Denoting the components of the $3\times3$ matrix $A \in O(3)$ as $a\_{ij}$, show that

$$ F: O(3) \rightarrow S^2, a\_{ij} \mapsto a\_{1j} $$

is a submersion. (The map is well defined since for $A \in O(3)$ it is true that $a\_{11}^2 + a\_{12}^2 + a\_{13}^2 = 1$.)

From what I understand to show that the map is a submersion, I have to show that the differential map

$$ dF: T\_AO(3) \rightarrow T\_{F(A)}S^2 $$

is onto for every $A \in O(3)$.

Now if I were given a tangent vector $X\in T\_AO(3)$ then from what I understand the map simply is $dF: x\_{ij} \mapsto x\_{1j}$, where $x\_{ij}$ denote the components of $X$. However, I fail to see that this is surjective at every point $A\in O(3)$.

Indeed at the identity $A = I$ I know that a basis of tangent vectors is given by the antisymmetric matrices with one non-zero number in the upper triangular part and that then the map is something like $X \mapsto (0, s, t)$ and arguably two parameters are enough to span the tangent space of $S^2$.

I fail to see how this is true for tangent vectors $X$ at an arbitrary point $A$ with the vector being given by

$$ X = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|\_{t=0} \gamma(t)$$

where $\gamma: \mathbb{R} \rightarrow O(3)$ with $\gamma(0) = A$.