Let a finite group $G$ acts on an orientable manifold $X$ freely. Denote $\pi:X\rightarrow Y=X/G$ be the quotient map. This covering map defines two maps between cohomology groups $\pi^*=H^\ast(\pi):H^*(Y,\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$ and $\pi_!:H^*(X,\mathbb{Z})\rightarrow H^*(Y,\mathbb{Z})$. The latter map is given by $\pi_!=(PD_Y)^{-1}\circ \pi_*\circ PD_X$, where $PD_X:H^\ast(X,\mathbb{Z})\to H_{n-\ast}(X,\mathbb{Z})$ denotes Poincare-duality and $\pi_\ast=H_\ast(\pi)$.

Is it true that $\pi_!\circ \pi^*=|G|\cdot id_{H^*(Y,\mathbb{Z})}$ and $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

transfer mapsthat satisfy your conclusion. I'm not sure your construction produces the transfer, and so I wouldn't immediately expect your conclusion to hold. (sorry I'm being lazy right now) $\endgroup$ – Chris Gerig Mar 7 '13 at 7:53