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I'm reading Atiyah's K-Theory book and in the section where he introduces the Grothendieck group, he gives two constructions. One of them is as follows:

Let $A$ be an abelian semigroup, let $\Delta:A\rightarrow\Delta\times\Delta$ be the diagonal map, and let $K(A)$ be the set of cosets of $\Delta(A)$ in $A\times A$.

Further down, he states the following, which I couldn't figure out and I hope to get some help with:

Let $X$ be a (topological) space, let $\mathrm{Vect}(X)$ denote the set of isomorphism classes of (complex) vector bundles over $X$, write $K(X)$ for $K(\mathrm{Vect}(X))$, and for a vector bundle $E$ over $X$, write $[E]$ for its image in $K(X)$. Then from the above construction, it follows that if $[E]=[F]$, then there exists a bundle $G$ such that $E\oplus G\cong F\oplus G$.

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An element $A\in K(X)$ is a pair $(A_1,A_2)$ for $A_1,A_2\in \text{Vect}(X)$ that is well-defined up to addition by an element of the form $(B,B)\in \Delta$ (where $\Delta$ is the diagonal of $\text{Vect}(X)$); i.e., $A$ is a coset of the diagonal in $\text{Vect}(X)\times \text{Vect}(X)$ and $(A_1,A_2)$ may be thought of as a representative of this coset.

You should think of such a pair $(A_1,A_2)\in K(X)$ as a formal difference $A_1-A_2$. The reason such a pair represents a coset of the diagonal (and isn't simply a single formal difference) reflects the fact that the formal differences $(A_1+B)-(A_2+B)$ should be equal for all $B\in \text{Vect}(X)$ (as you would expect). I encourage that you think about this in the case of the natural numbers (Exercise 1 below). Of course, remember that the goal of Grothendieck's K-construction is to formally add additive inverses to an abelian semigroup (or monoid). (It's easy enough to form an abelian monoid from an abelian semigroup (just artificially add an element denoted by $0$ to your abelian semigroup and define addition of $0$ with other elements in the obvious manner); it's slightly less trivial to form an abelian group from an abelian monoid (this is the purpose of Grothendieck's K-construction).)

The map $\text{Vect}(X)\to K(X)$ defined by the rule $E\to [E]$ is equivalent to the map defined by the rule $E\to (E,0)$ in the form described in the first paragraph. (Why?)

So, if $[E]=[F]$ for $E,F\in \text{Vect}(X)$, then $(E,0)=(F,0)$ in $K(X)$; i.e., $(E,0)$ and $(F,0)$ represent the same coset of $\Delta$ in $\text{Vect}(X)\times \text{Vect}(X)$. By definition (of coset equality in an abelian semigroup), this means that $(E,0)+(G,G)=(F,0)+(G,G)$ for some $(G,G)\in \Delta$, i.e., $(E+G,G)=(F+G,G)$; and the equality in this line is equality in $\text{Vect}(X)\times \text{Vect}(X)$. Of course, "is equal to" in $\text{Vect}(X)$ is the same thing as saying "is isomorphic to (as vector bundles)" so we have that $E\oplus G\cong F\oplus G$ as vector bundles over $X$.

To further your understanding, I recommend that you think about the following exercises:

Exercise 1: Let $\mathbb{N}$ be the abelian semigroup consisting of all natural numbers. What is $K(\mathbb{N})$ (as an isomorphism class of an abelian group) in the "cosets of the diagonal" description?

Exercise 2: What is $K(\mathbb{N}\cup \{0\})$ in the "cosets of the diagonal" description?

Exercise 3 (interesting!): Let $X$ be the set consisting of all nonnegative integers together with $\infty$. $X$ is an abelian semigroup where addition of nonnegative integers is as usual and we define $n+\infty=\infty=\infty+n$ for all nonnegative integers $n$. What is $K(X)$? (The answer might be surprising and it's a well-known abelian group!)

Solutions to Exercises (hover cursor over grey regions below):

Solution to Exercise 1:

$\mathbb{Z}$

Solution to Exercise 2:

$\mathbb{Z}$

Solution to Exercise 3:

$0$

I hope this helps!

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    $\begingroup$ Appreciate your detailed answer $\endgroup$ – cyc Aug 28 '13 at 0:00
  • $\begingroup$ You're welcome @learner! By the way, I've added the solutions to the Exercises above in case you're interested. (I haven't added explanations; only the solution to Exercise 3 might be genuinely surprising. Of course, I'm happy to add explanations or discuss these exercises further with you if you'd like.) Also, I'd like to add that if $X$ is a compact Hausdorff space in the context of your question, then we can in fact choose $G\in \text{Vect}(X)$ to be trivial! Indeed, there exists $H\in \text{Vect}(X)$ such that $G\oplus H$ is (isomorphic to) a trivial bundle. $\endgroup$ – Amitesh Datta Aug 28 '13 at 7:33

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