I am reading Dan Freed's notes on K-theory, where he defines a vector bundle on the set of Fredholm operators transverse to a fixed finite dimensional $W$. In other words, we have $$\mathcal{O}_W=\{T\in \mathrm{Fred}(H,H')|\,TH+W=H'\}$$being the open subspace of operators transverse to $W$ and he constructs $$K_W\rightarrow \mathcal{O}_W$$ a vector bundle whose fiber at $T$ is just the space $T^{-1}W\subset H$. The point is that the image of a compact $X$ into the space of Fredholm operators lands in one of these open subsets, so we only care about the bundles there.
However, these bundles don't behave nicely under e.g. inclusion: if $V\subset W$ is a subset, then the fibers of the same $T$ will be $T^{-1}V$ and $T^{-1}W$ in $K_V$ and $K_W$ respectively, which are related by the short exact sequence $$0\rightarrow T^{-1}V\rightarrow T^{-1}W\rightarrow W/V\rightarrow 0$$ This is the reason why we define the index of a map $F:X\rightarrow \mathrm{Fred}(H,H')$ by $F\mapsto F^*K_W-\underline{W}$, and not just pulling back $K_W$, i.e. a similar SES $$0\rightarrow F^*K_V\rightarrow F^*K_W\rightarrow \underline{W}/\underline{V}\rightarrow 0$$ of vector bundles over a space $X$ shows that the class is well-defined.
My question is whether there is some globally defined object, be it a bundle or a K-theory class, over the whole space $\mathrm{Fred}(H,H')$ which also gives us the index?
More precisely, I am thinking of this in comparison with cohomology and its classifying space $K(G,n)$, where we have a class $u\in H^n(K(G,n);G)$ such that $$\begin{gathered}{}[X,K(G,n)]\simeq H^n(X;G)\\ f\mapsto f^*u\end{gathered}$$
In our situation, we have that $\mathcal{F}=\mathrm{Fred}(H,H)$ classifies $K(X)$ so perhaps there is some class $t\in K(\mathcal{F})$ such that $$\begin{gathered}{}[X,\mathcal{F}]\simeq K(X)\\ F\mapsto F^*t\end{gathered}$$