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For the exactness at the left, you can use the long exact sequence in $K$ theory. Then every map involving (suspensions) of $i$ will split (why). Then the long exact sequence breaks down in short exact sequences.


Let $Vect^1(X)$ be isomorphism classes of complex line bundles on $X$. I assume you're familiar with the definition of topological $K^0(X)$. Let's examine the map $c_1: Vect^1(X) \to K^2(X)$, (recalling that, due to Bott periodicity, $K^0(X) \simeq K^2(X)$). The group operators are the tensor product of line bundles (in $Vect^1(X)$) and the tensor product ...


Let me just write $K$ for reduced and relative K-theory for readability. So far you agree that we can identify $K(X)$ with $K(X, A)$ and with $K(X, B)$, so the product can be identified with the product $K(X, A) \times K(X, B) \to K(X)$. Next you just need to agree that this product factors through $K(X, A \cup B)$, and this is because the corresponding ...

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