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For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set $\tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$. Then $\tilde S$ is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold remains connected after removing a finite set.

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set $\tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$. Then $\tilde S$ is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set has two singular points at the zero vectors at points $p=(1,0)$ and $q=(-1,0)$. Then the set you mentioned is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold.

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set $\tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$. Then $\tilde S$ is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold remains connected after removing a finite set.

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set has two singular points at the zero vectors at points $p=(1,0)$ and $p=(-1,0)$. So it is not a manifold.

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set has two singular points at the zero vectors at points $p=(1,0)$ and $q=(-1,0)$. Then the set you mentioned is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold.