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  "body": "\n## Proof\n\nWe choose $m$ coordinates in which the derivative is invertible, and apply the implicit function theorem.\n\n1. Let $U$, $m \\le n$, $F$ and $p$ be as in the statement. Let $J = J_F(p)$ be the $m \\times n$ [[Jacobian matrix|0YF0VXKXGB]], so $DF(p) v = J v$.\n2. The image of $DF(p)$ is the column space of $J$. Since $DF(p)$ is surjective, the column space is $\\R^m$, so the [[rank|0EWQEYRVA8]] of $J$ is $m$.\n3. A spanning set of $\\R^m$ contains a basis. So some $m$ columns of $J$, with indices $i_1 \u003c \\dots \u003c i_m$, are [[linearly independent|0EVHJFTG9Z]]. The $m \\times m$ matrix $C$ of these columns is invertible.\n4. Case $m = n$. Then $C = J$, and $DF(p)$ is invertible. By the [[inverse function theorem|0K9HSJFCMJ]], $F$ is injective on an open neighbourhood $W$ of $p$. So $\\set{x \\in W : F(x) = 0} = \\set{p}$. This is the graph of the constant function of the empty list of remaining coordinates.\n5. Case $m \u003c n$. For $x \\in \\R^n$, let $x'' = (x_{i_1}, \\dots, x_{i_m}) \\in \\R^m$, and let $x' \\in \\R^{n-m}$ list the other coordinates in their order. Let $P(x) = (x', x'')$.\n6. The map $P$ is a linear bijection that only reorders coordinates, so it preserves norms. Hence $\\Omega = P(U)$ is open, and $G = F \\circ P^{-1} \\colon \\Omega \\to \\R^m$ is of class $C^1$.\n7. Write $P(p) = (p', p'')$. By the [[chain rule|0EAAS2EHDR]], the partial derivatives of $G$ with respect to the coordinates of $x''$ at $(p', p'')$ are the columns $i_1, \\dots, i_m$ of $J$. So $D_{x''} G(p', p'') = C$, which is invertible.\n8. Also $G(p', p'') = F(p) = 0$. By the [[implicit function theorem|0BXBADR94C]], there are open sets $U' \\ni p'$ in $\\R^{n-m}$ and $V' \\ni p''$ in $\\R^m$ and a $C^1$ function $\\varphi \\colon U' \\to V'$ with this property. For $(x', x'') \\in U' \\times V'$, $G(x', x'') = 0$ if and only if $x'' = \\varphi(x')$.\n9. The function $G$ is evaluated on $U' \\times V'$ in this statement, so $U' \\times V' \\subseteq \\Omega$. Let $W = P^{-1}(U' \\times V')$. It is open, since $P$ is continuous, and $p \\in W \\subseteq U$.\n10. Let $x \\in W$. Then $(x', x'') = P(x) \\in U' \\times V'$ and $F(x) = G(x', x'')$. So $F(x) = 0$ if and only if $x'' = \\varphi(x')$.\n11. Hence $\\set{x \\in W : F(x) = 0}$ is the set of $x \\in \\R^n$ with $x' \\in U'$ and $x'' = \\varphi(x')$. This is the graph of the $C^1$ function $\\varphi$ of the $n - m$ coordinates in $x'$.\n12. So the coordinates $x_{i_1}, \\dots, x_{i_m}$ and the neighbourhood $W$ have the property in the statement.\n\n## Idea\n\nSurjectivity of $DF(p)$ means that the Jacobian matrix has $m$ independent columns. The coordinates of these columns play the role of $y$ in the implicit function theorem, and the remaining coordinates play the role of $x$. After this renumbering, the statement is the implicit function theorem.\n",
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