An example about trivial product of reduced K-theory This is example 2.13 in Hatcher's K-theory notes. Suppose $X$ is a pointed compact Hausdorff space and $X=A\cup B$, where $A$ and $B$ are compact contractible subspaces of $X$ containing the basepoint. The book claims that
the product $\tilde{K}^*(X)\otimes \tilde{K}^*(X)\to \tilde{K}^*(X)$ is identically zero since it is equivalent to the composition $\tilde{K}^*(X,A)\otimes \tilde{K}^*(X,B)\to \tilde{K}^*(X,A\cup B)\to \tilde{K}^*(X)$. I understand that since $A$ is contractible, $\tilde{K}^*(X)=\tilde{K}^*(X,A)$, and that $\tilde{K}^*(X,A\cup B)=0$. But I'm having trouble seeing the equivalence. Where do we use the fact that $A$ and $B$ both contain the basepoint? Should I think in terms of vector bundles?
 A: Let me just write $K$ for reduced and relative K-theory for readability. So far you agree that we can identify $K(X)$ with $K(X, A)$ and with $K(X, B)$, so the product can be identified with the product $K(X, A) \times K(X, B) \to K(X)$. Next you just need to agree that this product factors through $K(X, A \cup B)$, and this is because the corresponding relative diagonal map $X \to X/A \wedge X/B$ factors through $X/(A \cup B)$. 
The hypothesis that $A$ and $B$ both contain the basepoint is just the hypothesis you need for the maps $X \to X/A$ and $X \to X/B$ to be based maps, so that they define maps $K(X, A) \to K(X)$ and $K(X, B) \to K(X)$ at all. 
This argument has nothing to do with vector bundles in particular and applies without modification to any multiplicative cohomology theory. It may help to think about ordinary cohomology, and in the nicest cases so that it can be identified with the intersection product. Then the idea is that any two cocycles have trivial intersection product because you can homotope one so that it avoids $A$ and homotope the other so that it avoids $B$, and then they can't intersect. 
