# Quotient space and continuous linear operator.

Let $X$ and $Y$ be banach space let, $T\in B(X,Y)$. Define $N$ as the Kernel of $T$ and $M$ as the range of $T$ . We know that $\bar T : X/N\to M$ is well defined , ie if $(x+N) =(y+N)$ then $Tx$ and $Ty$ are equal , and this map is one to one , the kernel is the trivial kernel ie. $N$ .

Now to find out if $\bar T$ is continuous if we see $X/N$ as a quotient space . I wonder if its straight forward , I have a slight confusion with the above notation , $B(X,Y)$ means all the continuous operators from $X$ to $Y$ right ?? or these maps need not be continuous ?? If its continuous , then $T(x+N)=T(x)+T(N)$ but $T(N)=0$ because its kernel. So it follows that $\|T(x+N)\|=T(x)\le C \|x\|$ because $T$ is continuous from $X$ to $Y$. Am i wrong here ??

Can you tell me When a map is called topological ? couldn't find it in google . I want to find a condition on when $\bar T$ is topological , is it when $M$ is closed, but why , i would appreciate to see proof or some sketch of it ??

Yes i got that $B(X,Y)$ represents the bounded linear operators. But does my reasoning make any sense, if not i need help .

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Could you tidy up your question a bit? A linear operator is bounded if and only if it is continuous. In the second paragraph you are wondering whether elements in $B(X,Y)$ are continuous. In the last paragraph you say you understand that $B(X,Y)$ are bounded linear operators. ($B$ probably stands for bounded). Also, using more than one question mark doesn't make it any more of a question. Unless you're a chess player, perhaps. – Rudy the Reindeer Nov 21 '12 at 22:34

(2) You are right, $B(X,Y)$ usually denotes the continuous linear maps $X \to Y$.
$\def\norm#1{\left\|#1\right\|}$(3) To see that $\bar T$ is continuous when $T$ is, recall that the norm on $X/N$ is defined by $$\norm{x+N}_{X/N} = \inf_{n\in N} \norm x_X$$ So we have for $x + N \in X/N$ that \begin{align*} \norm{\bar T(x+N)}_Y &= \norm{Tx + 0}_Y\\ &= \inf_{n\in N} \norm{Tx + Tn}_Y\\ &\le \inf_{n\in N}\norm T \norm{x+n}_X\\ &= \norm T \norm{x+N}_{X/N} \end{align*} So $\bar T$ is continuous with norm $\norm{\bar T}\le \norm T$.
(4) If $\bar T$ is topological, $M$ is closed as $X/N$ is complete and hence the (then) isomorphic $M$ must be also, and therefore closed in $Y$. On the other side, suppose $M$ is closed, then $\bar T\colon X/N \to M$ is a continuous, bijective linear mapping between Banach spaces and therefore its inverse is continuous by the open mapping theorem.
why does $\bar T$ topological mean that $X/N$ complete ? – Theorem Nov 22 '12 at 19:24
$X/N$ is always complete (as $X$ is). As $\bar T$ is topological, $\bar T(X/N) = TX = M$ has to be also. – martini Nov 22 '12 at 21:20