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Homeomorphisms problems

Megus

New member
Oct 18, 2012
4
Let $H=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1+z^2\},$ and $C=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1\}.$ Prove that $H$ and $C$ as subspaces of $\mathbb R^3$ are homeomorph.

How to solve this analytically? I've seen a geometric solution but I don't see how to work it analytically.

Let $(M_1,d_1),\ldots,(M_n,d_n),$ and $M=M_1\times\cdots\times M_n.$ Prove the following:

$1_M: (M,d_e)\to(M,d_m)$ and $1_M: (M,d_e)\to(M,d_s)$ are homeomorphisms, where $d_e$ is the euclidian distance, $d_s=|x_1-y_1|+\cdots+|x_n-y_n|,$ and $d_m$ distance of the maximum.
$1_M$ is supposed to be the identity function, but I'm confused how to work with.

Let $f: (M_1,d_1)\to(M_2,d_2)$ and $G(f )=\{(x,f(x))|x\in M_1\}\subseteq M_1\times M_2$ and let $\tilde f: (M_1,d_1)\to(G(f ),d_p).$ Consider $\tilde f(x)=(x,f(x))$ and $d_p((x_1,x_2),(y_1,y_2))=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}$ for all $(x_1,y_1),(x_2,y_2)\in M_1\times M_2.$ Prove that $\tilde f$ is a homeomorphism.

In order to be a homeomorphism we need bijectivity for $\tilde f$ and continuity for $\tilde f$ and $\tilde f^{-1},$ is there a faster way to solve this?

Thanks!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Let $H=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1+z^2\},$ and $C=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1\}.$ Prove that $H$ and $C$ as subspaces of $\mathbb R^3$ are homeomorph.

How to solve this analytically? I've seen a geometric solution but I don't see how to work it analytically.
Define a map $f:H\to C$ by $f(x,y,z) = \Bigl(\frac x{\sqrt{1+z^2}},\frac y{\sqrt{1+z^2}},z\Bigr)$, and show that it is a homeomorphism.