Equivalence of two pictures of odd $K$-theory One can show that two functors $K^0$ and $K_0(C(-))$ from the category of compact topological spaces to the category of abelian groups are naturally equivalent. The first one is topological $K$-theory and the second is operator $K$-theory composed with the functor of "continuos functions". I wonder whether the same is true for $K^1$. Natural idea is to reduce this case to the even case using suspension functor, however there are two difficulties:
-first of all in the category of $C^*$-algebras suspension is defined by $SA=C_0(\mathbb{R}) \otimes A$ therefore it produces nonunital algebras and I don't know whether there is equivalence of functors $K^0$ and $K_0(C(-))$ in the locally compact case.
-secondly, this suspension as defined above is not the same as the ordinary topological suspension but it corresponds rather to crossing with $\mathbb{R}$.  
So to summarize, my question is 

Do we have natural equivalence of functors $K^1$ and $K_1(C(-))$ and do we have natural equivalence of $K^0$ and $K_0(C(-))$ in the locally compact case?

 A: The answer is basically "yes, because the definitions are rigged to make it so".  The point is that you have to be careful both with C*-algebra K-theory in the non-unital case and with topological K-theory in the non-compact case, and the standard ways of being careful in both areas are compatible.
Recall that the reduced K-theory of a C*-algebra (unital or not) is by definition:
$$\tilde{K}(A) = \ker(K(A^+) \to K(\mathbb{C})$$
where $A^+$ is the unitalization of $A$ (just $A \oplus \mathbb{C}$ in the unital case) and the map $K(A^+) \to K(\mathbb{C})$ is induced by $A^+ \to A^+/A \cong \mathbb{C}$.  Note that if $A = C_0(X)$ where $X$ is locally compact then $C_0(X)^+ \cong C_0(X^+)$ where $X^+$ is the one point compactification of $X$, and the map $C_0(X^+) \to \mathbb{C}$ corresponds to evaluation at the point at infinity, i.e. it is induced by the inclusion of infinity in $X^+$.  
Applying all of this to the suspension of a unital C*-algebra $A$, we get:
$$K_1(A) \cong \ker(K_0(C(S^1)) \otimes A \to K_0(\mathbb{C}))$$
where the map is induced by the inclusion of the point at infinity of $\mathbb{R}$ in $S^1$.
On the other hand the K-theory of the reduced suspension of a pointed compact space $(X, x_0)$ is isomorphic to the reduced K-theory of $S_1 \times X$, i.e. 
$$K^1(X) \cong \ker(K^0(S^1 \times X) \to K^0(\{\infty, x_0\}))$$
So the two definitions of degree 1 (or perhaps it should really be degree -1) K-theory agree.
