Why does $E_2: = (I-T)E_1$ with $(I-T)$ being injective imply that $E_2 \neq E_1$?

Question

In chapter 6 of Brezis: Functional Analysis, in the proof on the "Fredholm Alternative," there is the following:

$T$ is a compact operator on $E$ where $E$ is infinite dimensional. Assume $(I-T)$ is injective.

Let $E_1 := R(I-T) \neq E$.

Then $E_1$ is a Banach space and $T(E_1) \subset E_1$. Thus $T|_{E_1} $ is a compact operator on $E_1$, and $E_2:= (I-T)(E_1)$ is a closed subspace of $E_1$.

So far so good.

Then, it says: $E_2 \neq E_1$ since $(I-T)$ is injective. Why must this be true?

Answer 1

Suppose $E_2=E_1$. For $e\in E$ you have $(I-T)e\in E_1$. Since $E_2=E_1$, there exists $f\in E$ with $(I-T)^2f=(I-T)e$. As $I-T$ is injective you get that $$ e=(I-T)f, $$ contradicting the fact that $I-T$ is not surjective.