I'm trying to fill the details of the explanation of the interpretation of the reduced external product. I'll follow Hatcher second books, and I'll write part of his explanation in the gray boxes followed by my thoughts and comments.
For $a \in \tilde{K}(X)=\ker( K(X) \to K(x_0))$ and $b \in \tilde{K}(Y)=\ker( K(Y) \to K(y_0))$, the external product $a*b = p_1^*(a)p_2^*(b) \in K(X\times Y)$ has $p_1^*(a)$ restricting to zero in $K(Y)$ and $p_2^*(b)$ restricting to zero in $K(X)$
my explanation: Let $a= A_1-A_2$, $b=B_1-B_2$ formal differences of vector bundles. The fact that they lie in the kernel means that the rank of $A_1$ and $A_2$ in $x_0$ are equal to a fixed $m_A$ and similar for $B_1$ and $B_2$ ($m_B$). (we are considering the map "degree". Let $p_1 \colon X \times Y \to X$ the projection onto the first factor ($p_2$ onto the second one) So the pullback $$p_1^*(a)=p_1^*(A_1) - p_1^*(A_2)$$ is isomorphic to $$ A_1 \times Y - A_2 \times Y$$ (uniqueness of pullback, and according to my calculations) So if we restrict to $\{x_0\}\times Y$ we obtain the difference $$ \epsilon^{m_A} \times Y - \epsilon^{m_A}\times Y = 0$$ a similar approach for $p_2^*(b)$ leads to the conclusion.
Hence $p_1^*(a)p_2^*(b)$ restrict to zero in $K(X \vee Y)$
Here is my problem. We can see $X \vee Y$ as $X \times \{y_0\} \cup \{x_0\} \times Y$ and I've just proved that $a *b$ restrict to zero in $ X \times \{y_0\}$ and $\{x_0\} \times Y$, but i'm not so sure that I can conclude the thesis from here. Is looking at the transition functions a viable strategy for this problem?
The following is clear.
So I'm asking for some confirmations about the second gray box. Thanks in Advance!