Spaces with only even dimensional cells has zero $K^{-1}$ group?

Question

This is from D. Quillen's "On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field". Here $BU= \bigcup _1^\infty X_m$ is the union of finite subcomplexes with only even-dimensional cells. Why is $K^{-1} (X_m)=[X_m, U]=0$ in this case?

Answer 1

There is an Atiyah-Hirzebruch spectral sequence $$E_2^{p,q} = H^p(X; K^q(\mathrm{pt})) \Rightarrow K^{p+q}(X).$$ By evenness, $E_2^{p,q} = 0$ unless $p$ and $q$ are even, which implies $p + q$ is even. Therefore, there are no nontrivial terms contributing to $K^{\text{odd}}$, so $K^{\text{odd}}(X) = 0$.

Answer 2

You can also show it by induction on the dimension. If $Y$ is obtained from $X$ by attaching copies of $D^{2n}$ and $K^{odd}(X)=0$, then the long exact sequence for $(Y,X)$ gives $$0 = K^{odd}(\bigvee S^{2n}) \to K^{odd}(Y) \to K^{odd}(X) = 0.$$