Density of functions into the circle glueing

Question

Let $\{U_i\}_{i=1}^2$ be an open cover of $S^1$, with $U_i\cong \mathbb{R}$ (for example, $U_1$ is the lower arc of the circle and $U_2$ is the upper part). Let $\iota_i:U_i\hookrightarrow S^1$ be the cannonical inclusions and let $\iota_{i\star}$ be the push-forwards on the function spaces $C(\mathbb{R},U_i)$ defined by: $$ \begin{aligned} \iota_{i\star}: &C(\mathbb{R},U_i)\rightarrow C(\mathbb{R},S^1)\\ &f\mapsto \iota_i\circ f. \end{aligned} $$ If we equip $C(\mathbb{R},U_i)$, for $i=1,2$, and $C(\mathbb{R},S^1)$ with the compact-open topologies. Then, is it true that: $$ \overline{\bigcup_{i=1}^2 \iota_{i\star}\left[C(\mathbb{R},U_i)\right]} = C(\mathbb{R},S^1) ? $$

Note: If it simplifies the question, we may assume $U_i$ is diffeomorphic to $\mathbb{R}$ and ask if $\bigcup_{i=1}^2 \iota_{i\star}\left[C^{\infty}(\mathbb{R},U_i)\right] $ is dense in $C^{\infty}(\mathbb{R},S^1)$ with respect to the compact-open topology still. Then we conclude the original question by Weirestrass's theorem.

Answer 1

Maybe I don't understand your question. But isn't $\iota_{i*}[C(\mathbb R,U_i)]$ the set of continuous functions with values in $U_i$? Why should the set of functions with values in either of the subsets be dense? For example, $f(x)=e^{ix}$ is not in the closure of the union.