The bidual map on the circle and the integers

Given: The discrete group $\mathbb{Z}$ (The integers as equivalence classes of pairs of naturals, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and the circle $\mathbb{T}=\{z\in \mathbb{C}:\mathrm{\mid }z\mathrm{\mid }=1\}$ (The multiplicative unit circle is a compact metrizable topological abelian group).

A homomorphism $\gamma :\mathbb{Z}\to \mathbb{T}$ is determined by $\gamma (1)$, and every $z\in \mathbb{T}$ occurs; since $\mathbb{Z}$ carries the discrete topology every such homomorphism is continuous. Hence $z\mapsto (n\mapsto z^n)$ is an algebraic isomorphism $\mathbb{T}\to \widehat{\mathbb{Z}}$. Compact subsets of the discrete space $\mathbb{Z}$ are finite (their cover by singletons has a finite subcover), so the preimage of each compact-open subbasic set is a finite intersection of open conditions on powers of $z$; the inverse is evaluation at $1$, whose preimages of open sets are subbasic open sets. Thus both directions are continuous. (The integers as equivalence classes of pairs of naturals, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, The Pontryagin dual with the compact-open topology, Topological group: multiplication and inversion are continuous)

Every continuous homomorphism $\chi :\mathbb{T}\to \mathbb{T}$ is $\chi (z)=z^k$ for a unique $k\in \mathbb{Z}$. Indeed, writing $\mathbb{T}\cong \mathbb{R}\mathrm{/}\mathbb{Z}$ through the unit-circle isomorphism, $\chi$ corresponds to a continuous homomorphism $\psi :\mathbb{R}\to \mathbb{T}$ with $\psi (1)=1$; by the classification of characters of the line there is a unique $\xi \in \mathbb{R}$ with $\psi (t)=e^{2\pi i\xi t}$, and $\psi (1)=1$ forces $e^{2\pi i\xi }=1$, that is $\xi \in \mathbb{Z}$ by the kernel of the complex exponential. The dual of the compact circle is discrete by Compact groups have discrete duals and discrete groups have compact duals, so this bijection from the discrete group $\mathbb{Z}$ is a topological isomorphism. (The multiplicative unit circle is a compact metrizable topological abelian group, Continuous characters of the real line are exponentials, $\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$, The Pontryagin dual with the compact-open topology)

For every locally compact Hausdorff abelian group $G$ the evaluation map ${\mathrm{\Phi }}_G(x)(\gamma ):=\gamma (x)$ is an isomorphism of topological groups $G\to \widehat{\widehat{G}}$. (Pontryagin biduality: the evaluation map is a topological isomorphism)