# Defining the inverse of the Bott map

I hope this isn't too narrowly focused. I have a question concerning the inverse of the Bott map as defined in Atiyah's paper, Bott Periodicity and the Index of Elliptic Operators. On page 122 he defines it as the composition

$$K^{-2}(X) \to K(S^2 \times X) \stackrel{\text{index} \bar\partial}{\longrightarrow} K(X).$$

My question is: what is the first map? At first I thought it was $$K^{-2}(X) = \tilde K(S^2 \wedge X^+) \to \tilde K(S^2 \times X^+) \to \tilde K(S^2\times X) \subset K(S^2 \times X)$$ where the first map is the pullback of the projection and the second is the pullback of the inclusion. But from the 6 term exact sequence, for this first map to be injective (which it needs to be for this to be correct) we need $$K(S^2 \times X^+, S^2 \times X) \to \tilde K(S^2 \times X^+)$$ to be the zero map which doesn't seem correct since $S^2 \times X^+/S^2 \times X$ is identifiable with $S^2 \sqcup \{pt\}$ and if we pullback a non-trivial bundle over this space to $S^2 \times X^+$ it seems like it won't trivialize.

Thanks!

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Ah thanks. I don't know why I made the mistake of thinking the map $\tilde K(S^2\times X^+) \to \tilde K(S^2 \times X)$ had to be injective. – Eric O. Korman Apr 16 '11 at 20:21
Actually I think I found a better way to look at the map, which shows that it is injective. Consider the pair $(S^2 \times X, \{pt\} \times X)$. Then since $\{pt\} \times X$ is a retract of $S^2 \times X$, the 6-term sequence in $K$-theory splits to give a short exact sequence $$0 \to K(S^2 \times X, \{pt\} \times X) \to K(S^2 \times X) \to K(\{pt\} \times X).$$ But $(S^2 \times X) / (pt \times X) \simeq S^2 \wedge X^+$, giving the desired map.