Guillemin's proof of manifold boundary dimension

Question

I'm having trouble understanding Guillemin's proof that given a k-dimensional manifold $X$ with a boundary, $\partial X$'s a $(k-1)$ dimensional manifold without boundary.

I understand up to the point that we need to show $\phi(\partial U)=\partial V$. However, I am having a hard time understanding why showing $\phi(\partial U) \supset \psi(\partial W)$ holds for a different local parameterization $\phi:W\to V$ would imply $\phi(\partial U) \supset \partial V$.

Answer 1

The idea is to reduce the property $\phi(\partial U)\supset\partial V$ on a manifold into an equivalent property on the base space $\mathbb{H}^k$ (where the boundry $\partial\mathbb{H}^k$ is easier to understand). For this, one uses the fact that $X$ is a manifold, i.e. the fact that is covered by charts that behave well under composition:

Since $\psi(\partial W)\subset \partial V$ holds for all charts $\psi$ from $W\subset\mathbb{H}^k$ into $V\subset X$, and because additionally for each $x\in V$ there exists at least one chart $\psi$ with $x\in\psi(W)$, the property

$\phi(\partial U)\supset\partial V$

can be understood as

$\phi(\partial U)\supset \psi(\partial W)$ for all charts $(\psi,W)$ with $\psi(W)\subset V$.

This is then translated into $\partial U \supset (\phi^{-1}\circ\psi)( \partial W)$, which can be proven using the property that a composition $\phi^{-1}\circ\psi$ of two charts is a homoemorphism from $\mathbb{H}^k$ to itself which preserves open sets.