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Let $X$ and $Y$ be Banach spaces, and $T\in B(X,Y)$ such operator that $imT$ has a finite codimension in $Y$ i.e. there exists vector space $V\subseteq Y$ that $dimV$ is finite and $Y=imT\oplus V$. Let us define Banach space $X_1=X\oplus V$ with norm $\parallel (x,v)\parallel = \parallel x \parallel + \parallel v \parallel$ and operstor $S: X_1 \rightarrow Y$ $S(x,v)=Tx+v$, where $x\in X$ and $v\in V$. Prove that $S$ is open map and that $imT$ is closed in $Y$.

To prove that map is open I should prove that $S$ is bounded and that it is surjective. I have proved that it is bounded but I am stuck with surjectivity. To prove that $imT$ is closed is it enough to prove that $T$ is bounded below and is it possible to do it in this case?

I would appreciate some help with this problem. Or at least a hint to take me to solution.

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  • $\begingroup$ Surjectivity of $S$ comes for free from the fact that you can decompose $Y$ as $im T \oplus V$ $\endgroup$ – Tony S.F. Apr 25 '18 at 11:06
  • $\begingroup$ Yes, I see now thank you. $\endgroup$ – XYZ Apr 25 '18 at 12:18

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