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4a.)  Claim: $g:  R^2 \rightarrow R$, $g(x, y) = x + y^2$ is a 
quotient map.

If $x \in R$, $g({x}, 0) = x$.  Hence $g$ is onto.

Since $g(x, y) = \pi_1(x, y) + \pi_2(x, y)^2$ and $\pi_i$ are continuous, 
products and sums of continuous functions are continuous, $g$ is 
continuous.



Suppose $g^{-1}(U)$ is open in $R^2$.  We need to show $U$ is open in $R$.  
Take $z \in U$.  Then $({z}, 0) \in g^{-1}(z)$ and
there exists $(a, b) \times (c, d)$ such that $({z}, 0) \in (a, b)
\times (c, d)  \subset g^{-1}(U)$. 

Claim:  $z \in (a, b) \subset U$.

$({z}, 0) \in (a, b)     
\times (c, d) $ implies  $z \in (a, b)$

Suppose $w \in (a, b)$. Then $a < {w} < b$
Thus, $({w}, 0) \in (a, b) \times (c, d) \subset g^{-1}(U)$

Thus $w = g(({w}, 0)) \in g(g^{-1}(U)) = U$. Hence $(a, b) \subset U$.

Hence $U$ is open in R.  Thus $g$ is a quotient map.

${R^2/\sim} = \{g^{-1}\{z\}) ~|~ z \in R\}$ where 
${R^2/\sim}$ has the quotient topology from $p: R^2 \rightarrow 
{R^2/\sim}$.  Thus $g$ induces a bijective map $f: {R^2/\sim} 
\rightarrow R$ which is a homeomorphism since $g$ is a quotient 
map (cor 22.3).

Thus ${R^2/\sim}$ is homeomorphic to $R$ with standard topology.

4b.)  Claim: $g:  R^2 \rightarrow [0, \infty)$, $g(x, y) = x^2 + y^2$ is a 
quotient map.

If $x \in [0, \infty)$, $g(\sqrt{x}, 0) = x$.  Hence $g$ is onto.

Since $g(x, y) = \pi_1(x, y)^2 + \pi_2(x, y)^2$ and $\pi_i$ are 
continuous, 
products and sums of continuous functions are continuous, $g$ is 
continuous.


Suppose $g^{-1}(U)$ is open in $R^2$.  We need to show $U$ is open in $[0,
\infty)$.  Take $z \in U$.  Then $(\sqrt{z}, 0) \in g^{-1}(z)$ and
there exists $(a, b) \times (c, d)$ such that $(\sqrt{z}, 0) \in (a, b)
\times (c, d)  \subset g^{-1}(U)$.  Let $e = max\{0, a\}$ 

Claim:  $(e^2, b^2) \subset U$.


Suppose $w \in (e^2, b^2)$. Then since $e \geq 0$, $e < \sqrt{w} < b$.
Thus, $(\sqrt{w}, 0) \in (e, b) \times (c, d) \subset (a, b)
\times (c, d)  \subset g^{-1}(U)$

Thus $w = g((\sqrt{w}, 0)) \in g(g^{-1}(U)) = U$. Hence $(e^2, b^2) 
\subset U$.

If $z = 0$, then $a < \sqrt{z} = 0$ and hence $e = max\{0, a\} = 0$ .  
Since $z \in 
U$, $z \in [0, b^2) = [e^2, b^2) \subset U$ where $[0, b^2)$ is open in 
$[0, \infty)$.

If $z > 0$,  $\sqrt{z} \in (e, b) \subset (a, b)$.
$e < \sqrt{z} < b$ implies  $e^2 < z < b^2$ since $e \geq 0$.  Thus $z \in 
(e^2, b^2)$.  Hence $z \in (e^2, b^2) 
\subset U$ where $(e^2, b^2)$ is open in   
$[0, \infty)$.


Hence $U$ is open in $[0, \infty)$.  Thus $g$ is a quotient map.

${R^2/\sim} = \{g^{-1}\{z\}) ~|~ z \in [0 \infty)\}$ where 
${R^2/\sim}$ has the quotient topology from $p: R^2 \rightarrow 
{R^2/\sim}$.  Thus $g$ induces a bijective map $f: {R^2/\sim} 
\rightarrow [0, \infty)$ which is a homeomorphism since $g$ is a quotient 
map (cor 22.3).

Thus ${R^2/\sim}$ is homeomorphic to $[0, \infty)$.

\end