pervognsen/gist:11251717

## gistfile1.txt
Theorem (Smooth Invariance of Domain). Let f : R^n -> R^n be continuously differentiable and injective.
Then for every open set U, its image f(U) is an open set. In other words, f is an open mapping.

Proof. Let y_0 be a point in f(U) with y_0 = f(x_0) for some x_0 in U. We must show that we can fit
an open ball around y_0 that fits entirely within f(U).

Since U is open, there is some radius r > 0 such that the sphere S with center x_0 and radius r is
contained within U. Consider the function d(x) = |f(x) - y_0| defined on S. Because S is compact and
d is continuous, it attains its minimum m. As f(x_0) = y_0 and f is injective, it follows that d must
be strictly positive and hence m > 0. Then by construction we have |f(x) - f(x_0)| >= m for all x in S
so f(S) is strictly exterior to an open ball around y_0 of radius R < m.

If |y - y_0| < R then |f(x) - y| >= |f(x) - y_0| - |y - y_0| > m - R. For a given y, consider the
quadratic distance q(x) = |f(x) - y|^2 defined on the closed ball B with the sphere S as boundary. If
we choose R < m/2 then we're closer to y_0 than to f(S), so |y - y_0| < m/2 and |f(x) - y| > m/2, and
q(x_0) < (m/2)^2 at the center whereas q(x) > (m/2)^2 everywhere on the boundary. Hence h must assume
its minimum value at some point p in B's interior.

At this point, all that's left is basic calculus and linear algebra. The attainment of the minimum at the
interior point p means that 2 f'(p)^T (f(p) - y) = 0. An injective map has an injective derivative.
An injective linear map between spaces of the same dimension is surjective. Hence f'(p) is surjective.
The transpose of a linear map is injective if and only if the linear map is surjective. Hence f'(p)^T is
injective. The equation 2 f'(p)^T (f(p) - y) = 0 therefore lets us conclude f(p) = y.

This shows that every y such that |y - y_0| < R is the image of some point in B. Since B is a subset
of U, this means that the open ball centered at y_0 with radius R is contained within f(U).
	Theorem (Smooth Invariance of Domain). Let f : R^n -> R^n be continuously differentiable and injective.
	Then for every open set U, its image f(U) is an open set. In other words, f is an open mapping.

	Proof. Let y_0 be a point in f(U) with y_0 = f(x_0) for some x_0 in U. We must show that we can fit
	an open ball around y_0 that fits entirely within f(U).

	Since U is open, there is some radius r > 0 such that the sphere S with center x_0 and radius r is
	contained within U. Consider the function d(x) = \|f(x) - y_0\| defined on S. Because S is compact and
	d is continuous, it attains its minimum m. As f(x_0) = y_0 and f is injective, it follows that d must
	be strictly positive and hence m > 0. Then by construction we have \|f(x) - f(x_0)\| >= m for all x in S
	so f(S) is strictly exterior to an open ball around y_0 of radius R < m.

	If \|y - y_0\| < R then \|f(x) - y\| >= \|f(x) - y_0\| - \|y - y_0\| > m - R. For a given y, consider the
	quadratic distance q(x) = \|f(x) - y\|^2 defined on the closed ball B with the sphere S as boundary. If
	we choose R < m/2 then we're closer to y_0 than to f(S), so \|y - y_0\| < m/2 and \|f(x) - y\| > m/2, and
	q(x_0) < (m/2)^2 at the center whereas q(x) > (m/2)^2 everywhere on the boundary. Hence h must assume
	its minimum value at some point p in B's interior.

	At this point, all that's left is basic calculus and linear algebra. The attainment of the minimum at the
	interior point p means that 2 f'(p)^T (f(p) - y) = 0. An injective map has an injective derivative.
	An injective linear map between spaces of the same dimension is surjective. Hence f'(p) is surjective.
	The transpose of a linear map is injective if and only if the linear map is surjective. Hence f'(p)^T is
	injective. The equation 2 f'(p)^T (f(p) - y) = 0 therefore lets us conclude f(p) = y.

	This shows that every y such that \|y - y_0\| < R is the image of some point in B. Since B is a subset
	of U, this means that the open ball centered at y_0 with radius R is contained within f(U).