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!



Is $\Lambda:= \pi_2 \circ \pi_1:E \to L$ surjective?
A counter-example from [this](https://mathoverflow.net/a/447402/477203) 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.


