Connected, compact topological group, continuous group homomorphism, $f(g) = 1$?

Question

Let $G$ be a connected and compact topological group and let $f: G \to \mathbb{R}^*$ a continuous group homomorphism. Does it follow that $f(g) = 1$ for all $g \in G$?

Answer 1

Yes. To prove this, notice that $f(G)$ is a compact and connected subgroup of $\mathbb{R}^*$. Thus, it suffices to prove that any such subgroup if trivial.

Let $H$ be a connected and compact subgroup of $\mathbb{R}^*$. We can suppose $H$ is contained in $\{ x >0 \}$. If $x \in H$ and $x \neq 1$, then either $x < 1$ or $x > 1$. In the first case, the sequence $(x^n)$ is in $H$ but has no convergent subsequence, since it converges to $0 \not \in H$. In the second case, the sequence $(x^{-n})$ has no convergent subequence for the same reason. In any case, this contradicts the compactness of $H$. Thus, $H = \{ 1\}$, as claimed.