I am reading Differential topology by Guillermin, on one of the page the author made a statement. (I am paraphrasing)
suppose that $g_1,....g_l$ are smooth function from a manifold $X \rightarrow R$ The set $Z$ of common zero is a submanifold if $(0,.....0)$ is a regular value of $g = g_1 \times g_2.... \times g_l$ You can quickly verify that $dg_x: T_x(X) \rightarrow R^l$ is surjective if and only if the l functionals $d(g_1)_x.....d(g_l)_x$ are linearly independent on $T_x(X)$.
Seems to me the last statement is implying the derivative, jacobian is surjective if and only if the row vectors are linearly independent? So first question is, how does the linearly indpendent of the $d(g)i)_x$ translates to linearly independence of the row vectors?
Second question: If so, what's the proof of that (from linear algebra)? I know that column vectors of a linear map being linearly independent implies the map is injective, so I'm guessing the dual notion must be true (since for square matrices, column rank = row rank and a linear map is injective if and only if it is surjective).