Questions about the quotient map and constructing a homeomorphism from a quotient space. I have two questions I am completely stuck on. The answers have been given but I don't understand them. I would greatly appreciate any help you could offer.
Question 1: Let $p : X \to Y$ be a continuous map with the property that there exists a continuous
map $f : Y \to X$ such that $p \circ f$ equals the identity map of $Y$. Show that $p$ is a
quotient map.
Answer 1: Let $U ⊂ Y$ be such that $p^{−1}
(U) ⊂ X$ is open. Since $p◦f = Id(Y)$ and $f$ is a continuous
map we have
$$U = (p ◦ f)^{-1}
(U) = f^{−1}
(p^{−1}(U)) ⊂ X$$
is open.
Question 2: Define an equivalence relation on the plane X = $\mathbb{R}^2$ as follows:
$(x_1, y_1) ∼ (x_2, y_2)$, if $x_1 + y_1^{2} = x_2 + y_2^{2}$.
Show that the quotient space $X^* = \mathbb{R}^2 / \sim$ is homeomorphic to $\mathbb{R}$.
Answer 2: Define $p : \mathbb{R}^2 → \mathbb{R}$ by $p(x, y) = x + y^2$. 
Define also $f : R → \mathbb{R}^2$ by $f(x) = (x, 0)$.
Then $p$ and $f$ are continuous and $p ◦ f(x) = x$ for any $x ∈ R$. 
Thus by question 1, p
is a quotient map, i.e. the quotient space $\mathbb{R}^2/ ∼$ is homeomorphic to $\mathbb{R}$.
 A: From the comments I see. You have some misconceptions about quotient maps.
$p: X \to Y$ is quotient iff $p$ is onto and 
$$\text{1.} \forall U \subseteq Y: p^{-1}[U] \text{ is open } \leftrightarrow U \text{ is open }$$
Condition 1. Is not about the inverse of $p$ but about the pre-image of $p$: $p^{-1}[U] = \{x \in X:p(x) \in U \}$ for $U \subseteq Y$. This essentially says that every subset of $Y$ that could be open, without making $p$ discontinuous , is in fact open, i.e. $Y$ has the largest topology that makes $p$ continuous.
Now to see your question 1: Suppose we have $p$ continuous (part of the definition of being quotient (the right to left implication of the equivalence in condition 1.) and we have another continuous $f: Y \to X$ with $p \circ f = 1_Y$, then $p$ is onto, as when we have $y \in Y$, $p(f(y)) = y$ and so $f(y) $ maps to $y$ under $p$. So all points of $Y$ are in the image of $p$.
Also we need condition 1. to hold. We already know  from continuity of $p$ that $U$ open implies $p^{-1}[U]$ open. So only need to show that any $U$ with $p^{-1}[U]$ open, has $U$ open as well. So let $U$ be such that $p^{-1}[U]$ is open.
Then indeed $$U = (1_Y)^{-1}[U] = (p \circ f)^{-1}[U] = f^{-1}[p^{-1}[U]]$$ where the first equality is trivial, we then substitute the condition and use the general formula for the pre-images of composed functions (which should be familiar from standard set theory). But then $U$ has been written as the inverse image under $f$ of the supposedly open set $p^{-1}[U]$, so $U$ is open, as required, using here the continuity of $f$.
