If $A$ is a $C^*$-algebra, there is a notion of Hilbert $A$-module (which is something like Hilbert space but the inner product takes values in $A$). The standard example is $H_A:=\{(a_n)_{n=1}^{\infty}: \sum_{n=1}^{\infty}a_n^*a_n \ is \ norm \ convergent\}$ with $A$ valued inner product $(a,b):=\sum_{n=1}^{\infty}a_n^*b_n$. The algebra of all adjointable operators on Hilbert module turns out to be a $C^*$-algebra and there is a notion of $A$ compact operator: these are operators which are norm limits of linear combinations of operators of the form $\theta_{x,y}$ where $\theta_{x,y}(z)=x(y,z)$ (our modules are *right* modules). Having the notion of $A$ compact operator one can speak about $A$ Fredholm operators ($A$-linear). Let us take $A=C(X)$ (continuous functions on some compact space) and let $T:X \to Fred(H)$ be a continuous family of Fredholm operators on $H=\ell^2$.

Why is it true such that $T$ is $A$-Fredholm operator on $H_A$ for $A=C(X)$?

My guess is that if $T$ is a continuos family of Fredholm operators one can view $T$ as an operator $H_A \to H_A$ via $T\xi(x)=T_x\xi(x)$ where $\xi \in H_A$ is a sequence of continuous functions (so $\xi(x)$ makes sense as an element of $\ell^2$). However it is not obvious for me how to show that $T$ is now $A$-Fredholm. I think that the proof boils down to the following fact: if $T:X \to K(H)$ is a family of compact operators on $\ell^2$ then $T$ viewed as a $A$ linear operator in $H_A$ (as explained above) is $A$-compact.