Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

in the book "Transformation Groups" of tom Dieck on page 123 ff., an equivariant version of the Hopf classification theorem is developed. I extract the relevant data to state my question:

(U, B) is a relatively free $n$-dimensional W-CW complex, with W a finite group. M is a coefficient system which is isomorphic to $\mathbb{Z}$ as an abelian group. In addition, $H^n(U;\mathbb{Z})=\mathbb{Z}$ is assumed.

Now on page 124, he wants to show that the exact sequence $0\to H_n(U, B)\stackrel{i}{\to} H_n(U_n, U_{n-1})\stackrel{\partial}{\to} H_{n-1}(U_{n-1}, U_{n-2})$ is turned into an exact sequence (without the zero) when applying $Hom_W(\cdot, M)$. To show this, he chooses $\varphi:H_n(U_n, U_{n-1})\to M$ with $\varphi\circ i=0$ and a non-equivariant map $\psi:H_n(U_n, U_{n-1})\to H_n(U_n, U_{n-1})$ such that the norm-homomorphism $N$ satisfies $N(\psi)=id$. The $W$-action on $H_n(U_n, U_{n-1})$ is free, therefore such $\psi$ exists.

He wants to conclude that $\vert W\vert\cdot\varphi\circ\psi\circ i=N(\varphi\circ\psi\circ i)=\varphi\circ N(\psi)\circ i=\varphi\circ i=0$, so $\varphi\circ\psi\circ i=0$ and we find a non-equivariant homomorphism such that $\varphi\circ\psi=\psi'\circ\partial$. The norm of $\psi'$ is then the preimage of $\varphi$.

To carry out this argument, it is essential that the composition $\varphi\circ\psi\circ i$ is $W$-equivariant and I can't see the reason for this. Tom Diecks argument is that all self-maps of $H_n(U;\mathbb{Z})\cong\mathbb{Z}$ are $W$-equivariant, which is certainly true. But how does this help here? $\varphi\circ\psi\circ i$ is a map from $H_n(U, B)$ to $M$, the first is some free $\mathbb{Z}W$-module, the second need not be isomorphic to $H_n(U;\mathbb{Z})$ as a $\mathbb{Z}W$-module (so I think), and even if so, this does not seem to give the result.

Am I missing something very basic here, or are some assumptions missing? Sorry for stating such a long question with such relatively little interest. I would be grateful for answers nevertheless.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.