Equivariance of a map from tom Diecks book

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.

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