A homomorphism image need not be embedded

Let $i(t)=(e^{2\pi it},e^{2\pi i\alpha t})\mathrm{.}$ By [F1], it is an injective smooth homomorphism and immersion, and its image is dense in ${\mathbb{T}}^2$. Thus it is an immersed one-dimensional subgroup with intrinsic parameter $t\in \mathbb{R}$.

For each $j\ge 1$, the finite set $\{\parallel q\alpha \parallel :1\le q\le j\}$ has a positive minimum $d_j$. Choose an integer $N$ with $1\mathrm{/}N<\min (d_j,1\mathrm{/}j)$. Applying the finite pigeonhole principle to the $N+1$ fractional parts of $0,\alpha ,\dots ,N\alpha$ gives $1\le q\le N$ with $\parallel q\alpha \parallel <1\mathrm{/}N$. Such a $q$ must exceed $j$. Let $q_j$ be the least positive integer with $q_j>j$ and $\parallel q_j\alpha \parallel <1\mathrm{/}j$; taking the least witness avoids countable choice.

Then $q_j\to +\mathrm{\infty }$ in the intrinsic source $\mathbb{R}$, while $i(q_j)=(1,e^{2\pi i\alpha q_j})⟶(1,1)$ in the ambient torus and hence in the subspace topology on the image. If the image were embedded, the inverse $i^{-1}:i(\mathbb{R})\to \mathbb{R}$ would be continuous, forcing $q_j=i^{-1}(i(q_j))\to 0$, a contradiction.

Therefore a homomorphism image can be immersed but nonembedded. The failure is topological, not algebraic; compactness of the ambient torus alone would not prove it. The counterexample and sequence use no choice. Under countable choice, [F3] identifies the intrinsic structure just used with the canonical image structure.