Given a compact Hausdorff space $X$ we can construct its topological K-theory. On the other hand the continuous functions on $X$ give rise to a C-algebra $C(X)$ and we can consider the operator K-theory of this C-algebra. My question is how these two are related to each other.
By Swan's theorem we have $K^0(X)\cong K_0(C(X))$, i.e. an isomorphism between the zero-th topological K-group of $X$ on the left and the zero-th operator K-group of $C(X)$ on the right. Does a relation like this hold for the other higher K-groups as well?
Looking at the circle $\mathbb{S}^1$ in particular, I think we have \begin{align*} K^{-1}(\mathbb{S}^1) \cong & \; K^0(S\mathbb{S}^1) \cong K^0(\mathbb{S}^2) \cong \mathbb{Z}\oplus\mathbb{Z} \\ \tilde{K}^{-1}(\mathbb{S}^1) \cong & \; \tilde{K}^0(S\mathbb{S}^1) \cong \tilde{K}^0(\mathbb{S}^2)\cong \mathbb{Z} \\ K_1(C(\mathbb{S}^1)) \cong & \; K_1(C(\mathbb{R}^1)) \cong K_1(S\mathbb{C}) \cong K_0(\mathbb{C}) \cong \mathbb{Z} \\ \end{align*} where $\tilde{K}$ denotes the reduced K-theory and $S$ the suspension (both in the topological and C*-algebraic setting).
So it seems that we have $\tilde{K}^{-1}(X) \cong K_1(C(X))$. Is this true and if so, what is the proof? In particular I would like to see a proof involving the suspension operations and the result for $K^0$ and $K_0$. If there is a proof, does it generalise to $KO$-theory?
Also, I thought that for reduced cohomology theories one should use pointed topological spaces. So the $\tilde{K}$-groups should depend on the chosen basepoint, right? How does this choice then affect the above discussion?