Prove Inverse of Bijection function

  • Thread starter TheLegace
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  • #1
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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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956

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 [itex]\all b\in B, \exist a in A f(b)= a[/itex]. Again, that is true of any function from B to A and has nothing to do with "onto". You need to start "If [itex]a\in A[/itex]" and end "therefore [itex]\exist b\in B[/itex] 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.
 
  • #3
27
0
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 [itex]\all b\in B, \exist a in A f(b)= a[/itex]. Again, that is true of any function from B to A and has nothing to do with "onto". You need to start "If [itex]a\in A[/itex]" and end "therefore [itex]\exist b\in B[/itex] 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.
 
  • #4
Cyosis
Homework Helper
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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 [itex]\forall b \in B \;\; \exists a \in A , \text{such that} f(a)=b[/itex]. In words this means that a function's values span its whole codomain.

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

Using your definition of a surjection would allow you to conclude that this function, f, is a surjection. Seeing as for all values of [itex]x \in [0,1] [/itex] there is an [itex]y \in \{0,1\}[/itex] namely 1. Yet 0 never is reached for any [itex]x \in [0,1][/itex] thus it cannot be a surjection.
Now lets use the correct definition of a surjection. For [itex]y=1[/itex] there is always an [itex]x \in [0,1][/itex] such that [itex]f(x)=1[/itex], namely all values within [0,1]. But for y=0 there is not a single value for [itex]x \in [0,1][/itex] such that f(x)=0. So there exist an [itex]y \in \{0,1\}[/itex] such that [itex]f(x) \neq y[/itex], namely y=0. Therefore this function is not a surjection.
 
Last edited:
  • #5
27
0
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 [itex]\forall b \in B \;\; \exists a \in A , \text{such that} f(a)=b[/itex]. In words this means that a function's values span its whole codomain.

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

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

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