Homeomorphism between Three sets 
Which of the following are homeomorphic to the set
$$ \{(x,y) \in \mathbb{R}^2: xy=1\}$$
a) $ \{(x,y) \in \mathbb{R}^2: xy=0\}$
b) $ \{(x,y) \in \mathbb{R}^2: x^2-y^2=1\}$
c) $ \{(x,y) \in \mathbb{R}^2: x^2+y^2=1\}$

My attempt:
Given set is the hyperbola with two asymptotes.

Here also a) is the set with two axes:

Now it is connected whereas the given set is not, so a) is false
b) is the set like this:

I don't know whether it is homeomorphic or not ?
Finally, c) is our $S^1$:

It is compact and connected but the given set is not, so c) is false.
I need help to prove/disprove b).
Any help?
 A: You are correct for a), c) by using the property that homeomorphisms preserve connectedness.
For b) denote the sets $A = \{(x,y) \in \mathbb R^2: xy = 1\}$ and $B = \{(x,y) \in \mathbb R^2:x^2-y^2=1\}$. Now consider the functions $f:A \rightarrow B,\, g:B\rightarrow A$ given by $$f(x,y)=\bigg(\frac{x+y}{2},\frac{x-y}{2}\bigg),\, g(x,y)=(x+y,x-y)$$
Checking if these functions map to the correct spaces we have for $(x,y) \in A$,
$f(x,y)=\big(\frac{x+y}{2},\frac{x-y}{2}\big) =(z,w)$ we have:
$$z^2-w^2=\bigg(\frac{x+y}{2}\bigg)^2-\bigg(\frac{x-y}{2}\bigg)^2=\frac{x^2}{4}+\frac{xy}{2}+\frac{y^2}{4}-\bigg(\frac{x^2}{4}-\frac{xy}{2}+\frac{y^2}{4}\bigg)=xy=1 $$
Likewise for $(x,y)\in B$, $g(x,y)=(x+y,x-y)=(z,w)$ we have:
$$zw=(x+y)(x-y)=x^2-y^2=1 $$
So $f,g$ do map to the correct subspaces of $\mathbb R^2$ and the continuity of $f,g$ is clear as each is of the form $(P(x,y),Q(x,y))$ where $P,Q$ are polynomials in terms of $x,y$. Now if $f \circ g(x,y)=(x,y)$ and $g \circ f(x,y) =(x,y)$ we can conclude $f$ is a homeomorphism. We see:
$$ f\circ g(x,y)=f(x+y,x-y)=\bigg (\frac{(x+y)+(x-y)}{2},\frac{(x+y)-(x-y)}{2}\bigg)=(x,y)$$
Likewise:
$$ g \circ f(x,y)=g\bigg(\frac{x+y}{2},\frac{x-y}{2}\bigg)=\bigg(\frac{x+y}{2}+\frac{x-y}{2},\frac{x+y}{2}-\frac{x-y}{2} \bigg)=(x,y)$$
Thus we conclude the $f$ is a homeomorphism and $A \simeq B$.
