# Continuous linear operator / Proof

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

• Who is the $\pi$ mapping? – DanielC Oct 11 '17 at 10:36
• Probably the natural epimorphism $\pi:X\rightarrow X/M,x\mapsto [x]=x+M$, where $X/M$ is equipped with the quotient space topology. – Peter Melech Oct 11 '17 at 10:54
• $T^*$ is defined by $T^*(x + ker\,T) := Tx$. – amsmath Oct 11 '17 at 10:54
• Yes, $\pi$ is the natural epimorphism which @PeterMelech talks about – mathlife Oct 11 '17 at 11:04
• 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. – Peter Melech Oct 11 '17 at 11:10

Let $[x]=x+M$ , then $||[x]||= \inf\{||z||: z \in [x]\}$
Let $z \in [x]$ .
Then $||T^*[x]||=||Tz|| \le ||T||\cdot ||z||$,
hence $||T^*[x]||\le ||T||\cdot ||[x]||$.