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I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...

The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_o)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.

If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\[H\]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\[H-1\]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.

The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?

Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map $K(X)\otimes K(X)\to K(X\times X)\to K(X)$.

I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...

The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_0)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.

If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\left[H\right]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\left[H-1\right]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.

The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?

Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map $K(X)\otimes K(X)\to K(X\times X)\to K(X)$.

I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...

The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_o)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.

If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\[H\]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\[H-1\]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.

The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero.

Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map $K(X)\otimes K(X)\to K(X\times X)\to K(X)$.

I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...

The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_o)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.

If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\[H\]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\[H-1\]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.

The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?

Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map $K(X)\otimes K(X)\to K(X\times X)\to K(X)$.