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Can a non-invertible function be inverted by returning a set of all possible solutions?

How should I understand $f^{-1}(E):={xin A:f(x)in E}$?Show that each composite function $f_i circ f_j$ is one of the given functionsThe inverse function of f(x)=ln(x)/x.Find the derivative of $f^{-1}(x)$ at $x=2$ if $f(x)=x^2 + x + ln x$Inverse function of $x + x^q$ with rational $q$“Class” of functions whose inverse, where defined, is the same “class”Inverse derivative of a functionFinding inverse function of a function with multiple variableTerm for a function that handles all possible inputs?How to show $12^a cdot 18^b$ is injective

2

$begingroup$

Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?

For example:

$g(x) = x^2$

Would it be possible to create an inverse function $f(y)$ where, for example:

$f(4) = {-2,2}$

(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)

functions inverse-function

share|cite|improve this question

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If you define $f$ as a function from $mathbb{R}$ to the $textit{power set}$ of $mathbb{R}$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
  $endgroup$
  – Alexander Geldhof
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?

For example:

$g(x) = x^2$

Would it be possible to create an inverse function $f(y)$ where, for example:

$f(4) = {-2,2}$

(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)

functions inverse-function

share|cite|improve this question

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If you define $f$ as a function from $mathbb{R}$ to the $textit{power set}$ of $mathbb{R}$, then yes, that would be a function. However, $g circ f$ does not equal the identity anymore, which is a property you'd want/expect from an inverse.
  $endgroup$
  – Alexander Geldhof
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Is there a concept of inverting a non-invertible function by returning a set of the possible solutions?

For example:

$g(x) = x^2$

Would it be possible to create an inverse function $f(y)$ where, for example:

$f(4) = {-2,2}$

(I'm pretty sure I'm going about this wrong, but I'm still learning so I don't know *how* I'm wrong)

functions inverse-function

share|cite|improve this question

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 6 hours ago

Asaf Karagila♦

312k33446780

edited 6 hours ago

Asaf Karagila♦

312k33446780

asked 8 hours ago

Code SlingerCode Slinger

1164

New contributor

Code Slinger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Code SlingerCode Slinger

1164

5 Answers
5

active

oldest

votes

4

$begingroup$

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.

It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.

Finally, one could also view them simply as relations with a full domain.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Multivalued functions were the missing piece - thanks!
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

It is common to use the notation $f^{-1}(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set ${x : f(x) in A }$. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^{-1}(x)$ instead of $f^{-1}({x})$. So if in your case the context is clear, it is fine to write $g^{-1}(4) = {2, -2}$.

Also there are functions that are multivalued by default like the complex logarithm for example.

share|cite|improve this answer

answered 8 hours ago

LionCoderLionCoder

677315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $g^{-1}(4)$ could be the same as $g^{-1}({4})$ in a certain context? Would that make $g^{-1}$ a multivalued function?
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.

What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.

share|cite|improve this answer

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

$endgroup$

add a comment | 

0

$begingroup$

To give a function is the same to give the sets where the function is constant: $cmapsto f^{-1}(c)$. This would be a map $F:mathbb{R}to power(mathbb{R})$. The function $F$ is almost injective (since $f^{-1}(c)cap f^{-1}(c')=emptyset$ for $cneq c'$). It might not be injective if $f^{-1}(c)=f^{-1}(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:

$$F: Image(f)to power(mathbb{R})$$

is injective.

This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.

share|cite|improve this answer

answered 8 hours ago

Hudson LimaHudson Lima

664

$endgroup$

add a comment | 

-1

$begingroup$

We usually want functions to only take one value in value space.

If there are several different possible ones, we call them branches of a function.

For example $f^{-1}(t) = -sqrt{t}$ is one such branch that is inverse to $f(t) = t^2$

share|cite|improve this answer

answered 8 hours ago

mathreadlermathreadler

16k72263

$endgroup$

add a comment | 

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4

$begingroup$

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.

It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.

Finally, one could also view them simply as relations with a full domain.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Multivalued functions were the missing piece - thanks!
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.

It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.

Finally, one could also view them simply as relations with a full domain.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Multivalued functions were the missing piece - thanks!
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is a multivalued function (see especially the first example!), or multifunction, or set-valued function. A set-valued map, taking elements of $X$ and producing subsets of $Y$, is often denoted $f : X rightrightarrows Y$.

It can also be denoted more literally by $f : X to 2^Y$, as such maps can be thought of as (ordinary, single-valued) functions from $X$ to the power set of $Y$.

Finally, one could also view them simply as relations with a full domain.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

answered 8 hours ago

Theo BenditTheo Bendit

23.6k12359

4

$begingroup$

It is common to use the notation $f^{-1}(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set ${x : f(x) in A }$. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^{-1}(x)$ instead of $f^{-1}({x})$. So if in your case the context is clear, it is fine to write $g^{-1}(4) = {2, -2}$.

Also there are functions that are multivalued by default like the complex logarithm for example.

share|cite|improve this answer

answered 8 hours ago

LionCoderLionCoder

677315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $g^{-1}(4)$ could be the same as $g^{-1}({4})$ in a certain context? Would that make $g^{-1}$ a multivalued function?
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

It is common to use the notation $f^{-1}(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set ${x : f(x) in A }$. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^{-1}(x)$ instead of $f^{-1}({x})$. So if in your case the context is clear, it is fine to write $g^{-1}(4) = {2, -2}$.

