I am reading Vector Bundles and the Kunneth Formula in which Atiyah states (see second paragraph, page 1) that the external product $\mu : KO^*(X) \otimes KO^*(Y) \to KO^*(X \times Y)$ is not always injective. The counterexample he gives is $X = Y = \mathbb C \mathrm{P}^2$.
I'm trying to prove why $\mu$ is not injective in this case. I am trying to find non-trivial $a \in KO^*(X)$ and $b \in KO^*(Y)$ such that $\mu(a,b) = 0$. My guess is to take canonical line bundles $E_1, E_2$ over $\mathbb C\mathrm{P}^2$, considered as 2 dimensional real bundles.
Consider $E_1$ as an element of $KO(X)$ and $E_2$ as an element of $KO(Y)$. (I think that these are non-trivial classes, but I'm not sure.) Then $\mu(E_1 \otimes E_2) := p_1^*(E_1) \otimes p_2^*(E_2)$ where $p_1$ and $p_2$ are the projections of $X \times Y$ onto $X$ and $Y$.
In our case, I want to show that $p_1^*(E_1) \otimes p_2^*(E_2) = \epsilon^4$. We can compute the pullbacks $p_i^*(E_i) = \{(v,l_1,l_2) \in \mathbb C^3 \times \mathbb C \mathrm{P}^2 \times \mathbb C \mathrm{P}^2 : v \in l_i\}$.
The fibre of $p_1^*(E_1) \otimes p_2^*(E_2)$ is the tensor product of the fibres of $p_1^*(E_1)$ and $p_2^*(E_2)$. In $p_1^*(E_1)$, the fibre of a point $(l_1,l_2)$ in the base space is $l_1 \cong \mathbb C \cong \mathbb R^2$. So the fibre of $p_1^*(E_1) \otimes p_2^*(E_2)$ is $l_1 \otimes l_2 \cong \mathbb R^4$. I want to conclude that this shows that $p_1^*(E_1) \otimes p_2^*(E_2)$ is trivial.
Is this working right so far? Am I heading in the right track?