Let be $SO(\mathbb{R^{n+2}})$ the special orthogonal group of $\mathbb{R^{n+2} }$, manifold of dimension $\frac{(n+1)(n+2)}{2}$, and $\mathbb{S^{n+1}}$ the canonical unit sphere of $\mathbb{R^{n+2}}$. Consider the following map
$$F:SO(\mathbb{R^{n+2}})\to \mathbb{S^{n+1}}$$
that, for each $Z\in SO(\mathbb{R^{n+2}}),$ $F(Z)$ is the last line of the matrix $Z$.
I would like to show that $F$ is a submersion and I thought in this way:
Thinking $SO(\mathbb{R^{n+2}})$ like a subset of $\mathbb{R^{(n+2)^2}}$, I can see the map $F$ as the projection of last $(n+2)$-coordinates, then $F$ is obviously linear so $dF_Z=F$, for all $Z\in SO(\mathbb{R^{n+2}})$.
Then we have that $$dF_Z=F:\mathbb{R^{(n+2)^2}}\to \mathbb{R^{n+2}}$$
so $dF_Z$ is surjective. Is that right?
Because, after that I would like to conclude, for a special $c\in \mathbb{S}^{n+1}$, that the manifold $F^{-1}(\{c\})$ has dimension $\frac{n(n+1)}{2}$, and I have some doubts about the derivative above.
I.e., if I can use this $$dF_Z=F:\mathbb{R^{(n+2)^2}}\to \mathbb{R^{n+2}}$$ with the argument above, or I should use
$$dF_Z:T_Z SO(\mathbb{R^{n+2}})\to T_{F(Z)}\mathbb{S^{n+1}}$$ with another argument to show that $F$ is a submersion.
Thank some help.
EDIT: By the Andrew D. Hwang comments, I need to show that the restriction of $dF_z$ into $T_Z SO^+(\mathbb{R}^{n+2})$ is surjective.
As $$T_Z SO(\mathbb{R}^{n+2})=\{X\in \mathcal{M}_{n+2}(\mathbb{R}) \ | \ X^t+X=0\}$$ and $$T_{F(Z)}\mathbb{S}^{n+1}=\{y\in \mathbb{R}^{n+2} \ | \ \langle F(z), y\rangle=0\},$$
given $x\in T_{F(Z)}\mathbb{S}^{n+1}$ I need to find $X\in T_Z SO(\mathbb{R}^{n+2})$ such that $F(X)=x$. Someone can help me?