I've often heard people say that $K$-theory of $C^*$-algebras generalises topological $K$-theory, for the reason that, say if $M$ is a compact manifold, we have
$K_0(C(M)) = K^0(M)$,
where $C(M)$ denotes the continuous functions on $M$ with values in $\mathbb{C}$.
But the analogous relation doesn't seem to hold for $K_1$: let $M = pt$, then $K_1(C(pt)) = K_1(\mathbb{C}) = 0$, but $K^1(pt) = K^0(S^1 \wedge pt) = K^0(pt) = \mathbb{Z}.$
Have I made a mistake somewhere here?
It's not true that $K^1(pt)=K^0(S^1\wedge pt)$, or that $K^1(X)=K^0(S^1\wedge X)$ in general. This statement is only correct for reduced $K$-theory: $\widetilde{K}^1(X)=\widetilde{K}^0(S^1\wedge X)$ for any pointed space $X$ (or you could say $K^1(X)=\widetilde{K}^0(S^1\wedge X)$ since $\widetilde{K}^1$ is the same as $K^1$). So $K^1(pt)=\widetilde{K}^0(S^1\wedge pt)=\widetilde{K}^0(pt)$ which is indeed trivial.