How do I prove there is a homeomorphism between these two topological spaces?

Question

Let $ X=\{(x,y) \in \mathbb{R}^2 | |x| \leq 1 \} $with the induced euclidean topology $\tau$ and let $R$ be the equivalence relation $(x',y')R(x,y)$ if and only if $ y=y'=0$ or $(x',y')=(x,y)$. Tell if the quotient space$(X/R,\tau/R)$ is homeomorphic to $Y_k =\{(x,y) \in \mathbb{R}^2 | x^2-y^2 \leq k \}$ for $k=0$ and $k=1$

Can someone show how to do it?. I kow I have to use saturated sets somehow

This is the solution I was given, but there are many things I don't get.

The function $f:X\rightarrow Y_0 $ defined by $f((x,y))=(xy,y)$is totally compatible, surjective and continuous. Besides it maps saturated open sets in open sets: In fact, an open set A is saturated iff it does not intersect $(-1,1)X\{O\}$ or if it contains it. In the first case f is a homeomorphism of $ A$ with $f(A)$, in the second case there exist an $\epsilon$ such that $Q=(-1 ,1) X (-\epsilon , -\epsilon) \subseteq A $. Now $f(Q)$ is an open set containing $O$ and $f(A)$ is open and $f_R$ is an homeomorphism. All of this implies $(X/R-\{[O]\},\tau/R)$ is not connected, while $Y_1$ remains connected if one removes one point.

*note : I couldn't found an english traslation for the concept of "totally compatible function". I am doing a literal traslation here, but the definition is the following:

Let f be a function $f:X \rightarrow Y$, and $R$ an equivalence relation. f is said to be compatible with $R$ if $xRy$ implies $f(x)=f(y)$; $f$ is said totally compatible if the implication is in both ways. This allow to define a function $f_R: X/R \rightarrow Y$, so that $f=f_R o \Pi $, dove $\Pi$ is the projection to the quotient space $\Pi: X \rightarrow X/R$,

Then there is a theorem which is what they used in the proof: Let $f: X \rightarrow Y$ be compatible with $R$, then

1) $f_R$ is surjective iff $f$ is surjective

2)$f_R$ is injective iff $f$ is totally compatible

3)$f:(X,\tau)\rightarrow (Y,\tau')$ is continuous iff $f_R$ is continuous

4)f maps saturated open sets to open sets iff $f_R$ is an open function

what I don't understand in the proof:

1) How on earth could I have come up with the expresion for the function that maps $X$ to $Y_0$? I drew the sets, but still no clue

2) How do you prove f is surjective?

3) why is f(A) open? why are thet saying f is homeomorphism of A with f(A), we still need to prove that f is open

4) why does this imply $(X/R-\{[O]\},\tau/R)$ is not connected and why are they removing the class of the origin? by removing a point aren't I changing all of the previous conclusions?

5) The last sentence of the proof refers to the case k=1, but all of the previous work was done for k=0, why are they assuming it is still valid or how do I prove it doing something similar for k=1? with k=1 $Y_1$ is now the region between an hyperbole and the function f defined at the beginning is not valid anymore

Answer 1

Hint: the quotient of a compact space is compact.

However, the sequence $a_n := (n,n)$ for each $n \in \mathbb{N}$ is a subset of both $Y_0$ and $Y_1$, and so these spaces are not compact (here we are using that a subspace of $\mathbb{R}^n$ equipped with the metric topology is compact iff it is closed and bounded).