Prove that this mapping is a bijection

[schniefen]

Homework Statement

Prove, using the definition, that $\textbf{u}=\textbf{u}(\textbf{x})$ is a bijection from the strip $D=-\pi/2<x_1<\pi/2$ in the $x_1x_2$-plane onto the entire $u_1,u_2$-plane.

Relevant Equations

$u_1 = \tan{(x_1)}+x_2$
$u_2 = x_2^3$

How would one tackle this using the definition? (i.e. for some function ff that f(x)=f(y)⟹x=yf(x)=f(y)⟹x=y implies an injection and y=f(x)y=f(x) for all yy in the codomain of ff for a surjection, provided such x∈Dx∈D exist.)

One can solve the system of equations for x1x1 and x2x2 and that shows that u=u(x)u=u(x) has an inverse x(u)x(u) and that u(x(u))=xu(x(u))=x. This would only show that uu is surjective, correct?

 

[WWGD]

Yes,if you can find an inverse, you are done, no need to do anything else. But it must be a 2- sided inverse , you also need x(u(y))=y.

 

[schniefen]

Right. How does one find that $\textbf{u}(\textbf{x})$ also is injective, i.e. that $\textbf{x}(\textbf{u}(\textbf{x}))=\textbf{u}$?

 

[WWGD]

Right. How does one find that $\textbf{u}(\textbf{x})$ also is injective, i.e. that $\textbf{x}(\textbf{u}(\textbf{x}))=\textbf{u}$?

Either find the left inverse or show $f(x)=f(y) \rightarrow( x=y)$ analytically or otherwise.

 

[schniefen]

Either find the left inverse or show $f(x)=f(y) \rightarrow( x=y)$ analytically or otherwise.

How would one go about finding the left inverse? To show $f(x)=f(y) \rightarrow( x=y)$, do you mean to apply this to the separate functions within the vector $\textbf{u}$? Then $\tan{x_1}+x_2=\tan{y_1}+y_2$...

 

[WWGD]

Well, you can use it with the original to test if it is 1-1 . If your left inverse exists , it shows the original is 1-1.

 

[Mark44]

As @WWDG already said, you could find an inverse and be done.

Start by solving the equations
$u_1 = \tan(x_1)+x_2$
$u_2 = x_2^3$
for $x_1$ and $x_2$.

I added parentheses in the tangent function -- in LaTeX, braces have a different function than parentheses. I'm assuming you meant $\tan(x_1) + x_2$ rather than $\tan(x_1 + x_2)$.

 

[schniefen]

As @WWDG already said, you could find an inverse and be done.

Start by solving the equations
$u_1 = \tan{x_1}+x_2$
$u_2 = x_2^3$
for $x_1$ and $x_2$.

$x_1= \arctan{(u_1-u_2^{\frac{1}{3}})}$
$x_2= u_2^{\frac{1}{3}}$
This would imply $\textbf{u}$ is surjective.

 

[fresh_42]

Either find the left inverse or show $f(x)=f(y) \rightarrow( x=y)$ analytically or otherwise.

But injectivity alone isn't sufficient. We also need the surjective part:
$$
\mathbf{u}\, : \,D \stackrel{1:1}{\longrightarrow} \mathbb{R}^2
$$
a.) $\mathbf{u}(x)=\mathbf{u}(y) \Longrightarrow x=y$
b.) $\mathbf{u}(D) \subseteq \mathbb{R}^2$
c.) $\forall \,(v,w) \in \mathbb{R}^2 \,\exists \,(x,y)\in D\, : \,\mathbf{u}(x,y)=(v,w)$

 

[WWGD]

But injectivity alone isn't sufficient. We also need the surjective part:
$$
\mathbf{u}\, : \,D \stackrel{1:1}{\longrightarrow} \mathbb{R}^2
$$
a.) $\mathbf{u}(x)=\mathbf{u}(y) \Longrightarrow x=y$
b.) $\mathbf{u}(D) \subseteq \mathbb{R}^2$
c.) $\forall \,(v,w) \in \mathbb{R}^2 \,\exists \,(x,y)\in D\, : \,\mathbf{u}(x,y)=(v,w)$

Yes, i mentioned we need both right- and left- inverses. Schniefen had provided a leftr inverse and i suggested he find a left inverse too.