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Let $A_1$ and $A_2$ be unital $C^*$-algebras. If $p_1 \in M_{n_1}(A_1)$ and $p_2 \in M_{n_2}(A_2)$ are projections then $p_1 \otimes p_2 \in M_{n_1 n_2}(A_1 \otimes A_2)$ is also a projection, yielding a product $K_0(A_1) \times K_0(A_2) \to K_0(A_1 \otimes A_2)$. For general $A_1$ and $A_2$ the product on the unitalizations of $A_1$ and $A_2$ restricts to a product on $A_1$ and $A_2$ themselves. Combining this observation with the isomorphisms of suspension algebras $S^{p_1}(A_1) \otimes S^{p_2}(A_2) \cong S^{p_1 + p_2}(A_1 \otimes A_2)$, we obtain a product $K_{p_1}(A_1) \times K_{p_2}(A_2) \to K_{p_1 + p_2}(A_1 \otimes A_2)$.

I need help proving a well-known formula for the product $K_1(A_1) \times K_0(A_2) \to K_1(A_1 \otimes A_2)$ using unitary model of $K_1$: if $u$ is a unitary over $A_1$ and $p$ is a projection over $A_2$ then the class $[u] \times [p]$ is represented by the unitary $u \otimes p + 1 \otimes (1-p)$. Clearly the identification between the unitary model and the suspension model of $K_1$ is relevant here, but beyond that I was hoping this would be a simple matter of definition chasing. Still, I'm having trouble getting this to work out.

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