Assume $X$ is a Banach space, $\Omega \subseteq X$ is an open set, $K\in {C}^{1}( \overline{\Omega}, X)$ is a nonlinear compact map, I heard that $\operatorname{Id}-K$ is a proper map. Proper map means that the preimage of every compact set is also compact. But I don't know how to prove it? Could you give me some hints? And I often heard $\operatorname{Id}-K$ is a Fredholm operator when $K$ is a linear compact operator from my teacher. But I don't know how they are connected. Could you please explain it?
I'm trying to prove the first question but I met some difficulties. Here are my analysis. Take a compact set $M\subseteq X$, let $N={f}^{-1}(M)\cap \overline{\Omega}$. Here $f=\operatorname{Id}-K$. Take a sequence $\left \{ {x}_{n}\right \}$ from $N$, then there exists a sequence $\left \{ {y}_{n}\right \}$ from $M$, such that ${y}_{n}={x}_{n}-K({x}_{n})$. If $\left \{ {x}_{n}\right \}$ is bounded, I can see $\left \{ K({x}_{n})\right \}$ and $\left \{ {y}_{n}\right \}$ both have convergent subsequence, then $\left \{ {x}_{n}\right \}$ have convergent subsequence too, thus $f$ is proper. But I'm stuck in showing that $\left \{ {x}_{n}\right \}$ is bounded. How can it be shown?
I think maybe $f$ is not proper, because if X is a finite dimensional space, take $K=\operatorname{Id}$ is compact, then $f=0$, obviously it's not proper. Are there any mistakes? I think maybe it has something to do with the spectrum of the compact operator.