Bohr compactification

Remarks

The Bohr compactification exists for every topological group, and it is unique up to a unique isomorphism compatible with the maps from $G$. Example: if $G$ is a compact Hausdorff group, then $(G, \id_G)$ is a Bohr compactification, since each $f$ factors as $f = f \circ \id_G$ and the factor must be $f$. Non-example: $(\mathbb{T}, x \mapsto e^{2\pi i x})$ is not a Bohr compactification of $\R$. The continuous homomorphism $x \mapsto e^{2\pi i \sqrt{2} x}$ from $\R$ to $\mathbb{T}$ does not factor through $x \mapsto e^{2\pi i x}$. A factor $\tilde f$ would give $e^{2\pi i \sqrt{2}} = \tilde f(e^{2\pi i}) = \tilde f(1) = 1$, which is false. If $G$ is a locally compact abelian group, $bG$ is the Pontryagin dual of the group $\widehat{G}$ with the discrete topology, and $b(x)$ is evaluation at $x$. The map $b$ need not be injective, and its image is dense in $bG$.