Pullback of complex vector bundles along a retraction of compact Hausdorff spaces: a direct proof instead? Consider a pointed compact Hausdorff space $(X,x_0)$ and a closed pointed subspace $i:A\subset X$ such that there exists a continuous map $r:X\rightarrow A$ such that $r|_A=\text{Id}_A$. Set
$$q:(X,x_0)\rightarrow(X/A,A/A)$$
be the collapsing map and let $\tilde{K}^0$ denote the reduced (complex) topological $K$-group of a space, then there exists an exact sequence
$$\tilde{K}^0(X/A)\xrightarrow{q^*}\tilde{K}^0(X)\xrightarrow{i^*}\tilde{K}^0(A)\rightarrow0$$
where the right hand side is surjective because of the retraction. By the construction of negative $K$-groups and the long exact sequence of $K^i$, we know that the map $q^*:\tilde{K}^0(X/A)\rightarrow\tilde{K}^0(X)$ should be injective.

Is there a direct way to see the injectivity only involving vector bundles but not $K^{-1}$ or analytic computation like in Karoubi or Park?

Pick $\alpha\in\tilde{K}^0(X/A)$, we know by our setting that there exists a vector bundle $E$ over $X/A$ such that $[E]=\alpha$. Suppose that $q^*\alpha=0$, then $q^*E$ is stably trivial i.e. $q^*E\oplus H$ is trivial for some product bundle $H=X\times\mathbb{C}^n$.

Now it suffices to show that $E$ is also stably trivial. How can one see it directly?

 A: You may add $\mathbb C^n$ to $E$ so that in fact $q^\ast E$ has a trivialization $\phi$. Say that $E$ has rank $m$. The restriction of $q^\ast E$ to $A$ has also a trivialization $\psi$, simply because the restriction of $q$ to $A$ is a constant map. Comparing $\psi $ with the restriction of $\phi$ we get a map $u:A\to GL_m(\mathbb C)$. The composed map $u\circ r:X\to GL_m(\mathbb C)$ can now be used to modify $\phi$, making a trivialization of $q^\ast E$ that restricts to $\psi$ on $A$ and therefore comes from a trivialization of $E$ itself.
Alternatively, use classifying spaces. The problem is to show that if a map $i:X/A\to B$ is such that the composed map $i\circ q:X\to B$ is nullhomotopic then $i$ is nullhomotopic. (Here $B$ could $BU$ or any other based space.) A nullhomotopy of $i\circ q$ means an extension of $i$ to the mapping cone of $q$. The mapping cone of $q$ is homotopy equivalent to the suspension $\Sigma A$. A map $\Sigma A\to B$ extends to a map $\Sigma X\to B$ because of the retraction of $X$ to $A$. The composed map $X/A\to \Sigma A\to \Sigma X$ is nullhomotopic.
