Is there a sense in which $\tilde{K}$ is an exact functor?

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I'm going through Hatcher's K-Theory script for the first time and noticed that following theorem looked quite like a statement of the form “this functor is exact”:

If $X$ is compact Hausdorff and $A\subset X$ is a closed subspace, then the inclusion and quotient maps $A\stackrel{i}\to X \stackrel{q}\to X/A$ induce homomorphisms $\tilde K (X/A) \stackrel{q^\ast}\to \tilde K(X) \stackrel{i^\ast}\to \tilde K (A)$ for which the kernel of $i^*$ equals the image of $q^*$.

Here, $\tilde K(X)$ stands for the equivalence class of vector bundles over $X$ such that $E\sim F$ if and only if they are isomorphic after adding trivial bundles of potentially different dimension.

Together with the notion of a pullback of bundles, it is not hard to show that $\tilde K$ forms a contravariant functor from $\underline{\text{CompHaus}}$ to $\underline{\text{AbGrp}}$. But can we equip the first category (or some equivalent category) with an abelian structure so that we could say this functor is in fact exact?