Is $f:SO(n)\rightarrow S^{n-1}$, $f(A)=(A^n_i)_i$ a submersion? Let $f:SO(n)\rightarrow S^{n-1}$, $f(A)=(A^n_i)_i$, that is $f(A)$ is the last row of $A$. Show that $f$ is a submersion.
I'm not sure how to calculate $df$, because I only know how to calculate the differential using local charts, but I don't know how to parametrize $SO(n)$, so this is my attempt:
Let $F:M_n(\mathbb{R})\rightarrow\mathbb{R}^n$, $F(M)=(M^n_i)_i$. Then $F|SO(n)=f$. Since $F$ is linear, if $p\in SO(n)\subset M_n(\mathbb{R})$, $dF_p(v)=(v^n_i)_i$, and we conclude that $df_p(v)=(v^n_i)_i.$
So, this is right? And how can I show that df is surjective?

In fact, if $e_n$ is the $n-$th base vector $f(A)= Ae_n$. Let us extend this formula to the set of all matrices, $F(A)=A.e_n$. So $F$ is linear and is its own differential $F'(A)B= Be_n$. The derivative of $f$ at the point $A$ is the restriction of this map to the tangent space $  \cal {A}(n). A$ of $O(n)$ at $A$, where $\cal A$ is the set of antisymmetric matrices. So we are reduce to prove that for every vector $y$ perpendicular to $Ae_n =u$ we can find an antisymmetric matrix $B$ such that $y=B (Ae_n)=Bu$. But precisely ${\cal A} .u=u^{\perp}$, as one can check in an orthonormal  basis where $u$ is the first vector.
