The first singular cohomology group of double cover

Question

Suppose $M$ is a connected nonorientable manifold, consider it's orientable double cover: $\pi:\tilde{M}\rightarrow M$. Why is map $\pi^*: H^1(M;\mathbb{R})\rightarrow H^1(\tilde{M},\mathbb{R})$ injective ? Thanks for help.

Answer 1

Let $G \cong \mathbb{Z}/2\mathbb{Z}$ be the deck group of the cover $\pi: \tilde M \to M$. We'll define the transfer homomorphism $t: H^\ast(\tilde M;\mathbb R) \to H^\ast(M;\mathbb R)$, which will satisfy that $t \pi^\ast = |G| \cdot \mathrm{id}$. This shows $\pi^\ast$ is injective.

I will give three constructions. One suffices, but I think it demonstrates that the transfer homomorphism is a natural feature of algebraic topology.

Transfer via singular cochains:

Let's say we're working with singular cochains. Let $C_\bullet(M)$ and $C_\bullet(\tilde M)$ be the singular chains on $M$ and $\tilde M$. Then we have a map $s^\sharp: C_\bullet(M) \to C_\bullet(\tilde M)$ given by $$s(\sigma) = \sum_{g \in G} g\circ \tilde \sigma,$$ where $\tilde \sigma$ is any lift of $\sigma$ to $M$ (which exists and is unique up to the action of $G$ since $\tilde M \to M$ is a cover. This is chain map since the boundary of a lift is a lift of the boundary. At the chain level, $\pi^\sharp s^\sharp = |G| \cdot \mathrm{id}$.

Now let $t^\sharp: C^\bullet(\tilde M, \mathbb R) \to C^\bullet(M,\mathbb R)$ be the dual of $s^\sharp$. It follows that $t^\sharp$ descends to a map $t$ on cohomology, and that $t\pi^\ast = |G|\cdot \mathrm{id}$.

Transfer via de Rham cohomology:

The transfer map also has a nice description in the smooth setting with de Rham cohomology. We have $\pi^\ast: \Omega^\bullet(M) \to \Omega^\bullet(\tilde M)$ is injective on the nose, with image exactly the $G$-invariant forms: certainly the image of $\pi^\ast$ is invariant, and if $\omega \in \Omega^\bullet(\tilde M)$ is invariant, then we can define a form $\bar\omega$ on $M$ by $$\bar\omega_x = (\pi^{-1})^\ast\omega_{\tilde x},$$ where $\tilde x \in \pi^{-1}(x)$, and $\pi^{-1}$ is a local inverse to $\pi$ there. Now the transfer map is $t^\sharp: \Omega^\bullet(\tilde M) \to \Omega^\bullet(M)$ given by $t^\sharp(\omega) = \sum_{g \in G} g^\ast \omega$.

Transfer via Čech cohomology

Let $X$ be a space and $\pi: \tilde X \to X$ be a finite-sheeted cover. Let $\mathcal{F}$ be a sheaf of abelian groups on $X$. Let $\mathcal{U}$ be any cover of $X$ where every open set in $\mathcal U$ is evenly covered. Now define $\mathcal{V}$ to be the cover of $\tilde X$ by slices of preimages of elements of $\mathcal{U}$. Then we have a map $$ t^\sharp: \check{C}^\bullet(\mathcal{V}, \pi^{-1}\mathcal{F}) \to \check{C}^\bullet(\mathcal{U}, \mathcal{F})$$ which acts as follows: for $V_{\alpha_i} \in \mathcal{V}$ slices of $U_{\alpha_i} \in \mathcal{U}$, either $\cap_{i=0}^k V_{\alpha_i}=V_{\alpha_0\cdots\alpha_k}$ is empty, or it is a slice of $U_{\alpha_0\cdots\alpha_k}$. In the latter case, $\pi: V_{\alpha_0\cdots\alpha_k} \to U_{\alpha_0\cdots\alpha_k}$ is a homeomorphism, so we may set $$t^\sharp: \pi^{-1}\mathcal{F}(V_{\alpha_0\cdots\alpha_k}) \to \mathcal{F}(U_{\alpha_0\cdots\alpha_k})$$ to be the isomorphism given by $\pi$. The map on Čech complexes is defined by taking the product of these maps; it is a map of complexes. Now every cover of $\tilde X$ may be refined by a cover of the form $\mathcal{V}$ since $\pi$ is open, so taking the inverse limit gives the map $t$ on Čech cohomology. By construction, $t$ satisfies $t\pi^\ast = n \cdot \mathrm{id}$.