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added "orientable"
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Juan
Juan
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Let a finite group $G$ acts on aan 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 $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

Let a finite group $G$ acts on a 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 $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

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

added 126 characters in body; added 7 characters in body
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Chris Gerig
Chris Gerig
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Let a finite group $G$ acts on a 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^*(Y,\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$$\pi^*=H^\ast(\pi):H^*(Y,\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$ and $\pi_!:H^*(X,\mathbb{Z})\rightarrow H^*(Y,\mathbb{Z})$. The latter map is given by $\pi_!=PD_Y\circ \pi^*\circ PD_X$$\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 $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

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

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

Let a finite group $G$ acts on a 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 $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

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Juan
Juan
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A question on composites of pushforward and pullback

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

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