Is the Inverse of a Function Always Well-Defined?

Click For Summary
SUMMARY

The discussion centers on the properties of inverse functions, specifically whether the equations \(f^{-1}(f(x)) = x\) and \(f(f^{-1}(y)) = y\) hold true for all functions \(f\). The example function \(f(x) = x^2\) illustrates that the inverse is not well-defined for all inputs, as \(f^{-1}(1)\) yields both 1 and -1. The third question regarding \(f(f^{-1}(B))\) for subsets \(B \subset R\) raises complexity, indicating that the truth of the statement depends on the nature of \(f\) and its inverse.

PREREQUISITES
  • Understanding of inverse functions and their definitions
  • Familiarity with real-valued functions and their properties
  • Knowledge of function notation and operations
  • Basic concepts of set theory as applied to functions
NEXT STEPS
  • Study the properties of one-to-one and onto functions in relation to inverses
  • Explore the implications of non-injective functions on their inverses
  • Research the concept of function composition and its effects on inverses
  • Examine examples of piecewise functions to understand inverse behavior
USEFUL FOR

Mathematicians, educators, and students studying advanced algebra or calculus, particularly those interested in the properties of functions and their inverses.

OhMyMarkov
Messages
81
Reaction score
0
Hello everyone!

I have three questions:

(1) If $x\in R$, is it true that $f ^{-1} (f(x)) = x$?
(2) If $y\in R$, is it true that $f (f^{-1}(y)) = y$?
(3) If $B\subset R$, is it true that $f(f ^{-1} (B)$?

I think I have showed it for (3), but not sure of my proof. For (1) and (2), I considered the function $f (x) = x^2$. $f^{-1}(1)$ can be 1 and -1...

Thanks for the help!
 
Physics news on Phys.org
The answer may depend on what \(f\) is and on the precise definition of \(f^{-1}\). Also, in (3), \(f(f^{-1}(B))\) cannot be true or false.
 
OhMyMarkov said:
Hello everyone!

I have three questions:

(1) If $x\in R$, is it true that $f ^{-1} (f(x)) = x$?
(2) If $y\in R$, is it true that $f (f^{-1}(y)) = y$?
(3) If $B\subset R$, is it true that $f(f ^{-1} (B)$?

I think I have showed it for (3), but not sure of my proof. For (1) and (2), I considered the function $f (x) = x^2$. $f^{-1}(1)$ can be 1 and -1...

Thanks for the help!

As in Your example, the (1) supplies one and only one x if and only if $\displaystyle f^{-1} (*)$ is a single value function...

Kind regards

$\chi$ $\sigma$
 
I've returned to although I have seen this before, and I thought I was convinced.

Could you explain your answer, I don't think I understand...

Thank you.
 

Similar threads

Undergrad Question about convex property in Jensen's inequality
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Homemorphism in quotient topology
  • · Replies 5 ·
Replies
5
Views
1K
Graduate Question about lifting of a function
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad On commutativity of convolution
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Verifying pointwise convergence of indicator functions
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Exercise 2.23 in Folland's real analysis text
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad A concise "proof" of the Riemann-Lebesgue lemma
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Continuity of linear map from subspace of Euclidean space
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Regarding fibrations between smooth manifolds
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad No structure in ##(d,x)## in ##\textbf{Met*}(X)## is admissible
  • · Replies 0 ·
Replies
0
Views
2K