Is there any diffeomorphism from A to B that $f(A)=B$?

$A = \{(x,y) \in \mathbb{R}^2 : x \ge 0 \wedge y \ge 0\}$

$B = \mathbb{R}^2 \setminus \{(x,y) \in \mathbb{R}^2 : x>0 \wedge y>0 \}$

Is there any diffeomorphism $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$that $f(A)=B$?

My thoughts :

We need to “stretch” A set on three quadrants and then rotate it for $\frac{\pi}{2}$ angle. That “stretching” looks like complex numbers argument multiplication.

But, at the beginning of coordinate system derivative of this function is $0$.

Solutions Collecting From Web of "Is there any diffeomorphism from A to B that $f(A)=B$?"
Such a diffeo can’t exist. There is an obstruction at the corner (the origin). Imagine there were such a thing $f:{\mathbb R}^2\to{\mathbb R}^2$. Let $\alpha$ denote the half-axis $x\ge0,y=0$ and $\beta$ the half-axis $y\ge0,x=0$. Their union $Z$ is the boundary of both sets $A$ and $B$, hence $f(Z)=Z$. Since clearly the origin $0$ is the only point were $Z$ is not a smooth curve, the diffeomorphism fixes $0$. The connected components of $Z\setminus\{0\}$ are $\alpha\setminus\{0\}$ and $\beta\setminus\{0\}$, hence $f$ leaves them invariant or maps one onto the other. The argument is similar in both cases, hence suppose $f(\alpha)=\alpha$.
Now consider the regular curve $\gamma(t)=f(t,0)$. Note that $\gamma'(0)=\frac{\partial f}{\partial x}(0,0)$ is a not zero vector because $f$ is diffeo. The image of $\gamma$ has two pieces: $\alpha=\gamma\{t\ge0\}$ and $\sigma=\gamma\{t<0\}$, and they form a regular curve because $f$ is a diffeo. Note that $\sigma$ must be contained in the quadrant $x>0,y>0$ because $f({\mathbb R}^2\setminus A)={\mathbb R}^2\setminus B$ ($f$ is bijective). Now consider the two limits
$$\gamma'(0)=\lim_{t\to 0^-}\frac{f(0,0)-f(t,0)}{-t},\quad \gamma'(0)=\lim_{t\to 0^+}\frac{f(t,0)-f(0,0)}{t}$$
It’s not difficult to realise that the first limit must point into the third quadrant, and the second must be a vector $(a,0), a\ge0$. Thus we conclude $\gamma'(0)=(0,0)$, a contradiction.