Htykuut

I was reading Tu's Introduction to Smooth Manifolds and learned about the notion of manifolds with boundary. But there was a point which was not clear to me.

Here are the definitions(I will use the word smooth to mean that it is $C^{infty}$):

• Given two subsets $Ssubset mathbb{R}^n, Tsubset mathbb{R}^m$, we say that a map $f:Srightarrow T$ is smooth at a point $p$ if there is a neighborhood $U$ of $p$ in $mathbb {R}^n$ and a smooth map $tilde{f}:Urightarrow mathbb{R}^m$ that agrees with $f$ on $Ucap S$. If $f$ is smooth at every point in $S$, we say that $f$ is smooth on $S$. If $f$ is bijective, smooth on $S$, and has a smooth inverse, then $f$ is called a diffeomorphism.




• A topological space $M$ is said to be locally $mathcal{H}^n$ if every point in $M$ has a neighborhood that is homeomorphic to an open set of the upper half plane $mathcal{H}^n:={(x^{1},dots,x^{n})inmathbb{R}^{n}mid x^{n}geq0}$. If $M$ is alocally $mathcal{H}^n$, second countable, Hausdorff space, $M$ is called a topological $n$-manifold with boundary.




• A chart $(U,phi)$ on a topological $n$-manifold with boundary $M$ is a pair of an open set $U$ of $M$ and a homeomorphism $phi: Urightarrowphi (U)subset mathcal{H}^n$. A collection ${(U,phi)}$ of charts on $M$ is called a $C^infty $ atlas if the collection covers $M$ and for any two charts $(U,phi)$ and $(V,psi)$ in it, the transition map
  $$psi circ phi ^{-1}:phi (Ucap V)rightarrow psi (Ucap V)$$
  is a diffeomorphism. A $C^{infty}$ atlas $mathfrak{U}$ is said to be maximal if there is no other $C^{infty}$ atlas properly containing $mathfrak{U}$.




• A topological manifold with boundary together with a maximal $C^{infty}$ atlas is called a $C^infty$ manifold with boundary.

It is the last part that I am stuck with. For manifolds without boundary, we can prove that every atlas is contained in a unique maximal atlas. I figure that the situation is the same for manifolds with boundary, but I am having trouble proving it because of the somewhat complicated definition of smooth maps between two arbitrary subsets of Euclidean spaces. So my question is:

Given an atlas on a topological $n$-manifold with boundary, can we prove that the atlas is contained in a unique maximal atlas? If so, why?

The answer is positive. The maximal atlas containing an arbitrary atlas is obtained by adding all the charts that are compatible with the atlas. As pointed out by Gnampfissimo in the comment, the crux of the problem lies in showing that if two charts, say $(V_1,psi_1 )$ and $(V_2,psi_2 )$, are compatible with an atlas $mathfrak{U}$, then $(V_1,psi_1 )$ and $(V_2,psi_2 )$ are compatible with each other. So let us try to prove this.

Let $pin V_1cap V_2$. We would like to show that $psi_2circpsi_1^{-1}:psi_1(V_1cap V_2)rightarrowpsi_2(V_1cap V_2)$ is smooth at $psi_1(p)$. To prove this, choose a chart $(U,phi )$ in $mathfrak{U}$ that contains $p$. Then both $phicircpsi_1^{-1}:psi_1(Ucap V_1)rightarrowphi(Ucap V_1)$ and $psi_2circphi^{-1}:phi(Ucap V_2)rightarrow psi_2(Ucap V_2)$ are smooth. So there are

• neighborhoods $O_1$, $O_2$ of $psi_1(p)$ and $phi(p)$, respectively, in $mathbb{R}^n$,


• a smooth map $f:O_1rightarrowmathbb{R}^n$ that agrees with $phicircpsi_1^{-1}$ on $O_1cappsi_1(Ucap V_1)$, and


• a smooth map $g:O_2rightarrowmathbb{R}^n$ that agrees with $psi_2circphi^{-1}$ on $O_2capphi(Ucap V_2)$.

By taking the intersection of $O_1$ with $f^{-1}(O_2)$ if necessary, we may assume here that $O_1$ satisfies $f(O_1)subset O_2$. Then $gcirc f:O_1rightarrow mathbb{R}^n$ is a smooth map from a neighborhood of $psi_1(p)$ to $mathbb{R}^{n}$. So if we can show that $gcirc f$ agrees with $psi_2circpsi_1^{-1}$ on $O_1cappsi_1(V_1cap V_2)$, then we can say that $psi_2circpsi_1^{-1}:psi_1(V_1cap V_2)rightarrowpsi_2(V_1cap V_2)$ is smooth at $psi_1(p)$, and we are done.

For sure, $gcirc f$ agrees with $psi_2circpsi_1^{-1}$ on $O_1cappsi_1(Ucap V_1cap V_2)$. But what about on the set ${O_1cap psi_1(V_1cap V_2)}- {O_1cap psi_1(Ucap V_1cap V_2)}$? We have no information whatsoever about the behavior of $gcirc f$ on this set. What to do?

Here's a way to proceed: Since $psi_1(Ucap V_1 cap V_2)$ is open in $mathcal{H}^n$, there is an open set A in $mathbb{R}^n$ such that $Acap mathcal{H}^n=psi_1(Ucap V_1 cap V_2)$.
Then the three sets $(O_1 cap A) cap psi_1(U cap V_1)$, $(O_1 cap A) cap psi_1(V_1 cap V_2)$, and $(O_1 cap A) cap psi_1(Ucap V_1 cap V_2)$ are all equal.
So the replacement of $O_1$ by $O_1cap A$ fixes everything.