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I am using Hatcher's K-Theory book to work through the proof of the external product theorem:

$\mu:K(X) \otimes \mathbb{Z}[H]/(H-1)^2 \to K(X) \otimes K(S^2) \to K(X \times S^2)$ is an isomorphism

So far I have shown that $\mu$ is surjective. I am trying to work through the inverse function $\nu$. We use the notation $[E,f]$ where $f:E \times S^1 \to E \times S^1$ is the clutching function (I presume this is fairly standard notation, so i won't spend too much time on it!)

The proof of surjectivity follows the following rough process:

1) Show that $[E,f] \simeq [E,\ell]$ for $\ell$ a Laurent polynomial. Then reduce this to a product $\ell = z^{-n} q$, where $q$ is a polynomial (i.e. exponents $\ge 0$)

2) Reduce the polynomial clutching function to a linear clutching function $L^nq$

3) Further reduce the linear clutching function and show that there is splitting $E = E_+ \oplus E_-$

4) Expand out the original 'clutching data' and see that it lies in the image of $\mu$

To construct the inverse $\nu$ we define

$\nu([E,z^{-m}q])=((n+1)E)_- \otimes (H-1) + E \otimes H^{-m}$ for $n > \operatorname{deg} q$

To see this is well defined, part of the process is to see that the function $\nu$ is unchanged when $z^{-m}q$ is replaced with $z^{-m-1}(zq)$

According to Hatcher we also have the following two facts

a) $[(n+2)E,L^{n+1}(zq)] \simeq [(n+1)E,L^nq] \oplus [E,z]$

b) For $[E,z]$ $E_+=0$ and $E_-=E$

Now we come to my problem. There is some way to use (a) and (b) above (along with the inverse formula) to show that

$$\nu([E,z^{-m-1}(zq)]) = ((n+1)E)_- \otimes (H-1) + E\otimes (H-1) + E \otimes H^{-m-1}$$

My thoughts: We require $\deg q \geq n$. Since here we have multiplied by $z$ we have increased the degree of the polynomial by 1 (recall that $q$ is a polynomial in $z$).

I would have though we would have then said that

$$\nu([E,z^{-m}q])=((n+2)E)_- \otimes (H-1) + E \otimes H^{-m-1}$$

This is close, but not quite right I don't think. If I 'fudge' it to get the answer the line should read (I think!)

$$\nu([E,z^{-m}q])=[(n+2)E),L^{n+1}(zq)]_- \otimes (H-1) + E \otimes H^{-m-1}$$

I really need to get it in the form $[(n+2)E,L^{n+1}(zq)]$ so I can apply equation (a) above!

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Hi, I'm having a similar problem. I don't think you can "fudge" like you did. $[((n+2)E),L^{n+1}(zq)]_-$ is a vector bundle over $X \times S^2$ and not over $X$. – Riccardo Jul 10 '14 at 15:20

Well, I've put a bounty, but it seems that I've found a solution by myself. So here it is, for future reference :)

You'll need these facts, which I don't prove here.

Fact 1 The clutch construction over $X \times S^2$ preserve direct sum: $$[E_1 \oplus E_2 , f_1 \oplus f_2 ] \approx [E_1, f_1] \oplus [E_2,f_2]$$

Fact 2 the $\pm$splitting preserve direct sum. This is a bit tricky, maybe this is due to the notation of Hatcher which can lead to untrue statement. My advice is to use Husemöller's notation $L^n(E) := (n+1)E$. so the split is (in this case and with Husemöller notation): $$[L^{n+1}(E),L^{n+1}(zp)] \approx [\left( L^n(E,zp)\right)_+,Id] \oplus [\left( L^n(E,zp)\right)_-,z]$$ In fact the splitting heavily depends on the linear clutching function we are using.

So, let's start using a)

$$[(L^{n+1}(E),L^{n+1}(zq)] \approx [L^n(E),L^nq] \oplus [E,z] \approx [L^n(E) \oplus E, L^nq \oplus z]$$ thanks to Fact 1

then using the splitting at the left term we have: $$[L^{n+1}(E),L^{n+1}(zp)] \approx [\left( L^n(E,zp)\right)_+,Id] \oplus [\left( L^n(E,zp)\right)_-,z] $$ The splitting of the right term is $$[\left((L^n(E)\oplus E ,L^nq \oplus z)\right)_+ , Id]\oplus [\left((L^n(E)\oplus E ,L^nq \oplus z)\right)_- , z]$$ Now, the important thing is to note that by the uniqueness of such splitting, we must have that: $$ \left( L^n(E,zp)\right)_{\pm} \approx \left((L^n(E)\oplus E ,L^nq \oplus z)\right)_{\pm}$$ We are almost done!

Now it's turn to use Fact 2 and hint $b)$ in the last equality: $$ \left( L^n(E,zp)\right)_{-} \approx \left((L^n(E)\oplus E ,L^nq \oplus z)\right)_{-} \approx (L^n(E,L^nq))_- \oplus (E,z)_- \approx (L^n(E,L^nq))_- \oplus E $$

Hope it helps! I know this is a rather localized question, but nevertheless it's a question!

If someone has a better proof I'm more than happy to award to that proof the bounty!

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