Also there are functions that are multivalued by default like the complex logarithm for example.

share|cite|improve this answer

answered 8 hours ago

LionCoderLionCoder

677315

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So $g^{-1}(4)$ could be the same as $g^{-1}({4})$ in a certain context? Would that make $g^{-1}$ a multivalued function?
  $endgroup$
  – Code Slinger
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It is common to use the notation $f^{-1}(A)$, where $A$ is a subset of the value range of $f$, as a shorthand to describe the set ${x : f(x) in A }$. Furthermore, if $A$ is a set with only one value $x$ it is also somewhat common to just write $f^{-1}(x)$ instead of $f^{-1}({x})$. So if in your case the context is clear, it is fine to write $g^{-1}(4) = {2, -2}$.

Also there are functions that are multivalued by default like the complex logarithm for example.

share|cite|improve this answer

answered 8 hours ago

LionCoderLionCoder

677315

$endgroup$

share|cite|improve this answer

answered 8 hours ago

LionCoderLionCoder

677315

answered 8 hours ago

LionCoderLionCoder

677315

answered 8 hours ago

LionCoderLionCoder

677315

0

$begingroup$

It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.

What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.

share|cite|improve this answer

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

$endgroup$

add a comment | 

0

$begingroup$

It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.

What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.

share|cite|improve this answer

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

$endgroup$

add a comment | 

$begingroup$

It is possible to have a set-valued function just like your $f$. The only 'but' is that it is not a function $f: mathbb Rto mathbb R$ but actually $f:mathbb Rtomathcal P(mathbb R)$, i.e. its target is the set of parts (or power set) of $mathbb R$.

What you are actually describing by your $f$ is the level sets of $g$. Any more I would add is contained in @LionCoder's comment, so I invite you to continue reading his input instead of repeating what they have said.

share|cite|improve this answer

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

answered 8 hours ago

Sam SkywalkerSam Skywalker

54913

0

$begingroup$

To give a function is the same to give the sets where the function is constant: $cmapsto f^{-1}(c)$. This would be a map $F:mathbb{R}to power(mathbb{R})$. The function $F$ is almost injective (since $f^{-1}(c)cap f^{-1}(c')=emptyset$ for $cneq c'$). It might not be injective if $f^{-1}(c)=f^{-1}(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:

$$F: Image(f)to power(mathbb{R})$$

is injective.

This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.

share|cite|improve this answer

answered 8 hours ago

Hudson LimaHudson Lima

664

$endgroup$

add a comment | 

0

$begingroup$

To give a function is the same to give the sets where the function is constant: $cmapsto f^{-1}(c)$. This would be a map $F:mathbb{R}to power(mathbb{R})$. The function $F$ is almost injective (since $f^{-1}(c)cap f^{-1}(c')=emptyset$ for $cneq c'$). It might not be injective if $f^{-1}(c)=f^{-1}(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:

$$F: Image(f)to power(mathbb{R})$$

is injective.

This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.

share|cite|improve this answer

answered 8 hours ago

Hudson LimaHudson Lima

664

$endgroup$

add a comment | 

$begingroup$

To give a function is the same to give the sets where the function is constant: $cmapsto f^{-1}(c)$. This would be a map $F:mathbb{R}to power(mathbb{R})$. The function $F$ is almost injective (since $f^{-1}(c)cap f^{-1}(c')=emptyset$ for $cneq c'$). It might not be injective if $f^{-1}(c)=f^{-1}(c')=emptyset$, that is, if $c$ and $c'$ are not in the range of $f$. The result is:

$$F: Image(f)to power(mathbb{R})$$

is injective.

This map can never be surjective (by cardinality), so this is the best we can say. If you restrict an injective function to its image, then it becomes bijective.

share|cite|improve this answer

answered 8 hours ago

Hudson LimaHudson Lima

664

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Hudson LimaHudson Lima

664

answered 8 hours ago

Hudson LimaHudson Lima

664

answered 8 hours ago

Hudson LimaHudson Lima

664

-1

$begingroup$

We usually want functions to only take one value in value space.

If there are several different possible ones, we call them branches of a function.

For example $f^{-1}(t) = -sqrt{t}$ is one such branch that is inverse to $f(t) = t^2$

share|cite|improve this answer

answered 8 hours ago

mathreadlermathreadler

16k72263

$endgroup$

add a comment | 

-1

$begingroup$

We usually want functions to only take one value in value space.

If there are several different possible ones, we call them branches of a function.

For example $f^{-1}(t) = -sqrt{t}$ is one such branch that is inverse to $f(t) = t^2$

share|cite|improve this answer

answered 8 hours ago

mathreadlermathreadler

16k72263

$endgroup$

add a comment | 

$begingroup$

We usually want functions to only take one value in value space.

If there are several different possible ones, we call them branches of a function.

For example $f^{-1}(t) = -sqrt{t}$ is one such branch that is inverse to $f(t) = t^2$

share|cite|improve this answer

answered 8 hours ago

mathreadlermathreadler

16k72263

$endgroup$

share|cite|improve this answer

answered 8 hours ago

mathreadlermathreadler

16k72263

answered 8 hours ago

mathreadlermathreadler

16k72263

answered 8 hours ago

mathreadlermathreadler

16k72263