Criterion for $M$ being diffeomorphic to $\mathbb{R}^n$ I have two question concerning the proof of the following criterion:
Let $M$ be a paracompact manifold such that every compact subset is contained in an open set diffeomorphic to $\mathbb{R}^n$. Then $M$ itself is diffeomorphic to $\mathbb{R}^n$.
Proof: It is not difficult to show that $M$ is a monotone union of disks. That is, we can find submanifolds with boundary $W_1\subset W_2\subset W_3\subset...\subset M$ with union $M$ so that each $W_i$ is diffeomorphic to $D^n$ and so that each $W_i$ is contained in the interior of $W_{i+1}$. We wish to compare this sequence with the sequence $D_1^n\subset D_2^n\subset D_3^n\subset...\subset \mathbb{R}^n$ where $D_i^n$ denotes the disk of radius $i$ in $\mathbb{R}^n$. Start with any diffeomorphism $f_1:D_1^n\rightarrow W_1$. This can be extended to a diffeomorphism $f_2:D_2^n\rightarrow W_2$ and so on using a Lemma of Palais and Cerf. Finally, piecing together all these diffeomorphism $f_i$, we obtain the required diffeomorphism $\mathbb{R}^n\rightarrow M$.
My questions:


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*How do we see that $M$ is a monotone union of disks? My idea is to consider an exhaustion of $M$ by compact sets $K_i$, i.e. each $K_i$ is compact such that $\bigcup _i K_i=M$ and each $K_i$ is contained in the interior of $K_{i+1}$. Then
by assumption each $K_i$ is contained in a set $B_i$ which is diffeomorphic to an open disc in $\mathbb{R}^n$. My idea is to set $W_i:=\bar{B_i}$ and then the sets $W_i$ form the required exhaustion. But I'm not sure if $W_i\subset \mathring{W_i}$ holds for all $i$.

*What does "piecing together all these diffeomorphism" exactly mean? How can I write down the resulting diffeomorphism $\mathbb{R}^n\rightarrow M$ concretly ?
 A: 1). Let us construct the sequence $(W_i)_{i \ge 0}$ inductively, basing our construction on the existence of a compact exhaustion $\bigcup _{i \ge 0} K_i = M$.
Since $K_0$ is compact, there exists an open subset $K_0 \subset U_0 \subseteq M$. If $U_0 = M$ then $M$ is diffeomorphic to $\Bbb R^n$, and the criterion is proved. If $U_0 \ne M$, then let $h_0 : U_0 \to \Bbb R^n$ be the diffeomorphism, let $L_0 = h_0 (K_0)$ (compact, therefore bounded), pick some open ball $B_0 \subset \Bbb R^n$ around $L_0$ and let $W_0 = h_0 ^{-1}(\bar B_0)$. Note that since $\bar B_0$ is compact in $\Bbb R^n$ and $h_0 ^{-1}$ a homeomorphism, $W_0$ will be compact in $U_0$, therefore in $M$. It is easy to see that $W_0$ is a submanifold with boundary, with $\partial W_0 = h^{-1}(\partial \bar B_0)$. It is also clear that $K_0 \subset W_0$.
Assume $W_i \supset K_i$ constructed (as a compact submanifold with boundary) as above and let us construct $W_{i+1}$. Consider the $U_i$ from the construction of $W_i$: since we have constructed $W_i$, it means that $U_i \ne M$ (otherwise we would have stopped at $W_{i-1}$); pick then some $x_{i+1} \notin U_i$. The subset $W_i \cup K_{i+1} \cup \{x_{i+1}\} \subseteq M$ is compact, therefore there exists an open subset $U_{i+1} \supseteq W_i \cup K_{i+1} \cup \{x_{i+1}\}$ diffeomorphic to $\Bbb R^n$. Notice that $U_{i+1} \supsetneq U_i$ precisely because of that $x_{i+1}$. Let $h_{i+1} : U_{i+1} \to \Bbb R^n$ be this diffeomorphism. Since $L_{i+1} = h_{i+1} (W_i \cup K_{i+1} \cup \{x_{i+1}\}) \subset \Bbb R^n$ is compact (as the image of a compact through a homeomorphism), it will be bounded, so you may enclose it in an open ball $B_{i+1}$; let then $W_{i+1} = h_{i+1} ^{-1} (\bar B_{i+1})$. Then $W_{i+1}$ will be a compact submanifold with boundary and $K_{i+1} \subset W_{i+1}$.
It is easy to show that, under this inductive recipe, $W_i \subset \mathring W_{i+1}$. On the other hand, since $K_i \subset W_i$, it follows that $\bigcup _{i \ge 0} W_i \supseteq \bigcup _{i \ge 0} K_i = M$, so $\bigcup _{i \ge 0} W_i = M$. By construction, all the $W_i$ are diffeomorphic to closed balls in $\Bbb R^n$.
2). The essential (and difficult) fact is that $f_{i+1}$ extends $f_i$. If $x \in M$, since $\bigcup _{i \ge 0} W_i = M$ it follows that there exists $W_i$ such that $x \in W_i$. Define $f : M \to \Bbb R^n$ by $f(x) = f_i (x)$. What if there exists $W_j$ with $x \in W_j$ (and $i < j$)? Well, since $f_{i+1}$ extends $f_i$, and $f_{i+2}$ extends $f_{i+1}$, and so on until $f_j$ extends $f_{j-1}$, it follows that $f_j$ extends $f_i$, i.e. $f_j \Big| _{W_i} = f_i$, which implies that $f_j (x) = f_i(x)$, so $f(x)$ does not depend on the index $i$ and is therefore well defined.
