show the map is a bijection

**mathswannabe** · September 10th 2011, 08:05 PM

Let $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ , so:

$(u,v) \rightarrow (u^3,u+v^3,v)$

Show this is a homeomorphism:

To show injectivity: We want to show that:
$\phi(u_1,v_1)=\phi(u_2,v_2) \rightarrow (u_1,v_1)= (u_2,v_2)$

So if we set:

$u_{1}^{3}&=&u_{2}^{3}$

$u_1+v_1^3=u_2+v_2^3$

$v_1=v_2$

Then from the third equation we get $v_1=v_2$ , and can sub back in to find that $u_1=u_2$ .

The surjectivity is where I'm getting stuck. If we let any $(x,y,z)$ be a point in $\mathbb{R}^3$

Then we get $v=z$ and subbing in to the equation for y means that we get $u = y-z^3$ .

So I guess I go on to say that given any $(x,y,z)$ I can find a $(u,v)$ such that $\phi(u,v)=(x,y,z)$

However, I think this is wrong since what happens with the $x$ value in this case? I mean we also get that $u=x^{\frac{1}{3}}$

I'm lost here can someone help? I don't think what I did initially was enough?

Thanks!

**ModusPonens** · September 11th 2011, 01:23 AM

I knew something was fishy with this. Check and see that the point (2,2,2), for example, doesn't belong to the image of the function.

Math Help - show the map is a bijection

LinkBack

Thread Tools

Display

show the map is a bijection

Re: show the map is a bijection

Similar Math Help Forum Discussions

Bijection

Bijection

Bijection

Bijection

show this bijection is invariant and unbiased

Search Tags