$K$-theory exact sequence.

Question

Let $Y$ be a closed subspace of a compact space $X$. Let $i:Y \to X$ the inclusion and $r:X \to Y$ a retraction ($r \circ i = Id_Y$). I have to prove that exists this short exact sequence $$ 0 \to K(X,Y) \to K(X) \to K(Y) \to 0.$$ Then I have to verify that $K(X) \simeq K(X,Y) \oplus K(Y)$. How can I do it? I think that $K(X,Y) = \tilde{K}(X/Y).$ Thank you very much.

Answer 1

This is purely formal and relies on the fact that $K$ is a contravariant functor from topological spaces to abelian groups (actually to commutative rings but this is not needed here).
Since $r\circ i=Id_Y$ we get $i^*\circ r^*=Id_{K(Y)}$ so that $r^*:K(Y)\to K(X)$ is a section of $i^*:K(X)\to K(Y)$ and your exact sequence of abelian groups $ 0 \to K(X,Y) \to K(X) \stackrel {i^*}{\to} K(Y) \to 0$ splits , yielding the required isomorphism $K(X) \simeq K(X,Y) \oplus K(Y)$.

By the way, it is indeed true that $K(X,Y) = \tilde{K}(X/Y)$ but this fact is irrelevant to the question at hand.