A question on composites of pushforward and pullback

Question

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

Answer 1

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 $\{g\} \times Y$.

Answer 2

Yes: Let $\alpha$ be a cohomology class in $Y$. For simplicity, let's assume that $PD_{Y}\alpha$ is represented by an embedded cycle in $Y$. Then $PD_{X}\pi^{*}\alpha=\pi^{-1}PD_{Y}(\alpha)$. Therefore, $\pi\circ PD_{X}\pi^{*}\alpha = \pi (\pi^{-1}PD_{Y}\alpha)=|G|PD_{Y}\alpha$ as the map $\pi: \pi^{-1} PD_{Y}\alpha\rightarrow PD_{Y}\alpha$ is a $|G|$-to-1 covering. Now apply $PD_{Y}^{-1}$ to the previous equation to obtain $\pi_{!}\circ \pi^{*}\alpha=|G|\alpha$.