Let $H$ be a separable Hilbert space and let $\mathcal{B}(H)$ denote the algebra of linear bounded operators on $H$. Let $T \in \mathcal{B}(H)$ and let $M$ be a non-trivial closed invariant subspace for $T$. In general, you cannot expect $\sigma(T\mid_M) \subset \sigma(T)$.
For example, see the first answer in Spectra of restrictions of bounded operators. The spectrum of the operator of this example is the unit circle, but the spectrum of the restriction is the whole closed unit disk.
I've been looking for examples of this kind and everything I found satisfies $\sigma(T\mid_M) \subset \eta(\sigma(T)),$ where $\eta(\sigma(T))$ is the full spectrum, id est, $\sigma(T)$ together with the "holes" of $\sigma(T)$. Is it true that $\sigma(T\mid_M) \subset \eta(\sigma(T))$ for every operator $T \in \mathcal{B}(H)$? In particular, we would have $\sigma(T\mid_M) \subset \sigma(T)$ whenever $\sigma(T)$ is simply connected. If this is true, I would like to have a good reference to read the proof. If it is false, I would like to see a counterexample.
Thank you very much!