Let $S^1$ be the unit circle and for $x,y \in S^1$, let $x \sim y$ if and only if $x$ and $y$ are antipodal points of each other. So if $(\cos \theta, \sin \theta), (\cos \phi, \sin \phi) \in S^1$, then $(\cos \theta, \sin \theta) \sim (\cos \phi, \sin \phi) \iff \theta = \phi + \pi$.
So we construct the homeomorphism $\varphi: S^1 \to S^1 / \sim$ via
$$\varphi((\cos \theta, \sin \theta)) = [(\cos \theta, \sin \theta)]$$
Where $[(\cos \theta, \sin \theta)] = \{(\cos \phi, \sin \phi): \phi = \theta + \pi \}$ is the equivalence class of the point in question.
How can I show that $\varphi$ and $\varphi$ are continuous?
Recall that in an identification (decomposition) space, a set $\mathscr{F} \subset S^1 / \sim$ is open if and only if $\bigcup\{F: F \in \mathscr{F}\}$ is open in $S^1$.
To show $\varphi$ is continuous, let $\mathscr{F} \subset S^1 / \sim$ be open. Then by definition, $\bigcup\{F: F \in \mathscr{F}\}$ is open in $S^1$. We must show $\varphi^\leftarrow(\mathscr{F})$ is open. Note now that
$$\varphi^\leftarrow(\mathscr{F}) = \varphi^\leftarrow\Big(\displaystyle\bigcup_{F \in \mathscr{F}} F\Big) \displaystyle\bigcup_{F \in \mathscr{F}} \varphi^\leftarrow(F) = \dots ?$$
Is this headed in the right direction? What do I need to do to show $\varphi$ is continuous?
