Let $S^n$ be the $n$-sphere.
I'm trying to understand the "meaning" of the $\mathbb{Z}$ factors in $$ K_0(C(S^{2n+1}))\cong\mathbb{Z}$$ and $$ K_0(C(S^{2n}))\cong\mathbb{Z}\oplus\mathbb{Z}$$
So $S^n$ is the one-point compactification of $\mathbb{R}^n$, so that $C(S^n)$ is isomorphic to the unitization of $C_0(\mathbb{R}^n)$, which is in turn isomorphic to the $n$th suspension of $\mathbb{C}$: $$ C(S^n)\cong C_0(\mathbb{R}^n)^{\tilde{}}\cong S^n(\mathbb{C}) $$ We also know that $$ K_0(A^{\tilde{}}) \cong K_0(A)\oplus\mathbb{Z} $$ Thus we have $$ K_0(C(S^n))\cong K_n(\mathbb{C})\oplus \mathbb{Z}$$ at which point the result follow via Bott periodicity and the fact that $$ K_0(\mathbb{C})\cong\mathbb{Z}$$ and $$K_1(\mathbb{C})\cong0$$
While this is all nice and elegant, I don't really understand now what these $\mathbb{Z}$ factors "mean", in terms of equivalence of projections.
So given two projections $[p]_0$ and $[q]_0$ in $C(S^n)$, we know that one $\mathbb{Z}$ factor keeps track of their rank, so that clearly they cannot be equivalent if they have different rank. Now if they have the same rank, then we know that at least point-wise (points being on $S^n$) we can build a unitary isomorphism $u(x)$ between $p(x)$ and $q(x)$ for each $x\in S^n$.
Now intuitively I would say that since $\pi_n(U(m))$ for $m$ sufficiently large is exactly trivial or $\mathbb{Z}$ depending on whether $n$ is odd or even respectively, we get that the unitary $u(x)$ can or cannot define a continuous $u$ over the whole sphere, which determines whether there is another factor of $\mathbb{Z}$ or not.
Now my question is: is this "analysis" correct? Is there a more direct way to see this without using $\pi_n(U(m))$? Is there a general approach to understand the structure of the K-groups without explicitly computing them from scratch?
I suspect at least one step is the epimorphism $$\dim:K_0(C(X))\to\mathbb{Z}$$ for $X$ compact, connected which is given by the trace at a point (doesn't matter which) and is an isomorphism apparently for the $X=S^n$, $n$ odd cases. What is the other epimorphism for the $n$ even cases? Some sort of winding number? How is it defined (via some unitary?) Is there a general way to systematically generate these epimorphisms to $\mathbb{Z}$ or $\mathbb{Z}/m\mathbb{Z}$, given that any finitely generated Abelian group (that is, $K_0(A)$) is a direct sum of such factors?