# Prove Inverse of Bijection function

1. Jul 6, 2009

### TheLegace

1. The problem statement, all variables and given/known data

Suppose f is bijection. Prove that f⁻¹. is bijection.

2. Relevant equations

A bijection of a function occurs when f is one to one and onto.
I think the proof would involve showing f⁻¹. is bijective, by showing f⁻¹ is onto, and one to one, since f is bijective it is invertible.

3. The attempt at a solution

To start:

Since f is invertible/bijective
f⁻¹ is one-to-one:
f:A→B
f⁻¹:B→A

f(a)=b then f⁻¹(b)=a
if f(a)=b f(a)=b' then b=b'
So f⁻¹ is one-to-one

f⁻¹ is onto:
*for f to be onto:
f:A→B
∀a∈A,∃b∈B f(a)=b
*
f⁻¹:B→A
∀b∈B,∃a∈A f(b)=a
So f⁻¹ is onto.

Therefore f⁻¹ is bijective.
Would this be correct?
Thank You Very Much.

2. Jul 6, 2009

### HallsofIvy

Staff Emeritus
Yes, if f(a)= b and f(a)= b' then b= b'. But that is true of any function, not just one-to-one functions so it does NOT show that f-1 is one-to-one. You need to start from "if f-1(a)= f-1(b)" and finish "therefore a= b". I would recommend taking f of both sides.

3. Jul 6, 2009

4. Jul 6, 2009

### Cyosis

I very much doubt that the book you use defines a surjection like that, but if it does it's simply wrong. The definition of a surjection is that $\forall b \in B \;\; \exists a \in A , \text{such that} f(a)=b$. In words this means that a function's values span its whole codomain.

Example:
$$f:[0,1] \rightarrow \{1,0\}, f(x)=1$$

Using your definition of a surjection would allow you to conclude that this function, f, is a surjection. Seeing as for all values of $x \in [0,1]$ there is an $y \in \{0,1\}$ namely 1. Yet 0 never is reached for any $x \in [0,1]$ thus it cannot be a surjection.
Now lets use the correct definition of a surjection. For $y=1$ there is always an $x \in [0,1]$ such that $f(x)=1$, namely all values within [0,1]. But for y=0 there is not a single value for $x \in [0,1]$ such that f(x)=0. So there exist an $y \in \{0,1\}$ such that $f(x) \neq y$, namely y=0. Therefore this function is not a surjection.

Last edited: Jul 6, 2009
5. Jul 6, 2009

### TheLegace

Oh my apologies I made a mistake, you are right, I am working on some inverse function stuff and typing functions gets me confused very easily. But thank you.