Why is the K-theory of $X$ product of reduced $K$-theory and $\mathbb{Z}$

Question

The reduced $K$-theory of $\tilde{K}(X)$ of the based space $X$ is the kernel of $d:K(X)\to \mathbb{Z}$, where $d$ is induced by $d:Vect(X)\to\mathbb{Z}$ that sends a vector bundle to the dimension of its restriction to the component of the basepoint $*$. Why is it that $K(X)\cong \tilde{K}(X)\times \mathbb{Z}$?

Answer 1

If you have a surjective map of abelian groups $f:A\to\mathbb Z$, then $A\cong\mathbb Z\oplus\ker f$ simply because $\mathbb Z$ is a projective abelian group.