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Could someone help me with this question?

Let be $T\in{L(X,Y)}$ and $M=Ker(T)\neq{X}$ and $T^{*}:X/M\rightarrow{Y}$ the only linear operator such that $T=T^{*}\circ\pi$. Show that $ T^{*}$ is continuous.

Thanks

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    $\begingroup$ Who is the $\pi$ mapping? $\endgroup$ – DanielC Oct 11 '17 at 10:36
  • $\begingroup$ Probably the natural epimorphism $\pi:X\rightarrow X/M,x\mapsto [x]=x+M$, where $X/M$ is equipped with the quotient space topology. $\endgroup$ – Peter Melech Oct 11 '17 at 10:54
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    $\begingroup$ $T^*$ is defined by $T^*(x + ker\,T) := Tx$. $\endgroup$ – amsmath Oct 11 '17 at 10:54
  • $\begingroup$ Yes, $\pi$ is the natural epimorphism which @PeterMelech talks about $\endgroup$ – mathlife Oct 11 '17 at 11:04
  • $\begingroup$ The quotient space can be equipped with a norm $|[x]|_{X/M}=\inf_{m\in M}|x-m|_X$ that induces the quotient space topology. $\endgroup$ – Peter Melech Oct 11 '17 at 11:10
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Let $[x]=x+M$ , then $||[x]||= \inf\{||z||: z \in [x]\}$

Let $z \in [x]$ .

Then $||T^*[x]||=||Tz|| \le ||T||\cdot ||z||$,

hence $||T^*[x]||\le ||T||\cdot ||[x]||$.

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