Im working through the following theorem, and I don't understand the explanation.
Theorem: Let $T_{\varphi}: H^2(T) \longrightarrow H^2(T)$ be the Toeplitz operator with $T \subset \mathbb{C}$ the unit circle and $\varphi \in C(T)$. Then $\sigma(T_{\varphi})$ is connected.
Proof: Notice that $$\sigma(T_{\varphi}) = \varphi(T) \cup \{\lambda \in \mathbb{C} \: | \: \text{$T_{\varphi}-\lambda I$ is Fredholm of nonzero index}\}$$, which is the union of a connected compact set $\varphi(T)$. Hence $\sigma(T_{\varphi})$ is a compact set consisting of a connected compact set $\varphi(T)$ and some of its holes (some of the bounded components of $\mathbb{C} \setminus \varphi(T)$), thus $\sigma(T_{\varphi})$ is connected.
I understand the set equality, and obviously $\varphi(T)$ is connected and compact, but why dont we have to show that $$\{\lambda \in \mathbb{C} \: | \: \text{$T_{\varphi}-\lambda I$ is Fredholm of nonzero index}\}$$ is connected, and also has a limit point in $\varphi(T) = \sigma_e(T_{\varphi})$ (the essential spectrum) to show connectedness?
Since $$\{\lambda \in \mathbb{C} \: | \: \text{$T_{\varphi}-\lambda I$ is Fredholm of nonzero index}\}$$ is a subset of the unions of all holes of $\varphi(T)$, why is it the nessiarily case that these holes are 'filled' by the points in this set, and simply not isolated blobs contained inside the holes of $\varphi(T)$?
The set of complex numbers $\lambda $ such that $\text{ind}(T_{\varphi}-\lambda )\neq 0$ is not only
it is actually
In other words, whenever this set has a single point in a given hole, the entire hole is included!
To see this suppose that $\lambda _0$ and $\lambda _1$ lie is some connected component $\Omega \subseteq \mathbb C\setminus \varphi (T)$, and $\text{ind}(T_{\varphi}-\lambda_0 )\neq 0$. We may then choose a path $\lambda _t$ joining $\lambda _0$ and $\lambda _1$ within $\Omega $. Consequently $$ S_t:= T_{\varphi}-\lambda_t ,\quad t\in [0,1] $$ is a path of operators joining $T_{\varphi}-\lambda_0$ to $T_{\varphi}-\lambda_1$. Since $\lambda _t$ is never in the essential spectrum of $T$, then $S_t$ is Fredholm for all $t$. Now, using the fact that the index is continuous (hence locally constant), we see that the index of $S_t$ is constant with $t$, hence $$ \text{ind}(T_{\varphi}-\lambda_1) = \text{ind}(T_{\varphi}-\lambda_0) \neq 0, $$ so $\lambda _1$ lies in our set.