A map from the coinvariants of the dual to the dual of the invariants for a G-module Suppose $G$ is a group and $X$ is a $\mathbb{Z}[G]$-module.  Recall that the augmentation ideal $I \subset \mathbb{Z}[G]$ is generated by elements of the form $g - 1$ for $g \in G$, the coinvariants are defined as $X_G := X / IX$, and the invariants are given by $X^G := \{x \in X : gx = x, \forall g \in G\}$.
If $G$ is finite then the trace map $X_G \to X^G$ given by $x \mapsto \sum_{g\in G} gx$ is used in Tate cohomology, but this map doesn't make sense in the case that $G$ is infinite.
Suppose instead that we take a $\mathbb{Z}$-module $A$ and set $X^* := \operatorname{Hom}_{\mathbb{Z}}(X, A)$.  Then there is a well-defined map
\begin{align*}
\rho:(X^*)_G &\to (X^G)^* \\
[f] &\mapsto f|_{X^G}
\end{align*}
even when $G$ is infinite.  Is there a nice description of the kernel and cokernel of $\rho$?
 A: I will answer the question in the case that $G$ is cyclic or pro-cyclic.  Write $t$ for a (topological) generator of $G$.  Then we can describe $X^G$ and $X_G$ as the kernel and cokernel of $t-1$:
$$
0 \to X^G \to X \xrightarrow{t-1} X \to X_G \to 0.
$$
Write $Y = (t-1)X$ for the augmentation ideal.  We can split into two short exact sequences
\begin{align*}
&0 \to X^G \to X \to Y \to 0 \\
&0 \to Y \to X \to X_G \to 0.
\end{align*}
Note that the map $X \to Y$ is given by the action of $t-1$, while the map $Y \to X$ is just the inclusion.  Dualizing, we get sequences
\begin{align*}
&0 \to Y^* \xrightarrow{t-1} X^* \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)\\
&0 \to (X_G)^* \to X^* \longrightarrow Y^* \to \operatorname{Ext}^1_{\mathbb{Z}}(X_G, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A).
\end{align*}
Since the cokernel of $Y^* \xrightarrow{t-1} X^*$ is $(X^*)_G$ by definition, we get a sequence
$$0 \to (X^*)_G \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A).$$
The first map in this sequence is $\rho$, so $\rho$ is injective in this case, with cokernel equal to the kernel of $\operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)$.
If $A$ is divisible as an abelian group ($A = \mathbb{C}^\times$ for example), then $\operatorname{Ext}^1_{\mathbb{Z}}(-, A)$ vanishes and $\rho$ is an isomorphism.  In this case we also get an isomorphism $(X_G)^* \to (X^*)^G$.
