Which of the following subsets of $\Bbb R^2$ are homeomorphic to the set $\{(x, y) \in \Bbb R^2 \mid xy = 1\}$? 
Which of the following subsets of $\Bbb R^2$ are homeomorphic to the set $\{(x, y) \in \Bbb R^2 \mid xy = 1\}$?
$a. \{(x, y) ∈ \Bbb R^2 \mid xy − 2x − y + 2 = 0\}.$
$b. \{(x, y) ∈ \Bbb R^2 \mid x^2 − 3x + 2 = 0\}.$
$c. \{(x, y) ∈ \Bbb R^2 \mid  2x^2 − 2xy + 2y^2 = 1\}.$

$a.$ is only a single point $(2,2)$ and thus not homeomorphic to the set $\{(x, y) \in \Bbb R^2 \mid xy = 1\}$.
$b.$ is lines $x =1$ and $x = 2$ and thus not homeomorphic to the set $\{(x, y) \in \Bbb R^2 \mid xy = 1\}$.
$c.$ is an ellispe. Ellispe is a closed curve while rectangular hyperbola is not. Thus it is not homeomorphic to the set $\{(x, y) \in \Bbb R^2 \mid xy = 1\}$.
Is the logic above is correct?
 A: You’ve incorrectly identified the first set. Note that
$$xy-2x-y+2=x(y-2)-(y-2)=(x-1)(y-2)\;,$$
which is $0$ when $x=1$ or $y=2$, so the first set is actually the union of the vertical line $x=1$ and the horizontal line $y=2$. It’s true that this is not homeomorphic to the hyperbola $xy=1$, but you need a different reason. Here are two that work:


*

*This set is connected, and the hyperbola is not.  

*If you remove the point $\langle 1,2\rangle$ from this set, what’s left has four components; if you remove any single point from the hyperbola, what’s left has only three components.


You’ve correctly identified the second set, but it is homeomorphic to the hyperbola: the line $x=1$ is homeomorphic to the piece of the hyperbola in the first quadrant, say, and the line $x=2$ is homeomorphic to the piece of the hyperbola in the third quadrant. You should try to work out a homeomorphism.
It’s true that an ellipse is not homeomorphic to a hyperbola, but your reason isn’t convincing. Here again you can use the fact that one is connected and the other is not; you can also use the fact that one is compact and the other is not.
A: (Because errors in this post, I allow myself to be something "teaching". Sorry if some do not like this).
Let us denote the hyperbola $H$ and $S_a,S_b,S_c$ the three other sets; $P=(1,1)\in H$, $Q=(-1,-1)\in H$, $\mathcal F$ be the first quadrant and  $\mathcal T$ be the third quadrant.
►For $S_a$ you have $(x-1)(y-2)=0$ which, as a subset of $\mathbb R^2$ , is the union of the two straight lines $x=1$ and $y=2$. These lines intersect at the point $(1,2)$ so there cannot be a bijection between $S_a$ and $H$ (why?)  therefore  there cannot be a homeomorphism. (This is Important to understand what happen with $S_b$).
► No way to get a bicontinuous bijection between the closed curve $S_c$ (ellipse) and  a set of two disjoint arcs in $\mathbb R^2$  so non homeomorphism between $H$ and $S_c$ (otherwise $S_c$ is compact and conexe but $H$ is neither compact nor conexe, so there are not continuous function from $S_c$ to $H$).
► $S_b=\{(x, y) ∈ \Bbb R^2 \mid x^2 − 3x + 2 = 0\}=\{(x, y) ∈ \Bbb R^2 \mid (x-1)(x-2) = 0\}$ is the union of the two straight lines $x=1$ and $x=2$ which being parallel not intersect so you can put in bijection one of them with the arc of $H$ in the first quadrant and the other with the arc of $H$ in the third quadrant so a necessary condition is satisfied.
NOTE.- That $ S_b $ is homeomorphic to $H$ would have been very clear (by geometrical intuition) in the early days of topology, those of Analysis Situs (Poincaré) or "elastic geometry" or “deformations”.
HINT FOR A HOMEOMORPHISM.-  Let $F$ be the function from $H$ to $S_b$ in the figure below defined by
$F((1,1))=(2,2)$, $F((-1,-1))=(1,1)$ and
$$F((x,y))=\begin{cases}(2,2+l) \text{ where $l$ is the length of the arc }  \widehat{ZP},(x,y)\in\mathcal F\\(1,1+l)\text{ where $l$ is the length of the arc } \widehat{WQ},(x,y)\in\mathcal T\end{cases}$$
(One has for the length $l$ of the hyperbola, $$l=\int_Z^P\frac{\sqrt{x^4+1}}{x^2}dx$$ so we must keep in mind the change of sign of the integral according to the position of the point $Z$ with respect to $P$. On the other hand, this integral is not elementary (according to Wolfram, it depends on elliptic integrals of both first and second kind) so it is not matter of calculation here).
You can show that $F$ is a homeomorphism between $H$ and $S_b$ (the bicontinuity of the bijection is, besides the geometrical intuition which could be enough here, provable without explicit calculation of the integral).
(I apologize for a non improbable mistake and bad English).

