Prove Inverse of Bijection function

[TheLegace]

Homework Statement

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

Homework 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.

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.

 

[HallsofIvy]

Homework Statement

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

Homework 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.

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

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.

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.

There are some "special" symbols in there I can't see but I think you are saying $\all b\in B, \exist a in A f(b)= a$. Again, that is true of any function from B to A and has nothing to do with "onto". You need to start "If $a\in A$" and end "therefore $\exist b\in B$ such that f^-1(b)= a". That is, you have to start with a member of A and find a member of B so that is true. There is an obvious way to do that.

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

 

[TheLegace]

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.

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.

There are some "special" symbols in there I can't see but I think you are saying $\all b\in B, \exist a in A f(b)= a$. Again, that is true of any function from B to A and has nothing to do with "onto". You need to start "If $a\in A$" and end "therefore $\exist b\in B$ such that f^-1(b)= a". That is, you have to start with a member of A and find a member of B so that is true. There is an obvious way to do that.

The way onto is defined in my text and class is through quantification. For f, for every a in set A, there is some b in set B such that f(a)=b.
For the inverse for every b in B there is an element a in A, which defines the mapping of onto function.

Thank you for the input, I appreciate it.

TheLegace.

 

[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 let's 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.

 

[TheLegace]

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 let's 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.

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.