# How to prove Shapiro's lemma for homology of spaces?

Let $X$ be a nice space with basepoint $x_0 \in X$ and universal covering $\pi \colon \widetilde{X} \to X$.

For a left $\mathbb{Z}[\pi_1(X,x_0)]$-module $V$, the homology of $X$ twisted by $V$ is $$H^{tw}_n(X;V) = H_n\left(C_*(\widetilde{X}) \otimes_{\mathbb{Z}[\pi_1(X,x_0)]} V\right),$$ where $C_*(\widetilde{X})$ is given the right $\mathbb{Z}[\pi_1(X,x_0)]$-module structure via action $\widetilde{X} \curvearrowleft \pi_1(X,x_0)$.

How do I prove the following Lemma from Exercise 75 of Lecture notes in Algebraic Topology by Davis-Kirk, and what does the notation for the group ring mean when $U$ is not normal?

Shapiro's Lemma: For each subgroup $U$ of $\pi_1(X,x_0)$, there is a (natural?) isomorphism $$H_*^{tw}\bigl(X;\mathbb{Z}\bigl[\pi_1(X,x_0)/U\bigr]\bigr) \xrightarrow{\cong} H_*(\widetilde{X}/U).$$

## 1 Answer

For simplicity I'll denote $\pi:=\pi_1(X,x_0)$. The set of left cosets $\pi/U$ admits an obvious left action of $\pi$, and so $\mathbb{Z}[\pi/U]$ becomes a left $\pi$-module in a natural way. This is the meaning of the coefficients on the left-hand side of your isomorphism.

Now the left-hand side is the homology of the chain complex \begin{align*} C_*(\widetilde{X})\otimes_{\mathbb{Z}[\pi]} \mathbb{Z}[\pi/U] & \cong C_*(\widetilde{X})\otimes_{\mathbb{Z}[\pi]} (\mathbb{Z}[\pi]\otimes_{\mathbb{Z}[U]} \mathbb{Z}) \\ &\cong C_*(\widetilde{X})\otimes_{\mathbb{Z}[U]}\mathbb{Z} \\ & \cong C_*(\widetilde{X}/U). \end{align*}

This topological interpretation of Shapiro's Lemma (which is commonly used in group cohomology and homological algebra more generally) should be more widely known.