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.



Density of functions into the circle glueing
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.


