# Proof of external product theorem using K-theory

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 $\operatorname{deg} q > n$. 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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