Are $\{ x\geq 0, y\geq 0\}\subset \mathbb R^2$ and $\{x\leq 0 \text { or } y\leq 0\}\subset \mathbb R^2$ diffeomorphic? Consider the first quadrant $Q = [0,+\infty)\times [0,+\infty)\subset \mathbb R^2$ and the complement $X = \mathbb R^2\setminus \mathrm{int}(Q) = \{(x,y) \ |  x\leq 0\  \mathrm{ or } \ y\leq 0\}$.
Recall that a map $f:C\to \mathbb R^r$ defined on a closed subset  $C\subset \mathbb R^n$ is smooth if for any $x \in C$, exists  an open subset $U\subset \mathbb R^n$  and a smooth function $F:U\to \mathbb R^r$ such that $F|_{C\cap U} \equiv f|_{C\cap U}$.

Question: are $Q$ and $X$ diffeomorphic?

Clearly the two are homeomorphic, the problem is the differentiability.
This question is motivated by my other question Is the complement of a quadrant a manifold with corners?
 A: They are not diffeomorphic. The manifold $X$ satisfies the following property:

For every $x \in X$, there exists a smooth path in $X$ with nonvanishing derivative that passes through $x$.

For example, there is a smooth path in $X$ with nonvanishing derivative that passes through $x=(0,0)$, namely $\gamma(t)=(-t,t)$.
This property is invariant under diffeomorphism: if $X \subset \mathbb R^n$ is diffeomorphic to $Q \subset \mathbb R^m$, and if $X$ satisfies the above property, then so does $Q$.
However, $Q$ does not satisfy that property: for $q=(0,0)$ there does not exist any smooth path in $Q$ with nonvanishing derivative that passes through $q$. This can be proved using the local immersion theorem.
A: Edit: it seems my argument fails at the origin, and so serves only to prove a diffeomorphism between $X-\{0\}$ and $Q-\{0\}$. In the comments OP links to an answer to a different question which claims that a diffeomorphism between $X$ and $Q$ is impossible.
View $\mathbb{R}^2$ as the complex plane $\mathbb{C}$, then the complex function $f(z)=z^3$ maps $Q$ to the set $Y=\{x+i y\mid x\le0\;\textrm{or}\;y\ge 0\}$. If we then multiply by $i$ we rotate $Y$, transforming it into $X$. Therefore $g(z)=iz^3$ is a holomorphic function which maps $Q$ to $X$, which should be more than sufficient for proving a diffeomorphism.
