Let $T$ be the operator defined by $T:=I-A$ where $I:H\to H$ is the identity and $A:H\to H$ is a compact operator defined on Hilbert space $H$. In such a case, if we defined the chain of sets$$H\supset H^1\supset H^2\supset\ldots\supset H^k\supset \ldots$$by $H^k:=T(H^{k-1})$, there must be a $j\in\mathbb{N}$ such that $\forall k\ge j\quad H^{k+1}=H^k$ (p. 470 here).
In order to prove that $\ker T=\{0\}\Rightarrow \text{im }H=H$ (where $\text{im }H:=T(H)$), Kolmogorov and Fomin (p. 470) say that if $\ker T=\{0\}$ then operator $T$ will be injective and, consequentially, if $\text{im }H\ne H$, the sequence $H\supset H^1\supset H^2\supset\ldots$ will be composed of distinct subspaces, which contradicts the fact that $\exists j\in\mathbb{N}:$ $\forall k\ge j\quad H^{k+1}=H^k$.
Why, if $T$ is injective, must all $H^k$ be distinct? I see that $H^1\subsetneq H$, but I do not see why they are all distinct. I know that $H=\ker T\oplus\text{im }T^\ast$, but adjoint operator $T^\ast$ does not seem to be used here, to me, and I am not sure whether it is useful to understand this implication. Thank you so much for any explanation!
Since the $H^k$ are descending, having $H^m = H^n$ for some $n > m$ means $H^k = H^m$ for all $k \geqslant m$, as $H^m \supset H^{m+1} \supset \dotsc \supset H^n = H^m$ forces all these spaces to be equal, and then, for $k \geqslant m$ we have $H^{k+2} = T(H^{k+1}) = T(H^k) = H^{k+1}$ by induction.
Suppose that $H^1\subsetneq H$, but the $H^k$ are not all distinct. Then the sequence $(H^k)$ stabilizes at some point, say
$$H^{m-1} \supsetneq H^m = H^{m+1}.$$
Then look at some $x\in H^{m-1}\setminus H^m$. We have $Tx\in H^m$ by definition, and hence $Tx \in H^{m+1}$ by assumption. But that means there is an $y\in H^m$ with $Tx = Ty$. Then $x-y \in \ker T$, but by construction $y-x \in H^{m-1}\setminus H^m$, so $y-x \neq 0$, hence $\ker T \neq \{0\}$.
Thus, if the sequence stabilizes at some point, but is not constant from the beginning, then $T$ cannot be injective. By contrapositive, if $T$ is injective, either $H^1 = H$ or $H^{k+1} \subsetneq H^k$ for all $k$.