Let $p:\widetilde{X} \to X$ be an $n$-sheeted covering space. Consider a singular simplex $c:\triangle^k \to X$, because the simplex is simply-connected, there exist $n$ different lifts $\widetilde{c}$ of $c$ to $\widetilde{X}$.
Transfer Map: We define the transfer map $$ \begin{array}{rccl} \tau_k: & C_k(X) & \longrightarrow & C_k(\widetilde{X})\\ & c & \mapsto & \displaystyle \sum_{\widetilde{c} \,\mbox{lift of c}} \widetilde{c}. \end{array} $$
By dualizing, we get an induced map on cochain complexes $$ \begin{array}{rccl} \tau^k: & C^k(\widetilde{X}) & \longrightarrow & C^k(X)\\ & \phi & \mapsto & \displaystyle \sum_{\widetilde{c} \,\mbox{lift of c}} \phi(\widetilde{c}). \end{array} $$
Since $\tau^k$ is a chain map, we get the induced map $$ \begin{array}{rccl} \tau^*: & H^k(\widetilde{X}) & \longrightarrow & H^k(X)\\ & [\phi] & \mapsto & [\tau^k (\phi)] = [\phi \circ \tau_k] \end{array} $$ for any coefficient group $G$.
Now suppose that we have a $2$-sheeted covering space. And I want to know if $\tau^*$ is surjective, then I did this:
Let $[\psi] \in H^k(X;\mathbb{Z}_2)$. We want to see that exists $[\phi] \in H^j(\widetilde{X};\mathbb{Z}_2)$ such that $[\psi] = [\phi \circ \tau_k]$. Let's see at cochains level:
Choose a representative $\psi:C_k(X;\mathbb{Z}_2) \to \mathbb{Z}_2$.
Given a cochain $c \in C_k(\widetilde{X};\mathbb{Z}_2)$ we get a cochain $p \circ c \in C_k(X;\mathbb{Z}_2)$.
Then define $\phi: C_k(\widetilde{X};\mathbb{Z}_2) \to \mathbb{Z}_2$ by $c \mapsto \psi(p \circ c)$.
Then $\tau^*$ is...surjective? It is correct?
Thanks!
Let $p:\tilde X\rightarrow X$ be the projection of a $n$-covering map, $\tau^*\circ p^*$ is the multiplication by $n$. So the surjectivity of $\tau^*$ it depends of the characteristic of the coefficients. If the characteristic of the coefficient is relatively prime with $n$ it is surjective