Homeomorphism of two sets in $\Bbb{R}^2$ 
Prove or disprove: $$A=\{(x,y) \in \Bbb{R}^2: x+y \geq 0, xy=0 \}$$ is homeomorphic to $$B=\{(x,y) \in \Bbb{R}^2: x+y \geq 0, xy=1 \}$$

$x+y \geq 0$ represents the region above by the line $y=-x$ and together $xy=0$(the axes)
But $B$ represents the same region above by $y=-x$ together with the curve $xy=1$ .But $xy=1$ where $x>0,y>0$ is already covered in  $x+y \geq 0$ so the remaining portion(the other asymptote) is in the third quadrant
Therefore $B$ is not connected whereas $A$ is connected, so they are not homeomorphic
Am I right?
 A: *

*$A$ consists of the positive $x$-axis, positive $y$-axis, and the origin.

*$B$ consists of points on the first quadrant that falls on the line $y= \frac1x$.
Guide:  
To construct a bijection  continuous map from $B$ to $A$, 
$$g(x,y) = \begin{cases}  (0, y-x)&, y \ge x \\ (x-y,0)& x > y\end{cases}$$
Try to construct the corresponding inverse function.
Edit: To find the inverse function for $g$, 
Consider the point $(0,y')$, we want to find $(x,y)$ such that $g(x,y)=(0,y')$ where $xy=1$. Hence we have $y-x=y'$ and $xy=1$ which implies $$1-x^2=xy'.$$
We can then solve $x$ using the quadratic formula to obtain the inverse and then find the corresponding $y$.
Your proposed solution is more elegant.
A: $x+y \ge 0, xy = 0$ doesn't mean $x + y \ge 0$ or $xy = 0$ it means $x + y \ge 0$ and $xy = 0$. The set $A$ is the two positive parts of the $x$ and $y$ axes (an 'L' shape).
Similarly, the set $B$ is the portion of the hyperbola $xy = 1$ contained in the set $x + y \ge 0$. Namely, the portion in the first quadrant.
