The following comes from a 1971 paper by R. Langlands, "On the Functional Equation of the Artin L-functions"
There seem to be some formatting errors, for example $\mathcal S$ should be replaced by $\mathcal S_0$ in several places. I think $\mathcal S_1$ is the same thing as $\mathcal S_{0,1}$, and is meant to be a subcategory of $\mathcal S_0$, not just $\mathcal S$.
The main thing I am confused about is at the bottom of the page. I don't understand what the map $G_1/G_1^c \rightarrow \overline{G}/\overline{G}^c$ is. There is a natural map going the other way. But it is clear that Langlands wants to have a homomorphism $G_1/G_1^c \rightarrow G/G^c$. He says this is coming from the "transfer," but I don't understand what that is.
The "transfer" mentioned above is a group homomorphism that applies in a more general situation. I was unfamiliar with this notion until now. Let me answer my own question by giving the details.
Let $H$ be a subgroup of finite index of a group of $G$, and let $x_1, ... , x_n$ be a set of right coset representatives of $H$ in $G$. Let $\phi$ be a homomorphism of $H$ into an abelian group $A$ (for example, $A = H/[H,H]$).
For every $g \in G$, and each $1 \leq i \leq n$, there exists a unique $1 \leq i^g \leq n$ such that $Hx_ig = Hx_{i^g}$. Hence there exists an element $h_{i,g} \in H$ such that $ h_{i,g}x_ig =x_{i^g}$. Define a function $t: G \rightarrow A$ by
$$t(g) = \prod\limits_{i=1}^n \phi(h_{i,g})$$
1 . Well definedness:
Let $y_1, ... , y_n$ be another set of right coset representatives of $H$ in $G$. For $g \in G$ and $1 \leq i \leq n$, there exists a unique $1 \leq i_g \leq n$ such that $Hy_ig = Hy_{i_g}$. Hence there exists an element $_{i,g}h \in H$ such that $(_{i,g}h) y_ig = y_{i_g}$. We need to show that
$$\prod\limits_{i=1}^n \phi(h_{i,g}) = \prod\limits_{i=1}^n \phi(_{i,g} h)$$
The definition of $t$ clearly does not depend on the ordering of the coset representatives, so we may assume without loss of generality that $Hx_i = Hy_i$ for all $i$. Writing $h_ix_i = y_i$ for some $h_i \in H$ we have for all $i$ and any fixed $g \in G$:
$$h_{i^g} h_{i,g} x_i g = h_{i^g}x_{i^g} = y_{i^g} = (_{i,g}h)y_ig = (_{i,g}h)h_ix_ig$$ Thus $h_{i^g}h_{i,g} = (_{i,g}h)h_i$. This implies that
$$\prod\limits_{i=1}^n \phi(h_{i^g}) \phi( h_{i,g}) = \prod\limits_{i=1}^n \phi(_{i,g}h_i) \phi(h_i)$$ We are done, because
$$\prod\limits_{i=1}^n \phi(h_{i^g}) = \prod\limits_{i=1}^n \phi(h_i)$$
Homomorphism:
Let $g, g' \in G$. For $1 \leq i \leq n$, we have
$$x_{i^{(gg')}} = h_{i,gg'} x_i gg', x_{i^g} = h_{i,g} x_ig, x_{i^{g'}} = h_{i,g'} x_ig'$$
and we need to show that $\prod\limits_{i=1}^n \phi(h_{i,gg'}) = \prod\limits_{i=1}^n \phi(h_{i,g})\phi(h_{i,g'})$. For each $i$, we look at the integers $i^g$ and $(i^g)^{g'}$:
$$x_{(i^g)^{g'}} = h_{i^g,g'}x_{i^g}g' = h_{i^g,g'} h_{i,g}x_igg'$$
Therefore, $h_{i,gg'} = h_{i^g,g'}h_{i,g}$. This implies the result.
Topological Groups:
Of particular interest (to me) is when $G$ is a (Hausdorff) topological group, and $H$ is a closed subgroup of finite index. In that case, we let $A$ equal $H$ modulo the closure of $[H,H]$. Then $t$ is continuous.
To see this, it is enough to show that for each $1 \leq i \leq n$, the mapping $g \mapsto x_{i^g}$ is a continuous function $G \rightarrow G$. The image of this map is the finite set $\{x_1, ... , x_n\}$ with the discrete topology, so it is enough to show that the fiber of any $x_j$ is closed in $G$. For $g \in G$, we have $i^g$ is the unique number such that $x_igx_{i^g}^{-1} \in H$. We have $x_igx_j^{-1} \in H$ if and only if $g \in x_i^{-1}Hx_j$, which is closed in $G$ as required.
It is clear (without any topology) that the homomorphism $t: G \rightarrow A$ factors through the commutator subgroup $[G,G]$. This is the same as saying that $[G,G]$ is contained in the kernel of $t$. But since $t$ is continuous, $\overline{[G,G]}$ is also contained in the kernel of $t$, so we have an induced topological group homomorphism $G/\overline{[G,G]} \rightarrow H/\overline{[H,H]}$.