Is the induced map on the quotient space bounded?

Question

Let $T: V \rightarrow W$ be a continuous/bounded linear map between Banach spaces. Let $J \subseteq \ker(T)$ be a closed subspace. Then I know that this induces a unique linear map on the quotient $\bar{T}:V/J \rightarrow W$ s.t. $\bar{T}\circ \pi=T$. I also know from topology that $\bar{T}$ is continuous w.r.t. the quotient topology on $V/J$. We can also define a norm on $V/J$ via:

$$ \forall v \in J: \ \|v+J\| := \inf_{w \in J} \|v+w\| $$ In particular is $J$ itself again a Banach space with this norm. I am now wondering whether $\bar{T}$ is always continuous w.r.t. this norm (or is the topology induced by this norm exactly the quotient topology?).

I managed to come up with this proof and was wondering if it is correct or if there is another, faster way to see this (perhaps by showing that the norm topology is the quotient topology):

\begin{align*} &\sup \{\|\bar{T}(v+J)\|\ : \ \|v+J\| \leq 1\} \\ = & \sup \{\|T(v)\|\ : \ \|v+J\| \leq 1\} \end{align*}

Now by definition of the infimum $\|v+J\| \leq 1 \Rightarrow \forall \varepsilon > 0 \ \exists w \in J: \ \|v+w\| \leq 1 + \varepsilon$. Thus $\|T(v)\| = \|T(v+w)\| \leq \|T\|\|v+w\|\leq \|T\|(1+\varepsilon)$

Thus I get $\|\bar{T}\|\leq \|T\|(1+\varepsilon) \rightarrow \|T\|$ showing boundedness.

Answer 1

The quotient topology is topology induced by the quotient norm. The quotient topology is such that $U\subset V/J$ is open iff $Q^{-1}(U)$ is open, where $Q:V\to V/J$ is the quotient map $Qx=x+J$.

It suffices to show that $U\subset V/J$ is norm-open iff $Q^{-1}(U)$ is. The quotient map satisfies $\|Q\|\leqslant 1$, so it's Lipschitz continuous, and $U\subset V/J$ open implies $Q^{-1}(U)$ is open. On the other hand, if $Q^{-1}(U)$ is open and $U$ is non-empty, fix $y\in U$. Fix $x\in Q^{-1}(U)$ such that $Qx=y$. Since $Q^{-1}(U)$ is open, there exists $r>0$ such that $x+rB_V\subset Q^{-1}(U)$. Here $B_V=\{y\in V:\|y\|<1\}$ is the open unit ball. I claim that $Q(x+rB_V)=y+rB_{V/J}$, so that $$U\supset Q(Q^{-1}(U))\supset Q(x+rB_V)=y+rB_{V/J}.$$ Thus the open ball $y+rB_{V/J}$ is contained in $U$, and $U$ is open. Thus $U\subset V/J$ is open in the quotient topology iff it's open in the quotient norm topology.

To see why $Qx+rB_V=y+rB_{V/J}$, first fix $u\in V$ with $\|u\|<r$. Then $Q(x+u)=Qx+Qu$, and $\|Qx+Qu-Qx\|=\|Qu\|\leqslant \|u\|<r$. So $Q(x+u)\in y+rB_{V/J}$. Next, fix $w\in Q/J$ with $\|w\|<r$. Then by the definition of infimum, there exists $u\in V$ with $\|u\|<r$ such that $Qu=w$. Then $Q(x+u)=y+w$.