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 $\pi^\*\circ \pi\_!=|G|\cdot id\_{H^\*(X,\mathbb{Z})}$? 



A question on composites of pushforward and pullback
The first identity $\pi\_! \circ \pi^\* = \vert G \vert \cdot \mathrm{Id}$ holds, and follows from knowing that $\pi\_!$ is a $H^\*(Y)$-module map via $\pi^\*$, so
$$\pi\_!( \pi^\*(x)) = \pi\_!(1)\cdot x$$
and $\pi\_!(1) = \vert G \vert$ as may be seen from the $G$-cover over a point.


The second proposed identity $\pi^\* \circ \pi\_! = \vert G \vert \cdot \mathrm{Id}$ is false, as may be seen in the example $X= G \times Y$ and $G$ acting by translation on the first factor. A cohomology class $x$ supported on $\{e\} \times Y$ is sent by $\pi^\* \circ \pi\_!$ to its `$G$-invariantisation" $\sum\_{g \in G} g \cdot x$, which will never be a multiple of the original class, as it has support on each $\{g\} \times Y$.


