Forgetting the compact-open topology destroys Pontryagin duality

Identify $X=\mathrm{Hom}(\mathbb{Z},\mathbb{T})$ with $\mathbb{T}$ by [F1] and equip it with the discrete topology. Since a homomorphism on the discrete group $X$ is automatically continuous by [F3], the group of continuous characters ${\mathrm{Hom}}_{cts}(X,\mathbb{T})$ is exactly the group ${\mathrm{Hom}}_{alg}(\mathbb{T},\mathbb{T})$ of all algebraic homomorphisms. The evaluation map $\mathbb{Z}\to {\mathrm{Hom}}_{cts}(X,\mathbb{T})$ is $n\mapsto (z\mapsto z^n)$; with the compact-open topology on $\mathbb{T}$, [F1] says these power maps are exactly its continuous characters.

By [F2] choose a Hamel basis $B$ and expand $1$ in it. Choose a basis element $b_0$ whose rational coefficient $r={\mathrm{\Lambda }}_{b_0}(1)$ is nonzero, and put $T={\mathrm{\Lambda }}_{b_0}\mathrm{/}r:\mathbb{R}\to \mathbb{R}$. Then $T$ is $\mathbb{Q}$-linear with $T(1)=1$, hence $T(n)=n$ for every integer $n$. The complement clause of the Hamel-basis supplier gives another basis vector $b_1\mathrm{\ne }{b}_0$; then $T(b_1)=0$ while $b_1\mathrm{\ne }0$. Consequently $T$ cannot be multiplication by a real scalar: its value at $1$ would force that scalar to be $1$, contradicting its value at $b_1$. Since $T(\mathbb{Z})\subseteq \mathbb{Z}$, it induces the well-defined homomorphism $f_T(x+\mathbb{Z})=T(x)+\mathbb{Z}$ of $\mathbb{R}\mathrm{/}\mathbb{Z}$. This uses the supplied coordinate map and does not assume the arbitrary Hamel basis contains $1$.

Let $f_T$ be the homomorphism induced by $T$ of step 1.2. As a homomorphism from the discrete group $X$ it belongs to ${\mathrm{Hom}}_{cts}(X,\mathbb{T})$ by [F3]. If it were continuous for the compact-open topology on $\mathbb{T}$, then by [F1] there would be $k\in \mathbb{Z}$ with $f_T(z)=z^k$ for all $z$, that is $T(x)-kx\in \mathbb{Z}$ for all $x\in \mathbb{R}$. The map $G(x):=T(x)-kx$ is then $\mathbb{Q}$-linear with $G(\mathbb{R})\subseteq \mathbb{Z}$; for every $x$ and every $n\ge 1$ one has $G(x)=nG(x\mathrm{/}n)\in n\mathbb{Z}$, so $G(x)\in {\bigcap }_{n\ge 1}n\mathbb{Z}=\{0\}$ and $T=k\mathrm{i}\mathrm{d}$, contradicting the choice of $T$. Hence $f_T$ is an algebraic homomorphism continuous on the discrete $X$ but not continuous on the compact-open circle.

The homomorphism $f_T$ lies in ${\mathrm{Hom}}_{cts}(X,\mathbb{T})={\mathrm{Hom}}_{alg}(\mathbb{T},\mathbb{T})$ but is not one of the maps $z\mapsto z^k$, so it is not in the image of the evaluation map $\mathbb{Z}\to {\mathrm{Hom}}_{cts}(X,\mathbb{T})$, whose image is exactly $\{z\mapsto z^n:n\in \mathbb{Z}\}$ by [F1] and is a copy of $\mathbb{Z}$. The evaluation map is injective, since $z\mapsto z^n$ determines $n$, but it is not surjective: the discrete algebraic character group $X$ fails the biduality conclusion, and no topological constraint on $X$ beyond discreteness is available to repair it.

By contrast, with the compact-open topology the same group $X=\widehat{\mathbb{Z}}\cong \mathbb{T}$ is compact and [F4] gives ${\mathrm{Hom}}_{cts}(X,\mathbb{T})\cong \mathbb{Z}$ through the evaluation map. Thus it is the discarded compact-open topology, not the algebraic character group, that carries Pontryagin biduality; the statement refuted above is false.