# Final topology of surjective linear map on Banach space

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?

• Is the kernel of $L$ closed? – Narutaka OZAWA Apr 2 '14 at 17:11
• Not necessarily. Another phrasing for the question might be "Does the equivalence of the two topologies follow?" – jmk Apr 2 '14 at 17:31
• Is it clear that such a strongest norm exists? – Nate Eldredge Apr 2 '14 at 20:14

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 $B(y,\varepsilon)$.