Doubt in the iteration argument on Fredholm theorem's proof.

Question

In final of page 161 here:

By assumpuption I know that $E_1 \neq E$, because you're supposing that $I-T$ isn't surjective. However, I didin't understand why $E_2 \neq E_1$ (since $I-T$ is injective.)?

Answer 1

If $(I-T)E_1=E_1$, then the fact that $(I-T)E=E_1$ would result in a non-zero element $x\in E$ for which $(I-T)x=0$. To see this, choose $x\in E\setminus E_1$, note that $(I-T)x \in E_1=(I-T)E_1$ would give $y\in E_1$ such that $(I-T)y=(I-T)x$ or $(I-T)(x-y)=0$, even though $x\ne y$ because $x\notin E_1$ and $y\in E_1$.