Show that
$$M = \lbrace e^{it} X \in \mathbb{R}^{m+1} \times \mathbb{R}^{m+1}: t \in \mathbb{R}, \ X \in S^m \subset \mathbb{R}^{m+1} \rbrace$$
is a compact regular submanifold of $\mathbb{R}^{2m+2}.$ Is $M$ orientable?
My thoughts: let $f: S^1 \times S^m \rightarrow \mathbb{R}^{2 m+2}$ given by $f\left(e^{i t}, X\right)=e^{i t} X = (X\cos t, X \sin t)$. Note that $f\left(e^{i t}, X\right)=f\left(e^{i s}, Y\right)$ iff $\left(e^{i t}, X\right)=\left(e^{i s}, Y\right)$ or $s=t+\pi$ and $Y=-X$. Using universal property in quotient topology, $f$ induces a continuous bijection map $\tilde{f}: \mathbb{R}P^{m+1} \to M.$ Then we have that $M$ is a compact regular submanifold of $\mathbb{R}^{2m+2}.$
But I don't know how to verify if $M$ is orientable. If $\tilde{f}$ is an immersion, then $\tilde{f}$ is an embedding. Since $\mathbb{R}P^n$ is non-orientable, then $M$ is not orientable. But I don't know if $\tilde{f}$ is an immersion, because I have difficulty to compute $df(e^{it},X).$ Can you give me any suggestion?
