Is the Mapping h(x)=x3 + x Injective and Surjective?

[sunnyday11]

Homework Statement

Decide with proof whether the mapping is injective and/or surjective.

Let f: A-->B be a mapping.

h: C--> C; h(x)=x³ + x (complex field)

f: Z--> Z; h(x)=x³ + x (integer field)

Homework Equations

injective means f(a)=f(a') => a=a'
surjective means for all b belong to B, there exists a belong to A such that f(a)=b

The Attempt at a Solution

For injectivity, I sub a and a' into equations h and f but I have no idea how to equate them or to prove them false.

For subjectivity the same issue arises, I try to get a inverse of the equation since I think they are surjective and can't think of any example to contradict it.

Thank you very much!

 

[micromass]

When I first saw your question, I thought it might be wise to play a bit with the function to see if something cool props up.

First, I decided to find the roots of the functions. That is, find the x-values such that x³+x=0. Can you find these for me? Do these roots already tell us something about injectivity/surjectivity?

 

[sunnyday11]

The roots are 0 and i.

So for injectivity, h is not injective since both 0 and i lead to the same result; f on the other hand is injective since only 0 is in its field.

For surjectivity, I still can't think of anything to prove.

 

[gb7nash]

Well, fix a in C. Does there exist a point x such that x³ + x = a? Try moving everything to the same side. What theorem can we use?

 

[sunnyday11]

x³ + x - a = 0

Sorry, I am not quite familiar with the complex plane, can you give me a hint of theorem you mentioned?

 

[gb7nash]

I'm assuming you can use the fundamental theorem of algebra. If you can't, there's always the cubic formula.