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I am reading on K theory in Lawson and Michelson (Spin Geometry). One has the "exact sequence spaces" $L(X,Y)$ and there is the theorem that there is a unique equivalence of functors $\chi$ between $L$ and $K$ (the Euler characteristic) such that in the case $Y = \emptyset$ $$ \chi([V_0, \dots, V_n]) = \sum_{k=0}^n (-1)^k [V_k] \in K(X, \emptyset).$$ To give a more explicit characterization of the map in the case $n=1$, they make quite a complicated construction by defining a space $Z$ by gluying two copies of $X$ together along $Y$, defining bundles on $Z$ and using the isomorphism $K(Z, X_1) \cong K(X, Y)$. I compared to the original paper of Atiyah, Bott and Shapiro, and they make the same construction. However, I feel that one does not need all those surgical methods.

Instead, to define $\chi([V_0, V_1])$, look at the exact sequence $$0 \rightarrow K(X, Y) \stackrel{i}{\rightarrow} K(X) \stackrel{j}{\rightarrow} K(Y) \rightarrow 0$$ and notice that the element $[V_0] - [V_1] \in K(X)$ is in the kernel of $j$ since $V_0$ and $V_1$ are isomorphic when restricted to $Y$. Now define the Euler characteristic as the preimage of $[V_0] - [V_1]$ under $j$, which is well-defined by exactness.

Did I miss something or are the constructions that I referenced needlessly complicated?

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up vote 4 down vote accepted

You missed that the sequence is not exact at $K(X,Y)$ (neither is it exact at $K(Y)$, but that does not matter here). There is an ambiguity coming from $K^{-1} (Y)$, i.e. automorphisms of bundles. If $Y$ is a point, your construction works, and the purpose of the arguments in Atiyah-Bott-Shapiro is to extend it to $Y \neq \ast$.

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Ok, I see my mistake now. What I don't know is what the group $K^{-1}(Y)$ has to do with bundle automorphisms! Could you elaborate here? – Matthias Ludewig Oct 9 '12 at 8:09
You take pairs $(V,a)$ of vector bundles on $Y$, together with automorphisms. Consider the equivalence relation generated by isomorphism and homotopy (of the bundle automorphism). The trivial bundle, together with the identity isomorphism, is called trivial. Take the abelian semigroup of equivalence classes, take the Grothendieck group and mod out by the trivial elements. This is $K^{-1}(Y)$. – Johannes Ebert Oct 9 '12 at 8:53
Follow-up question: Does an exact sequence naturally determine an element of $K^{-1}$? – Matthias Ludewig Oct 9 '12 at 11:45

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