Let $L:X\rightarrow Y$ be a surjective linear map from Banach space $(X,||\cdot||\_X)$ to vector space $Y$ and denote with $\tau\_L$ the final topology on $Y$ induced by $T$.


Is $\tau\_L$ equivalent with the topology given by the strongest norm on $Y$ in which $L$ is bounded?



Final topology of surjective linear map on Banach space
There exists a norm on $Y$ making $L$ bounded if and only if the kernel of $L$ is closed.


However, $\|y\|\_Y = \inf\lbrace \|x\|: L(x)=y\rbrace$ is always a *semi-norm* (that is
$\|y\|\_Y$ may be $0$ for $y\neq 0$) which induces the final topology $\tau\_L=\lbrace A\subseteq Y: L^{-1}(A)$ is open in $X\rbrace$. This is simple: $\tau\_L$ is finer than
the $\|\cdot\|\_Y$-topology since $L$ is continuous w.r.t. $\|\cdot\|\_Y$. Conversely, if
$y\in A \in \tau\_L$ you find $x\in X$ with $L(x)=y$ and since $L^{-1}(A)$ is open there
is $\varepsilon>0$ with $T(B(x,\varepsilon)) \subseteq A$ which implies that $A$ 
contains $B(y,\varepsilon)$. 


