Let $\mathbb{S}^n=\{x\in\mathbb{R}^{n+1}\mid \langle x,x\rangle_\text{euc}=1\}$ and $\mathbb{S}^{(1,n+1)}:=\{x\in \mathbb{R}^{(1,n+2)}\mid \langle x,x\rangle=1\}$, where $(\mathbb{R}^{(1,n+2)},\langle\cdot,\cdot\rangle)$ is the Lorentzian space with $\langle\cdot,\cdot\rangle=-dx_1^2+dx_2^2+...+dx_{n+3}^2$. Define: \begin{align*} f: \mathbb{S}^n &\to \mathbb{S}^{(1,n+1)}\\ x &\mapsto (\psi(x),\psi(x),i(x)) \end{align*} where $\psi:\mathbb{S}^n\to\mathbb{R}$ is a smooth function and $i:\mathbb{S}^n\to \mathbb{R}^{n+1}$ is the natural embedding. Show that $f$ is an isometric immersion and compute its second fundamental form.
I could easily show that $f$ is an isometric immersion by using the fact that $i$ is an immersion and by verifying that $f^*(-dx_1^2+dx_2^2+...+dx_{n+3}^2)=i^*(dx_1^2+...+dx_{n+1}^2)=g_{\mathbb{S}^n}$.
I had the following idea for computing the fundamental form, which I'm not sure is correct:
If $\nabla, \nabla^\text{euc}$ are the Levi-Civita connections for $\mathbb{S}^n$ and $\mathbb{R}^{n+1}$ respectively, we can easily verify that $\nabla_XY(p)=\nabla^\text{euc}_xY(p)-\langle X,Y\rangle_\text{euc}p$. Similarly, if $\widetilde{\nabla}$, $\nabla^\text{lor}$ are the LC connections for $\mathbb{S}^{(1,n+1)}$ and $\mathbb{R}^{(1,n+2)}$ respectively, we find $\widetilde{\nabla}_XY(p)=\nabla^\text{lor}_XY(p)+\langle X,Y\rangle p$. The second fundamental form $\alpha$ of $f$ would therefore be given by: \begin{align*} \alpha(X,Y)(p)&=\widetilde{\nabla}_XY(p)-\nabla_XY(p)\\ &=\nabla^\text{euc}_XY(p)-\nabla^\text{lor}_XY(p)+(\langle X,Y\rangle+\langle X,Y\rangle_\text{euc})p\\ &=(\langle X,Y\rangle+\langle X,Y\rangle_\text{euc})p \end{align*}
Where in the last equality I tried to argue that, since Christoffel symbols vanish in Euclidean and Lorentzian spaces, then $\nabla^\text{euc}$ and $\nabla^\text{lor}$ look essentially the same.
I know I've used some cheating and language abuse, but I hope this is essentialy correct, or at least restorable.