Let $f$ be an invertible element of $C({\mathbb{T}}; C_b(r,1))$, that is, there exists a $f^{-1}\in C({\mathbb{T}}; C_b(r,1))$ such that for all $z\in {\mathbb{T}}$, $f(z)f^{-1}(z)=1$ in $C_b(r,1)$. Here $C_b(r,1)$ denotes the $C^*$-algebra of complex-valued bounded continuous functions on the open interval $(r,1)$, where $r$, fixed, belongs to $[0,1)$, and $\mathbb{T}$ denotes the unit circle with center $0$. Then $f$ induces a bounded operator $M_f$ on $L^2({\mathbb{T}}; L^2(r,1))$ in a natural manner: for each $z$ in $\mathbb{T}$, we have the multiplication operator corresponding to $f(z)$ in $C_b(r,1)$ going from $L^2(r,1)$ to $L^2(r,1)$. Using the projection $P:L^2({\mathbb{T}}; L^2(r,1)) \rightarrow H^2({\mathbb{T}}; L^2(r,1))$, we can consider the Toeplitz operator $T_f $ on $H^2({\mathbb{T}}; L^2(r,1))$ given by $T_f g = P (M_f g)$ for $g$ in $H^2({\mathbb{T}}; L^2(r,1))$.

My question is this: Is $T_f$ is Fredholm?