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Nigel Hitchin, in a paper on differentiable manifolds (https://people.maths.ox.ac.uk/hitchin/hitchinnotes/manifolds2012.pdf), he states the theorem:

Theorem 2.2 Let $F : U \rightarrow {\rm R}^m$ be a $C^{\infty}$ function on an open set $U \subseteq {\rm R}^{n+m}$ and take $c \in {\rm R}^m$. Assume that for each $a\in F^{-1}(c),$ the derivative

$$ DF_a : {\rm R}^{n+m} \rightarrow {\rm R}^m $$

is surjective. Then $F^{-1}(c)$ has the structure of an n-dimensional manifold which is Hausdorff and has a countable basis.

What does he mean by "surjective"? In the proof, he uses surjectivity of the derivative to infer that the n+m by m Jacobian matrix $\partial F_i / \partial x_j $ has rank m.

James Alexander
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    A function $f : X \to Y$ is surjective if, for every $y \in Y$, there is some $x \in X$ such that $f(x) = y$. If $f$ is a linear map between vector spaces, this is equivalent to saying that $f$ has full rank. – Qiaochu Yuan Aug 24 '15 at 22:28
  • Ah, sorry; what I meant to say was that if $f$ is a linear map $\mathbb{R}^{n+m} \to \mathbb{R}^m$ (above I'm missing the condition that the dimension of the source is at least the dimension of the target), then $f$ is surjective iff $f$ has rank $m$. – Qiaochu Yuan Aug 24 '15 at 23:31
  • Are you assuming linearity? The function F is smooth, not linear. Also, it is DF, not F, that is surjective. Furthermore, he says that the DF_a is surjective at the point a. What does it mean to be surjective at a point? – James Alexander Aug 25 '15 at 20:59
  • fortunately, it is $DF$ that is linear, not $F$. "At the point $a$" just means he's taking the derivative at $a$; it's not modifying "surjective." – Qiaochu Yuan Aug 25 '15 at 21:12
  • I got it, Thanks! – James Alexander Aug 25 '15 at 21:47

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The condition "surjective" means exactly that the rank of the matrix of $DF_a$ has rank $m$, once this matrix is a linear transformation between vector spaces. And this is what surjective means.

L.F. Cavenaghi
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