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  "body": "\n## Proof\n\nWe apply the inverse function theorem at each point.\n\n1. Let $U$ and $f$ be as in the statement, and let $a \\in U$. Since $Df(a)$ is invertible, its Jacobian determinant is not zero.\n2. By the [[inverse function theorem|0K9HSJFCMJ]] with $k = 1$, there are open sets $V \\subseteq U$ and $W \\subseteq \\R^n$ with $a \\in V$ and $f(a) \\in W$. The map $f$ restricted to $V$ is a bijection onto $W$. Its inverse $g \\colon W \\to V$ is of class $C^1$. So $f(V) = W$ is open.\n3. A function that is [[differentiable|0135XZZQ8M]] at a point $p$ is [[continuous|0S9EK5MQPE]] at $p$. Indeed, $\\norm{F(p + h) - F(p)} \\le \\norm{F(p + h) - F(p) - L h} + \\norm{L} \\norm{h}$ with $L = DF(p)$, and both terms tend to $0$ as $h \\to 0$.\n4. The maps $f$ and $g$ are of class $C^1$, so they are [[differentiable|0KXSK8TEEB]] at each point of their open domains. By step 3, $f$ restricted to $V$ and $g$ are continuous.\n5. So $f$ restricted to $V$ is a continuous bijection onto $f(V)$ with continuous inverse. That is, it is a [[homeomorphism|05GZ11PWRY]] onto the open set $f(V)$.\n6. Now let $O \\subseteq U$ be open, and let $b \\in f(O)$, say $b = f(a)$ with $a \\in O$.\n7. The restriction of $f$ to the open set $O$ is of class $C^1$ with invertible derivative at each point. By steps 1 to 5 applied to it, there is an open $V_a \\subseteq O$ with $a \\in V_a$ and $f(V_a)$ open.\n8. Then $b \\in f(V_a) \\subseteq f(O)$. So each point of $f(O)$ has an open neighbourhood inside $f(O)$, and $f(O)$ is open.\n9. Hence $f$ is an [[open map|0SQ2PC596Y]].\n\n## Idea\n\nThe inverse function theorem gives, around each point, an open set on which $f$ has a $C^1$ inverse. Differentiable maps are continuous, so this local inverse is continuous and $f$ is a homeomorphism there. An open set is a union of such neighbourhoods, so its image is a union of open sets.\n",
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