Why is this element invertible in the Calkin algebra?

Question

This question is coming from a section of Rordam's "K-theory for C*-algebras", section 9.4.2.

Let $H$ be an infinite dimensional separable Hilbert space, and let $T$ be a Fredholm operator in $B(H)$. Let $T=S|T|$ be the polar decomposition of $T$, where $S$ is a partial isometry and $|T|=(T^*T)^{1/2}$.

For $t\in[0,1]$, define the path $R_t=S(t|T|+(1-t)I)$. If $\pi$ is the projection of $B(H)$ into the Calkin algebra (modulo compact operators), then $\pi(R_t)=\pi(S)(t|\pi(T)|+(1-t)I)$. The book claims then for each $t$, this element is invertible in the Calkin algebra and hence $R_t$ is Fredholm.

Since $T$ is Fredholm, then $\pi(T)$ is invertible and hence $\pi(S)$ is unitary, so really I just need to see that $(t|\pi(T)|+(1-t)I)$ is invertible. Now $|\pi(T)|$ and $I$ are both invertible in the Calkin algebra, and so I'm basically creating the straight line between two invertible elements, but I don't immediately see why everything on that path must also be invertible.

Answer 1

It follows from this fact:

If you have a positive invertible operator, T, in a unital $C^*$-algebra, then the path $(1-t)T+tI$ consists of invertible (positive) operators.
Just apply the functional calculus to the element $T$, whose spectrum is contained in $[\epsilon, \infty)$ for some $\epsilon>0$. The function $(1-t)x+t1$ is strictly positive in this segment (greater than epsilon), and therefore invertible.