_X\le\|y\|_2\sum_{i=1}^n\|x_i\|_X\le \frac{\|y\|_Y\sum_{i=1}^n\|x_i\|_X}{c}<\frac{r_0\sum_{i=1}^n\|x_i\|_X}{c}=1$$ thus $x\in B_1(0)$, thereby $B_{r_0}(0)\subseteq A(B_1(0))$. Now, let $U\subseteq X$ be open so there exists $R_0>0$ s.t. $B_R(0)\subset U$ so $$A(B_R(0))\subseteq A(U)$$ but $A(B_R(0))=RA(B_1(0))$, so there exists $r_0>0$ s.t. $$B_{r_0}(0)\subset A(B_1(0))$$ – Arnulf Mar 12 '16 at 0:02
• $$\Leftrightarrow R_0B_{r_0}(0)\subset R_0A(B_1(0))\Leftrightarrow B_{R_0r_0}(0)\subset A(B_{R_0}(0))\subset A(U)$$ thus, $A(U)$ is open and thereby $A$ is open. – Arnulf Mar 12 '16 at 0:02