$\newcommand{\T}{\mathbb{T}}\newcommand{\R}{\mathbb{R}}$Suppose $p$ is a projection in $M_n(A^+)$. In order to get a unitary $u$ in $C(\T\to M_n(A^+))$, we only need a continuous function $f:\T\times \{0,1\} \to \mathbb {R}$ and to do the functional calculus on the second variable, that is, let $$u(z) =e^{if(z, p)}.$$
To make $u$ fall into $$K_1(SA)\cong \{[f]|f\in C(\T\to \mathcal U_n(A^+)), f(1)\sim_h 1\},$$
It is sufficient to set $$f(z,x)=|z-1|g(x),$$ where $g$ is a map from $\{0,1\}$ to $ \R$.
To make the map $B: p\mapsto u$ be invertible, we only need that the range of $e^{if(z,x)}$ does not equal to $\T$ and $g$ is invertible. Then the invert of $B$ is likely to be $B^{-1}: u\mapsto g^{-1}( \frac{1}{2 i}\operatorname{Ln} u(-1))$. Here we can simply take $g=\operatorname{id}$.

**Definition**. Suppose $(V_i)_{i\in I}$ is a family of linear spaces. The direct sum of $(V_i)_{i\in I}$ is defined as $\oplus_{i\in I}V_i=\{(v_i)_{i\in I}|\{i|v_i\neq 0\}\mbox{ is a finite set }\}$, and the tensor product of $(V_i)_{i\in I}$ is defined as a linear space $\otimes_{i\in I} V_i$ with a multilinear operator $\otimes:\oplus_{i\in I}V_i\to \otimes_{i\in I} V_i$ such that for any given linear space $W$ and a multilinear operator $f:\oplus_{i\in I}V_i\to W$, there is a unique linear operator $\tilde f:\otimes_{i\in I} V_i\to W$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
\oplus_{i\in I}V_i\ar[r]^{\otimes} \ar[dr]_f& \otimes_{i\in I} V_i\ar[d]^{\tilde f}\\
& W
}
\end{xy}
**Existence**. Let $L(\oplus_{i\in I} V_i)$ denotes the set of the multilinear operators on $\oplus_{i\in I} V_i$. Define $\otimes_{i\in I} v_i\in L(L(\oplus_{i\in I} v_i))$ by $$\otimes_{i\in I} v_i(f)=f(\oplus_{i\in I} v_i),$$ and define $\otimes_{i\in I} V_i$ to be the subspace generated by $\{\otimes_{i\in I} v_i|v_i\in V_i\forall i\in I\}$ in $L(L(\oplus_{i\in I} V_i))$.
For any given multilinear operator $f:\oplus_{i\in I}V_i\to W$, in order to make the diagram commutative $\tilde f$ must be defined by $$\tilde f(\otimes_{i\in I}v_i)=f(\oplus_{i\in I}v_i).$$
And $\tilde f$ is also well-defined in this way:
Suppose $\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k= \sum_{k=1}^nl'_i\otimes_{i\in I}{v'}_i^k$.
For any $\tau\in L(W)$, $\tau \circ f\in L(\oplus_{i\in I}V_i)$, so
$\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k(\tau\circ f)=\sum_{k=1}^nl'_i\otimes_{i\in I}{v'}_i^k(\tau\circ f)$, i.e.,
$\tau(f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k))=\tau(f(\sum_{k=1}^nl'_i\oplus_{i\in I}{v'}_i^k))$, hence
$f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k)=f(\sum_{k=1}^nl'_i\oplus_{i\in I}{v'}_i^k)$.
**Remark**. The proof seems the same as the finite case~

