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?



Fredholmness of an operator-valued Toeplitz operator
This is too long for a comment; but maybe it can help. (I am not an expert in $C^\*$-algebras, though, so it is very possible that I am all wet. If anyone can clean this up I would be grateful). Per the comments, we have that $T\_f$ is not Fredholm but $f$ does have a winding number. This winding number *may* have the following $K$-homology interpretation: let us write simply $C(X)$ for $C\_b(r,1)$. Then $C(\mathbb{T}; C(X))$ may be identified with the algebra $C(\mathbb{T})\otimes C(X)$ (these algebras are commutative, hence nuclear, so there is only one $C^\*$ norm on the tensor product). Likewise, it *seems* (though this should be checked carefully) that the $C^\*$ algebra generated by the $T\_f$'s should be isomorphic to $\mathcal T\otimes C(X)$, where $\mathcal T$ denotes the usual Toeplitz algebra (continuous symbols). On the other hand, since $C(X)$ is nuclear, it is exact, and hence preservers exact sequences under tensoring. Thus by tensoring the usual exact sequence for the Toeplitz algebra with $C(X)$, we have an exact sequence
$$
0\to \mathcal{K}\otimes C(X) \to \mathcal T\otimes C(X)\to C(\mathbb{T})\otimes C(X)\to 0.
$$
This sequence represents an element of the $K$-homology group $Ext(C(\mathbb{T})\otimes C(X), \mathcal{K}\otimes C(X))$. If there is any justice in the world, the quotient map here should be your "symbol map" $T\_f\to f$, and it seems likely that this extension generates a copy of $\mathbb{Z}$ in the $Ext$ group; the pairing of this $Ext$ element with $K^1(C(\mathbb T)\otimes C(X))$ should recover the winding number you describe in the comments. All of this needs to be checked of course, and I am not sure if the non-seperability of $C(X)$ breaks anything. 


**EDIT**: On further reflection, it seems like getting the pairings to work out is really a job for $KK$-theory, which is beyond me; additionaly the non-seperability may be an issue there.


