Explaining the $K$ functor isomorphism $K(X) \cong \tilde{K}(X) \oplus \Bbb Z$

Question

It seems unclear to me what the splitting means in page 40, of Hatcher's for a $K$ functor.

1. There is a natural homomorphism, $\varphi:K(X)\rightarrow \tilde{K}(X)$, sending $[E-\epsilon^n] \rightarrow [E]$ and also $$ \ker \varphi \cong \Bbb Z$$
2. There is an inclusion map $i:\{x_0\} \rightarrow X$ thus inducing a map $K(X) \rightarrow K(x_0)$ which is the restriction bundle. $$[E-\epsilon^n] \mapsto [i^*(E)-i^*(\epsilon^n) ]= [E|_{x_0} - \epsilon^n|_{x_0} ]$$ also gives $$ K(x_0) \cong \Bbb Z$$

This induces a splitting $$K(X) \cong \tilde{K}(X) \oplus \Bbb Z.$$

So how does this induce a splitting? What are the explicit maps?

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EDIT: after correction.

When $i^*$ restricted induces isomoprhism $\psi : \ker \varphi \rightarrow K(x_0)$ Thus, there is a well -defined inverse

$$ 0 \rightarrow K(x_0) \xrightarrow{\psi^{-1}} K(X) \xrightarrow{\varphi} \tilde{K}(X) \rightarrow 0 $$

The left splitting representation is thus given by $$ ([E-E']-\psi^{-1}i^*[E-E']) \oplus \psi^{-1}i^*[E-\epsilon^n] = ([E-E'] - [\epsilon^m-\epsilon^n]) + [\epsilon^m - \epsilon^n]$$ where $m,n$ are the ranks of the bundles $E$, $E'$ over $x_0$ respectively.

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Addendum question: The above method, and that shown in the reply produces a splitting that is actually independent of the choice of base point $x_0$?

Answer 1

The splitting $K(X)\to\ker\varphi=\mathbb{Z}$ of the short exact sequence $$ 0\to\mathbb{Z}\xrightarrow{\iota} K(X)\xrightarrow{\varphi}\tilde{K}(X)\to 0 $$ is the second map $i^*\colon K(X)\to K(x_0)\cong\mathbb{Z}$. You can check that this is indeed a splitting (i.e., $i^*\iota$ is the identity on $\ker\varphi$).

Explcitly, the isomorphism $K(X)\cong\tilde{K}(X)\oplus\mathbb{Z}$ is $[E]_{\approx_s}\mapsto ([E]_{\sim},r)$ where $r$ is the virtual rank of $E$ (at $x_0$), i.e., if $E\approx_s E_1-E_2$ where $E_1,E_2$ are vector bundles over $X$, then $r=\operatorname{rank} E_1-\operatorname{rank} E_2$ (taken at the fiber over $x_0$).

Addendum

The splitting is dependent on the component of $x_0$, but that is the cost Hatcher is paying for allowing the base $X$ to be disconnected here. Expanding that a little, the other splitting map $\tilde{K}(X)\to K(X)$ is determined by the component of $x_0$ because we want that copy to have virtual rank $0$ at $x_0$. In fact, you can split off a copy of $\mathbb{Z}$ for each component of $X$, mirroring the possibility of different rank on different components as discussed in page 39.