Definition 16.20.
The reduced K-theory of a based space \(X\) is
\begin{equation*}
\widetilde{K}(X) \;:=\; \ker\bigl(d \colon K(X) \to \mathbb{Z}\bigr).
\end{equation*}
Section 16.6 Reduced K-Theory
The rank of a bundle is information that every space carries, even a point. As with reduced homology, we remove it to isolate what depends on the shape of \(X\text{.}\) The reduced theory is also the one that behaves well under suspension, which is how we will define K-theory in other degrees in the next chapter.
For a based space \((X, x_0)\) define \(d \colon \mathrm{Vect}(X) \to \mathbb{N}\) by \(d(E) = \dim E_{x_0}\text{,}\) the rank of \(E\) on the component of \(x_0\text{.}\) This is a monoid homomorphism, so it descends to a homomorphism \(d \colon K(X) \to \mathbb{Z}\text{.}\) Equivalently, \(d\) is the map \(K(X) \to K(x_0) = \mathbb{Z}\) induced by the inclusion of the basepoint.
The constant map \(X \to x_0\) induces a section \(\mathbb{Z} = K(x_0) \to K(X)\) of \(d\text{,}\) sending \(n\) to \(n[\varepsilon^1]\) (which is \([\varepsilon^n]\) for \(n \geq 0\)). This gives a natural splitting
\begin{equation*}
K(X) \;\cong\; \widetilde{K}(X) \oplus \mathbb{Z},
\end{equation*}
analogous to the splitting of unreduced into reduced (co)homology plus the coefficient at the basepoint.
Example 16.21.
If \(X\) is paracompact and contractible then \(\widetilde{K}(X) = 0\) by Example 16.18. By Example 16.19, \(\widetilde{K}(S^2) \cong \mathbb{Z}\text{,}\) generated by \([\gamma] \ominus [\varepsilon^1]\text{,}\) and the product of any two elements of \(\widetilde{K}(S^2)\) is zero.
For compact spaces, reduced K-theory has a description that needs no formal differences at all. Hatcher takes it as the definition of \(\widetilde{K}(X)\) (Vector Bundles and K-Theory, Prop. 2.1). It is the form in which \(\widetilde{K}\) will be classified by a space in the next chapter.
Proposition 16.22.
Let \(X\) be a based compact Hausdorff space. Call bundles \(E\) and \(E'\) stably equivalent if \(E \oplus \varepsilon^a \cong E' \oplus \varepsilon^b\) for some \(a, b \geq 0\text{.}\) Then
\begin{equation*}
\mathrm{Vect}(X)/(\text{stable equivalence}) \;\xrightarrow{\;\cong\;}\; \widetilde{K}(X), \qquad E \;\longmapsto\; [E] \ominus [\varepsilon^{d(E)}],
\end{equation*}
is a bijection.
Proof.
The map lands in \(\widetilde{K}(X)\text{,}\) since \(d([E] \ominus [\varepsilon^{d(E)}]) = 0\text{.}\) It is constant on stable equivalence classes, because \(E \oplus \varepsilon^a\) maps to \([E] \oplus [\varepsilon^a] \ominus [\varepsilon^{d(E) + a}] = [E] \ominus [\varepsilon^{d(E)}]\text{.}\) It is surjective, since by Remark 16.17 every element of \(\widetilde{K}(X)\) has the form \([H] \ominus [\varepsilon^n]\text{,}\) and \(d = 0\) forces \(n = d(H)\text{.}\) For injectivity, suppose \([E] \ominus [\varepsilon^{a}] = [E'] \ominus [\varepsilon^{b}]\) with \(a = d(E)\) and \(b = d(E')\text{.}\) Then \([E \oplus \varepsilon^b] = [E' \oplus \varepsilon^a]\) in \(K(X)\text{.}\) By Remark 16.17 we get \(E \oplus \varepsilon^{b+n} \cong E' \oplus \varepsilon^{a+n}\) for some \(n\text{,}\) so \(E\) and \(E'\) are stably equivalent.
Unlike the stable isomorphism of Remark 16.17, stable equivalence allows \(a \neq b\text{,}\) and so it forgets the rank. For example, all trivial bundles are stably equivalent to each other, which matches \(\widetilde{K}(\mathrm{pt}) = 0\text{.}\)
