Let $A$ be a commutative unital $C^*$ algebra. Then $A=C(X)$ for some compact Hausdorff space $X$. Topological $K$-theory group (namely $K_0$) is defined in terms of vector bundles as a Grothendieck group of the semigroup of isomorphism classes of vector bundles (with respect to the Whitney sum of bundles). On the other side operator $K$-theory ($K_0$ group) is defined in terms of *projections* (self adjoint idempotents) namely one takes classes of projections in $M_{\infty}(A)$ with respect to relation $\sim$ where $x \sim y$ if there is a homotopy of projections in $M_{\infty}(A)$ connecting $x$ and $y$ (this is equivalent to the unitary equivalence and also to the Murray-von Neumann equivalence when we admit arbitrary large matrices). There is also the theorem of Serre-Swan which states that each projective finitely generated $C(X)$ module is of the form $\Gamma(E)$ for some vector bundle $E \to X$. In fact this theorems gives the equivalence of categories of finitely generated projective $C(X)$ modules and vector bundles over $X$. Usually authors say that thanks to this theorem the two aproaches: (classiccal) topological and operator theoretic are the same. However there are some details which (at least for me) are not quite clear:

First problem: each vector bundle gives rise to a finitely generated projective $C(X)$ module which further gives rise to someidempotentin $M_n(C(X))$. This idempotent needs not to be self adjoint. So how to constructprojectiongiving an operator theoretic $K$-theory class, out of the vector bundle?

Suppose that we somehow overcame this diffulty and constructed the isomorphism of $K^0(X) \cong K_0(C(X))$. Still it is not obvious that this isomorphism behaves nicely at the level of maps:

Second problemIs this isomorphism natural? In other words: given continuous $f:X \to Y$ one has a morphism $T_f:C(Y) \to C(X)$ $T_f(g)=g \circ f$ (and each (unital, star) morphism between $C(Y)$ and $C(X)$ is of this form) and further one has $f^*:K^0(Y) \to K^0(X)$ given by the pullback bundle construction and $(T_f)_*:K_0(C(Y)) \to K_0(C(X))$ acting on the each entry of the (class of) projection. Then the question is whether the obvious diagram commutes?

Maybe thinks would become more clear if I would see the explicit formula for the isomorphism.

**Apologies** I'm pretty convinced that this is not a research level question-however it seems to me that this is a kind of *folklore* which is hard to find carefully and properly explained in the literature. So my hope is that somebody will find useful to have a careful explanation of this issue.

statesthat the Gelfand transform induces the isomorphism) $\endgroup$ – truebaran May 6 '16 at 23:16