Duals of finite products and of discrete direct sums

Part (1), the map $\mathrm{\Phi }$ is a bijective group homomorphism: it is a homomorphism because both sides multiply pointwise, $\mathrm{\Phi }({\gamma }_1{\gamma }_1^{\mathrm{\prime }},\dots ,{\gamma }_n{\gamma }_n^{\mathrm{\prime }})(x)={\prod }_j{\gamma }_j(x_j){\gamma }_j^{\mathrm{\prime }}(x_j)=(\mathrm{\Phi }(\gamma )\mathrm{\Phi }({\gamma }^{\mathrm{\prime }}))(x)$; it is injective because ${\gamma }_j$ is recovered from $\chi =\mathrm{\Phi }(\gamma )$ by restricting $\chi$ to the $j$-th coordinate axis, and it is surjective because every character $\chi$ of the product gives characters ${\gamma }_j(x_j):=\chi (0,\dots ,0,x_j,0,\dots ,0)$ with $\chi (x)={\prod }_j{\gamma }_j(x_j)$ for every $x$, by multiplicativity of $\chi$ and the decomposition of $x$ into its coordinate vectors; ${\gamma }_j$ is a continuous homomorphism because the coordinate inclusion $x_j\mapsto (0,\dots ,x_j,\dots ,0)$ is continuous.

Part (2), the restriction map: let $G={⨁}_i{G}_i$ carry the discrete topology and let ${\iota }_i:G_i\to G$ be the coordinate inclusion, a homomorphism and continuous because its source $G_i$ is discrete: every open target set has an open preimage, as every subset of $G_i$ is open. The map $\rho :\widehat{G}\to {\prod }_i\widehat{G_i}$, $\rho (\chi ):=(\chi \circ {\iota }_i{)}_i$, is a group homomorphism, and it is bijective: injective because a homomorphism on the direct sum is determined by its values on the summands, and surjective because for any family $({\gamma }_i)$ the formula $\chi (x):={\prod }_i{\gamma }_i(x_i)$ is a finite product over the support of $x$, is a homomorphism, is continuous because every open target set has an open preimage in the discrete source $G$, and satisfies $\chi \circ {\iota }_i={\gamma }_i$.

Part (1), $\mathrm{\Phi }$ is continuous at the identity: let $K\subseteq G_1\times \cdots \times G_n$ be compact and $W\subseteq \mathbb{T}$ open with $1\in W$; the projections $K_j:={\pi }_j[K]$ are compact, and by [F5] choose an open neighbourhood $V$ of $1$ with $V^n\subseteq W$, so that whenever ${\gamma }_j[K_j]\subseteq V$ for all $j$ and $x\in K$ one has ${\prod }_j{\gamma }_j(x_j)\in V^n\subseteq W$. Hence $\mathrm{\Phi }(S(K_1,V)\times \cdots \times S(K_n,V))\subseteq S(K,W)$, and the product is an open neighbourhood of the identity of $\widehat{G_1}\times \cdots \times \widehat{G_n}$ by [F1] and [F2]; since $\mathrm{\Phi }$ is a homomorphism of topological groups and translations are homeomorphisms, continuity at the identity gives continuity everywhere.

$\rho$ is continuous: each component $\chi \mapsto \chi \circ {\iota }_i$ is the pullback along the continuous homomorphism ${\iota }_i$, hence continuous by [F3]; a map into the product ${\prod }_i\widehat{G_i}$ is continuous exactly when all its components are.

Part (1), ${\mathrm{\Phi }}^{-1}$ is continuous at the identity: let $K_j\subseteq G_j$ be compact and $U_j\subseteq \mathbb{T}$ open with $1\in U_j$, and put $K:=(K_1\cup \{0\})\times \cdots \times (K_n\cup \{0\})$, a compact subset of the product; if $\chi \in S(K,{\bigcap }_j{U}_j)$ and ${\gamma }_j(x_j):=\chi (0,\dots ,x_j,\dots ,0)$ for $x_j\in K_j$, then ${\gamma }_j(x_j)\in {\bigcap }_j{U}_j\subseteq U_j$, so ${\mathrm{\Phi }}^{-1}(\chi )\in S(K_1,U_1)\times \cdots \times S(K_n,U_n)$; hence ${\mathrm{\Phi }}^{-1}$ maps a subbasic identity neighbourhood into a basic identity neighbourhood and is continuous at the identity, hence everywhere. Since $\mathrm{\Phi }$ is a continuous bijective homomorphism with continuous inverse, it is an isomorphism of topological groups, completing (1).

Both sides of $\rho$ are compact Hausdorff: $\widehat{G}$ is compact by [F2] because $G$ is discrete, each $\widehat{G_i}$ is compact by [F2], and the product ${\prod }_i\widehat{G_i}$ is compact by Tychonoff's theorem by [F4]; both are Hausdorff being duals of topological groups by [F2] and products of Hausdorff spaces. Therefore the continuous bijection $\rho$ from the compact space $\widehat{G}$ onto the Hausdorff space ${\prod }_i\widehat{G_i}$ is a homeomorphism by [F4], so $\widehat{G}$ is topologically isomorphic to the product of the duals; this completes (2).

Parts (1) and (2) are steps 3.1 and 3.2 respectively, so the lemma is proved.