Atiyah–Jänich's theorem says that $$ \left[X\to\mathcal{F}\left(\mathcal{H}\right)\right] = K_0\left(X\right) $$where $\mathcal{H}$ is any separable complex Hilbert space, $\mathcal{F}\left(\mathcal{H}\right)$ is the set of Fredholm operators on $\mathcal{H}$, $X$ is any topological space, and $K_0$ is the topological-$K_0$ functor which gives roughly speaking stable isomorphism classes of vector bundles on $X$.
My question is, how can one complete the following analogous equation for $K_1$:$$ \left[X\to???\right] = K_1\left(X\right) $$ (or in fancy language, what is the classifying space of the $K_1$ group?
Via suspensions, we can translate the question back to $K_0$: $K_1\left(X\right)=K_0\left(S X\right)$, so that $$ \left[SX\to\mathcal{F}\left(\mathcal{H}\right)\right] = K_1\left(X\right) $$However, this is unpleasant, because then you get homotopy classes of maps from $SX$ on the L.H.S. instead of homotopy classes of maps from $X$. Additionally, one would like an answer that takes into account the more intrinsic nature of $K_1$ (as opposed to $K_0$--not via suspensions): $K_1$ classifies unitaries, at least in algebraic K-theory.
The closest thing I could find is pp. 11 of Higson, Connes, Baum which mentions self-adjoint Fredholm operators which apparently define a class in $K_1$. However, I thought those were always trivial?
The question is answered in the comments: $[SX,\mathcal{F}]\cong [X,\Omega(\mathcal{F}(\mathcal{H}))]\cong [X,\Omega(\mathbb{Z}\times BU)]\cong [X,\Omega(BU)]\cong [X,U]$.
If you want a more operator theoretic definition of this, you can prove that there exists a homotopy equivalence between $U$ and the following subgroup $GL_c(\mathcal{H})$ of $GL(\mathcal{H})$
$$ GL_c(\mathcal{H}):=\{A\in GL(\mathcal{H})\,|\, A=1+K,\quad \text{with}\quad K\quad \text{a compact operator}\} $$
These are the invertible operators that differ from the identity by a compact operator.
This is probably discussed in the paper: R. S. Palais, On the homotopy type of certain groups of operators, Topology 3 (1965), 271–279.
Edit: I don't think it is too surprising that $GL_c(\mathcal{H})$ has the right topology. Recall that $GL(\mathbb{C}^\infty)$ deformation retracts to $U$ ($=U(\mathbb{C}^\infty)$). You can view a $A\in GL(C^\infty)$ as a matrix of the form $$ A=\left(\begin{array}{cc}B&0\\0&1\end{array}\right) $$ where $B\in GL(n)$ and $1$ represents the identity on the second part of $\mathbb{C}^\infty\cong \mathbb{C}^n\times \mathbb{C}^{\infty-n}$. If one embeds $\mathbb{C}^\infty$ into a complex Hilbert space (via a choice of basis) $B$ can be viewed as a finite rank operator. Limits of finite rank operators are compact operators, hence I would understand $GL_c(\mathcal{H})$ as a closure of $GL(C^\infty)$.