Given two vector spaces $V$ and $W$, an *alternating multilinear operator* from $V^k$ to $W$ is a multilinear map
$ f: V^k \to X $
such that whenever $v_1,\cdots,v_k$ are linearly dependent vectors in $V$, then
$f(v_1,\ldots, v_k)=0$.
The *kth exterior power* of a linear space $V$, is a linear space $\Lambda^k(V)$ with alternating multilinear operator $\Lambda:V^k\to \Lambda^k(V)$ such that for any given linear space and any alternating multilinear operator $f:V^k\to W$, there is a unique linear map $\tilde f$from $\Lambda^k(V)$ to $W$ such that $\tilde f\circ \Lambda=f$ as indicated by the following commutative diagram:
\begin{xy}
\xymatrix{
V^k\ar[r]^{\Lambda} \ar[dr]_f & \Lambda^k(V)\ar[d]^{\tilde f}\\
& W
}
\end{xy}
**Construction**. Let $I$ be the linear subspace generated by $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$ in $\otimes^k V$. Then $\otimes^k V/I$ is a kth exterior power.
*Proof*. Let $\bar f:\otimes^k(V)\to W$ be the linear map such that $\bar f\circ \
\otimes= f$, $q:\otimes^k(V) \to \otimes^k V/I$ be the quotient map, and $\tilde f([z])=\bar f(z)$.
\begin{xy}
\xymatrix{
V^k\ar[r]^{\otimes} \ar[dr]_f & \otimes^k(V)\ar[d]^{\bar f}\ar[r]^q & \otimes^k(V)/I\ar[dl]^{\tilde f}\\
& W
}
\end{xy}
If $z\in I$ then $z$ is a linear combination of element in $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$, so $\tilde f([z])=\bar f(z)= \sum_{i=1}^n k_if(x_1^i,\cdots,x_n^i) = 0$, so $\tilde f$ is well-defined.
>Then what about the infinite exterior power?

- How to compute $V(\mathbb{B}/\mathbb{K})$. Suppose $\alpha:A\to B$ is surjective, is there a lift for each projection $p\in B$?
~~- the limit of normal elements, the universal property of Borel functional calculus.~~
~~- u^t~~

$\newcommand{\P}{\operatorname{Proj}}$$\newcommand{\qCP}{\operatorname{CP}}$
Suppose $\theta$ is a projection orthoisomorphism between the set $\qCP(A)$ and $\qCP(B)$ of all q-closed projections of $A=C(X,\mathcal M_m)$ and $B=C(Y,\mathcal M_n)$.
We know that [it preserves 0, order and rank-one projections.][1]
**Theorem 1.** $\theta$ can be extended to a projection orthoisomorphism between
$\ell^\infty(X,\P(A))$ and $\ell^\infty(Y,\P(B))$.
*Proof*.
For each projection $p\in \ell^\infty(X,\P(\mathcal M_m))$, denote by $p_1$ the set of all the rank-one projections majorised by $p$, and define $\tilde\theta(p)\in \ell^\infty(Y, \P(\mathcal M_n))$ by $\tilde\theta(p)=\sup_{m\in p_1}\theta(m)$. This extends $\theta$.
Since any projection in $\mathcal M_m$ is the supremum of rank-one projctions majorised by it,
any projection in $\ell^\infty(X,\P(\mathcal M_m))$ is the supremum of rank-one projctions majorised by it.
From the definition we konw that $\tilde\theta$ preserves order, hence $\{\theta(m)|m\in p_1\}$ consists of all the rank-one projections majorised by $\tilde \theta(p)$. Therefore, $\tilde\theta$ is a bijection.
Since $\tilde\theta$ preserves rank-one projections and order, and a projection $p$ is orthogonal to another projection $q$ in $\ell^\infty(X,\mathcal M_m)$ iff $rs=0$ for all rank-one projection $r$ majorised by $p$ and $s$ majorised by $q$, we have $\tilde\theta$ preserves orthogonality.
By Dye's theorem, we have
**Corollary.** $\theta$ is implemented by a Jordan *-isomorphism between $\ell^\infty(X,\mathcal M_m)$ and $\ell^\infty(Y,\mathcal M_n)$.
[1]: https://math.liveadvances.com/c-ding/properties-of-a-projection-orthoisomorphism-on-the-set-of-closed-projections/