First part of the restriction map $H_{j}(G,A)\to H_{j}(H,A)$ on group homology Let $G$ be a group, $H$ be a subgroup of finite index and $A$ be a $G$-module.
I'm trying to understand the restriction map on homology $H_{j}(G,A)\to H_{j}(H,A)$ explicitly.
The first step would be to introduce a map $H_{j}(G,A)\to H_{j}(G,\operatorname{Ind}^{G}_{H}(A))$ and then use Shapiro.
We would like to induce this map functorially from a $G$-morphism
\begin{equation}
\tau\colon A\to\mathbb{Z}[G]\otimes_{\mathbb{Z}[H]}A=\operatorname{Ind}^{G}_{H}(A).
\end{equation}
Recall that the latter is a $G$-module by the general procedure of extending scalars.
According to Brown's Cohomology of Groups, the map $\tau$ should be
\begin{equation}
\tau(a)=\sum_{x\in G/H}x\otimes_{\small{H}}x^{-1}a
\end{equation}
(The notation here means $x$ varies over a fixed chosen system of representatives for $G/H$.)
Now, my question is: why is this a $G$-morphism? One the one hand, we have
\begin{equation}
\tau(ga)=\sum_{x\in G/H}x\otimes_{H}x^{-1}(ga).
\end{equation}
On the other hand, we have
\begin{equation}
g\tau(a)=\sum_{x\in G/H}g(x\otimes_{H}x^{-1}a)=\sum_{x\in G/H}gx\otimes_{H}x^{-1}a.
\end{equation}
Why are these two equal? I don't get it. Any help will be greatly appreciated. Thanks!

The morphism $\tau(a) = \sum_{x\in G/H}x\otimes_H x^{-1}a$ is independent of the choice of representatives $\{x \in G/H\}$ of $G/H$. Clearly if we have chosen another system of representatives $\{x'\in G/H\}$ then $x = x'h$ for some $h \in H$ and we have $\tau(a) = \sum_{x'\in G/H}x'\otimes_H x'^{-1}a = \sum_{x\in G/H}xh\otimes_H h^{-1}x^{-1}a = \sum_{x\in G/H}x\otimes_H x^{-1}a$. 
On the other hand, if $\{x\in G/H\}$ is a system of representatives then $\{gx\in G/H\}$ is also a system of representatives for any $g\in G$. Therefore,
$\tau(gx) = \sum_{x\in G/H}x\otimes_H x^{-1}(ga) = \sum_{x\in G/H}gx\otimes_H (gx)^{-1}(ga) = g\sum_{x\in G/H}x\otimes_Hx^{-1}a = g\tau(a)$.
