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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?

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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.

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    $\begingroup$ Your definition of reduced $K$-theory is not quite right. Operator algebraic K-theory is covariant in the algebra, so the unit inclusion $\mathbb{C} \to A$ would give you a group homomorphism $K_*(\mathbb{C}) \to K_*(A)$, which is not what you want. What you do instead is use the homomorphism $A^+ \to \mathbb{C}$ you have from the unitisation, which corresponds to the evaluation at the point at infinity in the compactification of a space. $\endgroup$
    – Ulrich Pennig
    Commented Sep 24, 2016 at 22:47
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    $\begingroup$ You're quite right - I modified the answer accordingly. Thanks! $\endgroup$
    – Paul Siegel
    Commented Sep 24, 2016 at 23:44
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    $\begingroup$ Thank you for the answer. I've got some questions to your answer: first, isn't $ker(K(A^+) \to K(\mathbb{C})$ just the ordinary definition of the $K_0$ group for a general $C^*$-algebra (not reduced K-theory)? See for example page 111 in Wegge Olsen book. Second I understand that $K_1(A)$ is isomorphic to $K_0(SA)=K_0(C(\mathbb{R} \otimes A)$ and therefore $K_1(A) \cong ker(K_0((SA)^+) \to K_0(\mathbb{C})$. How did you get $K_1(A) \cong ker(K_0(C(S^1) \otimes A) \to K_0(A))$? $\endgroup$
    – truebaran
    Commented Sep 25, 2016 at 12:20
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    $\begingroup$ First of all, this is sort of what I mean when I say that the definitions are rigged to make everything work out - naively there is no obstruction to defining $K(A)$ to be equivalence classes of projections for non-unital C*-algebras, but the definitions are chosen to make C*-algebra K-theory agree with reduced topological K-theory because it is reduced K-theory that has the right functorial properties, is compatible with smash products, satisfies Bott periodicity, etc. Some authors emphasize this point of terminology, and others do not. $\endgroup$
    – Paul Siegel
    Commented Sep 25, 2016 at 12:42
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    $\begingroup$ Second, I goofed in the right-hand sides of the kernels - they should be $K(\mathbb{C})$ and $K(\text{point})$, respectively. I edited again. See my remarks about unitalization and one point compactification for the rest. $\endgroup$
    – Paul Siegel
    Commented Sep 25, 2016 at 12:45

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