On the euclidean plane $\mathbb{R}^2$ we define an equivalence relation by $(x,y) \sim (x',y')$ iff $x^2 -y = x'^2 - y'$. In other words, I consider the set $$ \mathbb{R}/{\sim} := \{ [(x,y)] \mid (x,y) \in \mathbb{R}^2, x^2 - y = c \}. $$ I need to draw a picture of this quotient set, because I have difficulties of getting a proper intuition how it might look. Then I have to find a subset $V \subseteq \mathbb{R}$, such that $\mathbb{R}^2/{\sim} \cong V$, as topological spaces.
The first of my problems is to understand better the quotient set. I think I can elaborate the rest, using the universal property for quotients. I kindly appreciate explanations and elaborations on this problem. Thank you in advance!
I find it more natural to map the quotient space to the more familiar space to establish a homeomorphism (the is the inverse of what's suggested in the comments). Begin by constructing a map $f: \mathbb{R}^2 \to \mathbb{R}$ by $f(x, y) = x^2 - y$. Now, verify that the preimage of each point is $c \in \mathbb{R}$ contains the equivalence class $$ [(0, c)] = \{ (x, y) \in \mathbb{R}^2 \mid x^2 - y = c \}. $$ Thus, $f$ factors through the quotient map $p: \mathbb{R}^2 \to \mathbb{R}^2/{\sim}$. In other words, $f = \bar{f} \circ p$, where $\bar{f}: \mathbb{R}^2/{\sim} \to \mathbb{R}$ sending $[(x, y)] \mapsto y$ is well-defined. $$\require{AMScd} \begin{CD} \mathbb{R}^2 @>f>> \mathbb{R} \\ @V {p\,} V V @| \\ \mathbb{R}^2/{\sim} @>\smash[t]{\bar{f}}>> \mathbb{R} \end{CD} $$ In fact, for each $c \in \mathbb{R}$, we have $f^{-1}(c) = [(0, c)]$ on the nose, not just containment, so $\bar{f}$ is injective. Surjectivity is pretty straightforward. So, this map is a (continuous) bijection.
You have to convince yourself that its inverse is continuous, as well, but that's not too bad, considering the definition of quotient topology.