"Defining a smooth structure on a topological manifold with boundary" 
This is a proposition for topological manifolds. However, I am wondering if this prop holds when $M$ is a topological manifold "with boundary." Thus, can the expression like "a smooth atlas on a topological manifold with boundary defines(determines) a smooth structure on it" be used based on (a) of this proposition? I have a suspicion that my guess is true. However I am not perfectly sure. Am I correct?
 A: The proof of the cited proposition should go something like this. Notice that it's very general. The argument I give here, due to the generality I use for smoothness of a map should apply word for word to manifolds as well as manifolds with boundary, and generalize naturally to other similar situations where a structure is defined by atlases of compatible charts.
We say two smooth atlases $A$ and $B$ are compatible if their union is a smooth atlas.
Note that this relation is trivially symmetric and reflexive, since union is symmetric and idempotent. Also note that if $A\subseteq B$, then $A\cup B = B$, so $A$ and $B$ are compatible. Hence all atlases containing $A$ are compatible with $A$. We'll use $A\sim B$ to mean that $A$ and $B$ are compatible.
To prove (a), we need that if $\{B_i\}$, $i\in I$ is a family of smooth atlases compatible with $A$, then $\bigcup_{i\in I} B_i$ is also a smooth atlas. Then the union of all atlases compatible to a given atlas $A$ is maximal among atlases compatible to $A$ (which includes all atlases containing $A$), and is in fact the unique maximal compatible atlas, since it contains all other compatible atlases. Let $M_A$ denote the smooth structure of $A$.
Then to prove (b), one just needs to check that compatibility is transitive. Then if $A$ and $B$ are compatible, $M_A\sim A\sim B \sim M_B$, so $M_A\sim B$, which implies $M_A\subseteq M_B$ and $M_B\sim A$, which implies $M_B\subseteq M_A$. Hence $M_B=M_A$. Conversely, if $M_A=M_B$, then $A\sim M_A=M_B\sim B$, so $A\sim B$.
I've broken the argument up like this, because the same general argument can be applied in any case where one defines a structure on an object by a family of compatible charts that cover the object to argue that there is a unique maximal chart for that structure.
Now we've reduced the argument to verifying two properties of atlases:


*

*If $A\sim B_i$ for all $i\in I$, then $A\sim \bigcup_{i\in I} B_i$, and

*if $A\sim B$ and $B\sim C$, then $A\sim C$.


Now $A$ and $B$ are compatible if and only if $A\cup B$ is a smooth atlas if and only if for all pairs of charts $\phi\in A$, $\psi\in B$, $\phi$ and $\psi$ are compatible charts, i.e. $\phi\circ\psi^{-1}$ and $\psi\circ\phi^{-1}$ are smooth maps. Property 1 follows immediately from the elementwise criterion for compatibility, since $A$ is compatible with $\bigcup_{i\in I} B_i$ if and only if every chart in $A$ is compatible with every chart in every $B_i$ if and only if $A$ is compatible with every $B_i$.
Thus property 1 holds in the case of manifolds or manifolds with boundary, or in more general situations where other kinds of atlases are defined as families of compatible charts. In particular (a) always holds in such situations. 
Thus one only needs to verify (b) by verifying property 2. This is the only part of the proof that depends on the definition of compatibility of charts. One needs to show if every chart of $A$ is compatible with every chart of $B$, and if every chart of $B$ is compatible with every chart of $C$, then every chart of $A$ is compatible with every chart of $C$. The proof goes as follows. 
First recall that if $X\subseteq\mathbb{R}^n$ is an arbitrary subset, then a map $\phi:X\to \mathbb{R}^m$ is smooth if for each point $x\in X$, there exists an open neighborhood $U$ of $x$ in $\mathbb{R}^n$ and a smooth map (in the usual sense) $\psi : U\to \mathbb{R}^m$ such that $\phi|_{U\cap X}=\psi|_{U\cap X}$. Then clearly restrictions of smooth maps are still smooth. Moreover, compositions of smooth functions are smooth, because if $\phi: X\to Y$, and $\psi : Y\to Z$, then if $x\in X$, there is an nhood $V$ of $\phi(x)$ on which $\psi$ can be extended to a smooth function, $\psi'$, and since $\phi$ is smooth, an nhood $U$ of $x$ such that $\phi(U)\subseteq V$, and on which $\phi$ can be extended to a smooth function, $\phi'$. Restricting $U$ further if necessary, so that $\phi'(U)\subseteq V$, we then have that $\psi'\circ\phi'$ is a smooth extension of $\psi\circ \phi$ to an nhood of $x$. Also since smoothness is a local condition, the gluing of smooth maps is smooth.
Let $(U,\phi)\in A$, $(V,\psi)\in B$, $(W,\tau)\in C$ where $U$, $V$, and $W$ are the open subsets of $M$ that are the domains of $\phi$, $\psi$, and $\tau$ respectively.
Then $\phi\circ \psi^{-1} : \psi(U\cap V) \to \phi(U\cap V)$ is smooth, and $\psi\circ\tau^{-1} : \tau(W\cap V) \to \psi(W\cap V)$ is smooth, so
$\phi\circ\psi^{-1}\circ\psi\circ\tau^{-1}=\phi\circ\tau^{-1}|_{\tau(W\cap V\cap U)} : \tau(W\cap V\cap U) \to \phi(W\cap V\cap U)$ is smooth, since compositions of smooth maps are smooth. Then as $\psi$ varies over all possible charts in $B$, since the corresponding domains $V$ cover $M$, $W\cap V\cap U$ covers $W\cap U$, and $\tau(W\cap V\cap U)$ covers $\tau(W\cap U)$. Since the map $\phi\circ\tau^{-1}$ is locally smooth on each $\tau(W\cap V\cap U)$, it's therefore smooth on all of $W\cap U$. By symmetry, $\tau\circ\phi^{-1}$ is also smooth. Hence $\phi$ and $\tau$ are compatible. Since $\phi$ and $\tau$ were arbitrary, $A\sim C$.
