Let $A$ and $B$ be unital $C^\*$-algebras. Let $u\in M\_n(A)$ be a unitary representing an element $[u]\in K\_1(A)$ and $p\in M\_m(B)$ be a projection representing an element $[p]\in K\_0(B)$. Then the unitary $$u\otimes p+1\_A\otimes (1\_B-p)$$ represets the product $[u]\times [p]\in K\_1(A\otimes B)$ (see in Nigel Higson's and John Roes book "Analytic K-homology", proposition 4.8.3). This gives us a description of a product map $K\_1(A)\times K\_0(B)\to K\_1(A\otimes B)$. 


Now, let $A$ non-unital and $B$ unital, but $u$ and $p$ as above, where $u\in K\_1(A)\cong K\_1(A^+)$ with $A^+$ the unitization of $A$. 


My question is: 
Is it possible to give an explicit description for $K\_1(A)\times K\_0(B)\to K\_1(A\otimes B)$ as above for this case to map $[u]$ and $[p]$ to $K\_1(A\otimes B)$ naturally? 


I guess that this is not possible. My first try was to consider $$u\otimes p+1\_{A^+}\otimes (1\_B-p),$$ which represents an element in $K\_1(A^+\otimes B)$. However, since $A\otimes B$ is non-unital, the product map should map to $K\_1(A\otimes B)\cong K\_1((A\otimes B)^+)$, but $ K\_1((A\otimes B)^+)$
is not isomorphic to $K\_1(A^+\otimes B)$. For $A$ nonunital and $B$ unital and nuclear, there is the following relation: there is a commutative diagram


$$\require{AMScd}\def\Z{\mathbb{Z}}
\begin{CD}
0 @>>> K\_1(A)\otimes K\_0(B) @>>> K\_1(A^+)\otimes K\_0(B) @>>>> K\_1(\mathbb{C})\otimes K\_0(B) @>>> 0 \\
@. @VV V @VV\alpha V @VV\alpha V @. \\
0 @>>>K\_1(A\otimes B) @>>> K\_1(A^+\otimes B) @>>> K\_1(\mathbb{C}\otimes B) @>>> 0 \\
\end{CD}
$$
where $K\_1(\mathbb{C})=0$, $K\_1(\mathbb{C}\otimes B)\cong K\_1(B)$ and $K\_1(A)\otimes K\_0(B) \to K\_1(A^+)\otimes K\_0(B)$ is an isomorphism. If $B$ is nonnuclear, then the sequence in this diagram is not exact at $K\_1(A\otimes B)$.



explicit description of the product map in K-theory
I believe the correct formula is $$(u-1\_{A^+}) \otimes p + 1\_{(A \otimes B)^+}$$ which makes sense if you work this out in the following special case. Let $A$ be $C\_0(0,1)$ and let $B$ be $C(\{0,1\})$, and take $u(s)=e^{2\pi is}$ and $p(t)=t$. In this case $$((u-1\_{A^+}) \otimes p + 1\_{(A \otimes B)^+})(s,t)= (e^{2\pi i s}-1)t + 1$$ which is constantly one on $(0,1)\times\{0\}$ and loops around the unit circle on $(0,1)\times\{1\}$. Check that this is plus-or-minus the correct answer for this special case. Now use the universal properties of $C\_0(0,1)$ and $C(\{0,1\})$ to get maps from these to a generic $A\_1$ and $B\_1$ and use the fact that the product map is natural to show this is the correct formula in general.


