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Could you help me to construct a mapping from a triangle onto a cone?

Let $X$ be a triangle: $\{(x,y): |x| \le y \le 1\}$ with the subspace topology in $\mathbb{R}^2$.

Let $Y$ be a cone: $\{(x_1, x_2, x_3): x_1^2 + x_2^2 = x_3^2 \le 1\}$ with the subspace topology in $\mathbb{R}^3$.

Construct a quotient mapping from $X$ onto this $Y$.

Thank you.

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1 Answer 1

The basic idea is to roll the triangle into a cone, though a little stretching is also involved. The top edge of the triangle (the intersection of $X$ with the line $y=1$) will become the circle at the top of the cone (the intersection of $Y$ with the plane $z=1$). More generally, each horizontal 'slice' through $X$, (i.e., each intersection of $X$ with a line $y=c$ for some $c$ with $0\le c\le 1$), will wrap around to become the corresponding circular horizontal 'slice' through $Y$ (i.e., the intersection of $Y$ with the plane $z=c$).

For $0\le c\le 1$ let $X_c=\{\langle x,y\rangle\in X:y=c\}$ and $Y_c=\{\langle x,y,z\rangle\in Y:z=c\}$; we want to map each $X_c$ to the corresponding $Y_c$. In doing this we'll obviously have to identity the endpoints of $X_c$, wrapping the line segment into a circle of radius $c$. The $x$-coordinates of points in $X_c$ range from $-c$ to $c$. One natural way to wrap the interval $[-c,c]$ into a circle of radius $c$ is to multiply it by $\frac{\pi}c$ (for $c>0$, of course!) and think of it as the $\theta$-coordinate in polar coordinates; the corresponding $x$- and $y$-coordinates are then $c\cos\frac{\pi}c$ and $c\sin\frac{\pi}c$, respectively. The resulting map is then

$$f:X\to Y:\langle x,y\rangle\mapsto\begin{cases} \langle 0,0,0\rangle,&\text{if }y=0\\\\ \left\langle y\cos\frac{\pi}y,y\sin\frac{\pi}y,y\right\rangle,&\text{otherwise}\;. \end{cases}$$

I'll leave it to you to check that this $f$ really is a quotient map.

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Why do it this way, rather than identifying the two edges of the triangle that are the same length, and making the hypotenuse the circular boundary? –  MJD May 14 '12 at 3:33
    
@Mark: That’s what I did. The hypotenuse is $\{\langle x,1\rangle:-1\le x\le 1\}$. –  Brian M. Scott May 14 '12 at 3:35
    
Thanks. I forgot about the part of $X$ where $x\le0$! –  MJD May 14 '12 at 3:38
    
@Mark: Phew! You had me worried for a minute. :-) –  Brian M. Scott May 14 '12 at 3:39

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