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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 .

Thanks for your 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.2012-11-21

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