Let's $f\in \mathcal{L}(\mathbb{R^n},\mathbb{R^p})$. Prove that $f$ is surjective if and only if the image of each open set is an open set.
My attempt :
- The second implication is easy. If the image of each open set is an open set, it's especially true for $\text{Im} f$, and so it's $\mathbb{R}^p$ because the only open subvectorial space that is not empty is the vectorial space (because each open ball centred on $0$ contains a base of $\mathbb{R}^n$).
- I've got problems proving the direct implication. Here is what I tried :
If $U$ is an open set, $y\in U$, $x$ such that $f(x)=y$, then the only thing I noticed is for all $r$, $f(B(x,r)) = f(x+rB)=y+rf(B)$ where $B$ is $B(0,1)$, so that I just need to show $f(B)$ is a neighbourhood of $0$ to conclude.
Can someone help me?
Thanks in advance.
Thanks to @user3482749 's hint, I can answer my own question :
$f(B)$ contains a basis of $\mathbb{R}^n$. Let's not $B=(e_1,...,e_n)$. We can suppose without loosing generality that $(f(e_1),...f(e_p))$ is a basis of $\mathbb{R}^p$. Then $B$ contains all $\lambda_1 f(e_1) + \cdot\cdot\cdot \lambda_p f(e_p)$, with the $\lambda_i\in ]-1,1[$ which is clearly a neighbourhood of $0$.