# Prove that this mapping is a bijection

• schniefen
In summary, the conversation discusses finding a solution to a system of equations and determining if a function is both injective and surjective. One method is to find a two-sided inverse, while another is to show that the function is one-to-one and onto. The conversation also touches on the importance of both right- and left- inverses in determining the injectivity and surjectivity of a function.
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?

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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}##?

schniefen said:
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.

WWGD said:
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##...

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.

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)##.

Mark44 said:
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.

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WWGD said:
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)##

fresh_42 said:
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.

## 1. What is a bijection mapping?

A bijection mapping is a function that maps each element of one set to a unique element in another set. This means that each element in the first set has a corresponding element in the second set, and vice versa. In other words, the mapping is both one-to-one and onto.

## 2. How can I prove that a mapping is a bijection?

To prove that a mapping is a bijection, you need to show that it is both one-to-one and onto. This can be done by showing that each element in the first set has a unique corresponding element in the second set, and that every element in the second set has at least one corresponding element in the first set.

## 3. What is the importance of bijection mappings in mathematics?

Bijection mappings are important in mathematics because they allow us to establish a one-to-one correspondence between two sets, which can help us understand the relationship between those sets. They are also useful in proving theorems and solving problems in various branches of mathematics, such as algebra, analysis, and topology.

## 4. Can a mapping be both a bijection and a function?

Yes, a bijection is a type of function that is both one-to-one and onto. In fact, every bijection is a function, but not every function is a bijection. A function can be one-to-one but not onto, or it can be onto but not one-to-one.

## 5. How can I check if a mapping is a bijection?

To check if a mapping is a bijection, you can use the horizontal line test. Draw horizontal lines across the graph of the mapping and see if each line intersects the graph at exactly one point. If this is true for every line, then the mapping is a bijection. Additionally, you can also check if the mapping has an inverse, which is another method of proving that a mapping is a bijection.

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