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Let $E$ be a Banach space. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map. Let $T:E \to E$ be a compact (bounded linear) operator. Then Fredholm alternative tells us that $$ \dim N(I-T)=\dim N(I-T^*) < \infty. $$

Then there are closed subspaces $G$ and $L$ of $E$ such that $$ N(I-T) \oplus G =E= R(I-T) \oplus L. $$

Let $\pi_1 : E \to N(I-T)$ be the (continuous linear) projection map. Then $\pi_1$ is surjective with $N (\pi_1) = G$. Let $\pi_2 : E \to L$ be the (continuous linear) projection map. Then $\pi_2$ is surjective with $N(\pi_2) = R(I-T)$.

I would like to ask if $\Lambda:= \pi_2 \circ \pi_1:E \to L$ is surjective.

Thank you so much for your elaboration!

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    $\begingroup$ No. Try $1-T=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ (and take $L=L(e_2)$). $\endgroup$ Commented May 22, 2023 at 15:06
  • $\begingroup$ @ChristianRemling Could you explain the meaning of the notation $L(e_2)$? $\endgroup$ Commented May 23, 2023 at 6:40
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    $\begingroup$ $L(e_2)=\textrm{span }(e_2)$ ("L" as in linear, has nothing to do with the other $L$). $\endgroup$ Commented May 23, 2023 at 15:05

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A counter-example from this answer by Iosif Pinelis also works.

Indeed, suppose e.g. that $E=\mathbb R^2$ and $$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$ here we will identify the linear operators with their matrices in the standard basis of $\mathbb R^2$.

Then (assuming $G$ and $L$ are the orthogonal complements of $N(I-T)$ and $R(I-T)$ respectively) we have $$\pi_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$ $$\pi_2=\frac12\begin{bmatrix}1&1\\ 1&1 \end{bmatrix};$$ $$\Lambda=\begin{bmatrix}0&0\\ 0&0 \end{bmatrix};$$

Clearly, $\Lambda$ is not surjective.

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