- Define an equivalence relation on the plane \(X = \mathbb{R}^2\) as follows:
\[ x_0 \times y_0 \sim x_1 \times y_1 \quad \text{if } x_0 + y_0^2 = x_1 + y_1^2. \]
Let \(X^*\) be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? [Hint: Set \(g(x \times y) = x + y^2\).]
- Repeat (a) for the equivalence relation
\[ x_0 \times y_0 \sim x_1 \times y_1 \quad \text{if } x_0^2 + y_0^2 = x_1^2 + y_1^2. \]
Solution
Define \[ g:\mathbb R^2\to\mathbb R,\qquad g(x,y)=x+y^2. \] Its fibers are exactly the equivalence classes. The map \[ H:\mathbb R^2\to\mathbb R^2,\qquad H(x,y)=(x+y^2,y) \] is a homeomorphism with inverse \(H^{-1}(u,v)=(u-v^2,v)\), and \[ g=\pi_1\circ H. \] Since the projection \(\pi_1\) is an open quotient map, so is \(g\). Therefore the induced bijection \[ \bar g:X^*\to\mathbb R \] is a homeomorphism. Thus \(X^*\cong\mathbb R\).
Now let \[ h(x,y)=x^2+y^2. \] Its fibers are the circles centered at the origin, with the origin as the radius-zero fiber, so the equivalence classes are exactly its fibers. The norm map \[ r(x,y)=\sqrt{x^2+y^2}:\mathbb R^2\to[0,\infty) \] is continuous, surjective, and open onto \([0,\infty)\): for \(z\) with \(\abs z=r_0\), the disk \(B_\varepsilon(z)\) contains the points \(z+tu\) with \(\abs t<\varepsilon\), where \(u=z/r_0\) if \(r_0>0\) and \(u\) is any unit vector if \(r_0=0\), and their norms \(\abs{r_0+t}\) cover \((r_0-\varepsilon,r_0+\varepsilon)\cap[0,\infty)\). Hence \(r\) is quotient. Squaring is a homeomorphism \([0,\infty)\to[0,\infty)\), so \(h=r^2\) is quotient. Consequently the induced map identifies the quotient with \[ [0,\infty). \]