Projective finitely-generated modules, and Serre-Swan Theorem.

Solution:

For the first part, we do the case where there are finitely many trivializations \(\{ (U_i, \psi _i ) \}_{i =1} ^m \) with \(M = \cup _{i =1 }^m U_i\) (this always happens for example if \(M\) is compact).

Consider \(\{ \rho _i \} _{i = 1 }^m\) a partition of unity subordinate to the cover. Let \(N = m k\) and define

\[ \Phi : E \to M \times \mathbb {R}^N = M \times \mathbb {R}^k \times \cdots \times \mathbb {R}^k \]
as
\[ \Phi ( (x, v ) ) = ( x , \rho _1 (x) \psi _1 (v) , \ldots , \rho _m (x) \psi _m (v) ) . \]
Note that while many of the entries in the above expression are not well defined (due to \(v\) not being in the domain of \(\psi _i\) for many \(i\)’s), for such \(i\)’s we have \(\rho _i (x) = 0\) so we define that entry simply as \(0\).

For each \(x \in M\), we have \(\psi _i (x) \neq 0\) for some \(i\), so the map \(\Phi : E_x \to \{ x \} \times \mathbb {R}^N\) is injective due to the \(i\)-th coordinate being non-zero for all non-zero \(v \in E_x\). Then we can define

\[ Q' : = \{ s : M \to \mathbb {R}^N \, \vert \, s(x) \perp \Phi (E_x) \text { for all }x \in M \} . \]
It is then not hard to check that the map
\[ \lambda : \Gamma (M) \oplus Q ' \to C^{\infty }(M ; \mathbb {R}^N) \]
given by
\[ \lambda ( t , s ) (x) : = ( \Phi ( t(x) ) + s (x) ) \]
is an isomorphism of \(C^{\infty }(M)\)-modules.

To deal with the general case, one needs to find a locally-finite collection of trivializations with bounded intersection number and follow a similar strategy.

A discussion regarding the second part and a much deeper conversation can be found in the book by Bruce Blackadar: "\(K\)-theory for operator algebras".

Solution:

(We only give an outline). First assume \(Q=A^N\). For each \(i \in \{ 1, \ldots , N \}\), let \(e_i \in Q\) be the element \((0, \ldots , 1, \ldots , 0)\) that has \(1\) in the \(i\)-th coordinate and \(0\) everywhere else. Also let \(\eta _i \in Q ^{\ast } \) be the projection onto the \(i\)-th coordinate.

For any \(f \in Hom _A (Q,Z)\) we define

\[ \zeta (f) : = \sum _{i = 1} ^N \eta _i \otimes f(e_i) . \]
It is then easy to check that \( \tilde {\theta } \circ \zeta = Id \text { and } \zeta \circ \tilde {\theta } = Id \) on their respective domains (For the second one, check first on elements of the form \(\lambda \otimes y\)). This shows that \(\tilde {\theta }\) is an isomorphism.

For the general case, notice that if \(Q \oplus Q ' = A ^N\), then we have splittings:

\begin{gather*} (A^N)^{\ast } \otimes Z = (Q ^{\ast } \otimes Z ) \oplus ((Q')^{\ast } \otimes Z) \\ Hom _A (A^N , Z) = Hom _A(Q, Z) \oplus Hom_A (Q' , Z). \end{gather*}
From the first part, we have an isomorphism
\[ (Q ^{\ast } \otimes Z ) \oplus ((Q') ^{\ast } \otimes Z) = Hom _A(Q, Z) \oplus Hom_A (Q' , Z) . \]
By how the isomorphism was constructed, it is block diagonal (it respects the sum). Therefore we have ismorphisms
\begin{gather*} Q^{\ast } \otimes Z = Hom _A(Q, Z) \\ (Q')^{\ast } \otimes Z = Hom _A(Q', Z) \end{gather*}