# Poincaré Duality as a $G$-morphism

In Brown's Cohomology of Groups book, at page 211 there is the following statement (Prop. 8.2):

Let $Y$ be a compact $d$-dimensional $K(G,1)$-manifold (possibly with boundary). Let $X$ be its universal cover and let $\Omega$ be the corresponding orientation module. Then there are $G$-module isomoprhisms for all $i$ $$H^i(G,\mathbb{Z}G) \cong \tilde{H}_{d-i-1}(\partial X)\otimes \Omega$$

The proof uses Poincaré Duality ($X$ is orientable being simply connected) $$H^i_c(X) \cong H_{d-i}(X,\partial X)$$ and then the author claims

This however is not canonical; it depends on a choice of orientation of $X$. In particular, it commutes or anti-commutes with the action of an element $g\in G$ according as $g$ preserves or reverses the orientation of $X$. Consequently, we have a $G$-module isomorphism $$H^i_c(X) \cong H_{d-i}(X,\partial X)\otimes \Omega$$

But I really don't the understand the "behaviour" of the $G$-action: according to me we should show something like $$g_*((\sigma\frown [X])\otimes \xi) = (g^*\sigma \frown [X])\otimes \xi)$$ (my naive interpretation of $G$-map. Then, under my hypothesis that $G$ acts diagonally on the LHS I get $$g_*(\sigma\frown [X])\otimes g_*(\xi) = (g^*\sigma \frown [X])\otimes \xi)$$

But I really don't understand how to prove such identity. I mean, the only naturality of cap product which comes to my mind is far from this equality and it is $$f_*(f^*\sigma \frown \alpha)=\sigma \frown f_*\alpha$$

Anyone can help me solving this doubt?

Cohomology is seen as a right $G$-module, therefore the naturality of the cap product is exactly what we need