Solving Surjective Functions Homework

SD123
Messages
3
Reaction score
0

Homework Statement


i) Show explicitly that any non-injective function with a right inverse has another right inverse
ii) Give an example of a function which has (at least) two distinct left inverses

Homework Equations


i) I don't believe there are any.
ii) "

The Attempt at a Solution


i) Since it says "explicitly" I doubt that my attempt would count as a solution even if it were correct, but here it is;

Since the function is non-injective and has a right inverse, it must therefore be surjective as it cannot be bijective, and in order for this function to be surjective it must have at least one more right inverse.

I also tried showing first that f(x) = x^3 - x is non-injective and then showing that the values x = -1,0,1 map to f(x) = 0 that I would be showing this is true but I am not sure if it is correct.

ii) For this I am really not sure if there is an easier way to do it apart from trial and error but I understand the theory behind it.

Thanks in advance,
SD123
 
Physics news on Phys.org
Try something as simple as this: f:{a, b, c}--> {x, y} defined by f(a)= x,f(b)= x, f(c)= y. g:{x, y}-->{a, b, c}, defined by g(x)= a, g(y)= c is a right inverse.

Think about that example to prove (i). However, note that an example is NOT a general proof.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top