Intuitively explain why $U: X \rightarrow X$ is invertible if it is close enough to the identity operator. Let $X$ be a Banach space. The Neumann theorem states that an operator $U: X \rightarrow X$ is invertible if it is close enough to the identity operator. This is the theorem.

If $U: X \rightarrow X$ is bounded and $\|I-U\|<1$, then $U$ is invertible, and
$$
U^{-1}=\sum_{k=0}^{\infty}(I-U)^k
$$
Furthermore,
$$
\left\|U^{-1}\right\| \leq \frac{1}{1-\|I-U\|}
$$

I'm not interested in the proof of this theorem, but why $I$ is so important for it? I mean, can I (for example) substitute the operator $I$ with another one and then obtain a similar theorem?
 A: If $T$ is any invertible operator and $\|T-U\|<\|T^{-1}\|^{-1}$, then
$$
\|I-T^{-1}U\|=\|T^{-1}(T-U)\|\leq\|T^{-1}\|\,\|T-U\|<1,
$$
and then by the result quoted in the question you haveh $T^{-1}U$ invertible, which then implies that $U$ is invertible. What this shows is that the set of invertible operators is open; that is, any operator sufficiently close in norm to an invertible operator is invertible.
On an intuitive level, an invertible operator is bounded below: $\|Tx\|\geq c\|x\|$ for all $x$. If $\|T-U\|<c$, then $\|U-T\|<\delta<c$ and
$$
\|Ux\|=\|Tx+(U-T)x\|\geq\|Tx\|-\|(U-T)x\|\geq c\,\|x\|-\delta\,\|x\|=(c-\delta)\,\|x\|
$$
and $U$ is bounded below (hence injective). Being bounded below $U$ has closed range. If $U$ is not surjective, given $y\in (\operatorname{ran}U)^\perp$ there exists $x$ such that $Tx=y$. By scaling if necessary, we may assume that $\|x\|=1$. Then $Tx\perp Ux$, and so
$$
\|T-U\|\geq\|Tx-Ux\|=\sqrt{\|Tx\|^2+\|Ux\|^2}\geq\|Tx\|\geq c,
$$
a contradiction. Hence $U$ is bijective, and so invertible by the Inverse Mapping Theorem.
In summary, if $\|T-U\|<c$ then $U$ is invertible.
