Set theory, functions and inverses
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I'm doing an intro course on set theory and have the question if the inverses of the surjective functions in the sets A=｛a, b｝and B= {c, d, e} are also functions.
So far, I thought that the inverse of a function f(x) for example, is f⁻¹(y), meaning that every function also has an inverse (which is also a function).
Given that this would make the question rather redundant, I'm not quite sure in my assumption anymore, so I would be glad, if someone could verify or falsify (and explain it properly) it.
functions elementary-set-theory inverse-function
asked Nov 19 at 15:04
K. Meyer
11
add a comment |
up vote
0
down vote
favorite
I'm doing an intro course on set theory and have the question if the inverses of the surjective functions in the sets A=｛a, b｝and B= {c, d, e} are also functions.
So far, I thought that the inverse of a function f(x) for example, is f⁻¹(y), meaning that every function also has an inverse (which is also a function).
Given that this would make the question rather redundant, I'm not quite sure in my assumption anymore, so I would be glad, if someone could verify or falsify (and explain it properly) it.
functions elementary-set-theory inverse-function
asked Nov 19 at 15:04
K. Meyer
11
Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm doing an intro course on set theory and have the question if the inverses of the surjective functions in the sets A=｛a, b｝and B= {c, d, e} are also functions.
So far, I thought that the inverse of a function f(x) for example, is f⁻¹(y), meaning that every function also has an inverse (which is also a function).
Given that this would make the question rather redundant, I'm not quite sure in my assumption anymore, so I would be glad, if someone could verify or falsify (and explain it properly) it.
functions elementary-set-theory inverse-function
asked Nov 19 at 15:04
K. Meyer
11
I'm doing an intro course on set theory and have the question if the inverses of the surjective functions in the sets A=｛a, b｝and B= {c, d, e} are also functions.
So far, I thought that the inverse of a function f(x) for example, is f⁻¹(y), meaning that every function also has an inverse (which is also a function).
Given that this would make the question rather redundant, I'm not quite sure in my assumption anymore, so I would be glad, if someone could verify or falsify (and explain it properly) it.
functions elementary-set-theory inverse-function
functions elementary-set-theory inverse-function
asked Nov 19 at 15:04
K. Meyer
11
asked Nov 19 at 15:04
K. Meyer
11
asked Nov 19 at 15:04
K. Meyer
11
asked Nov 19 at 15:04
K. Meyer
11
asked Nov 19 at 15:04
K. Meyer
11
11
Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11
add a comment |
Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11
Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11
Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11
add a comment |
1 Answer
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A relation is a set of ordered pairs, and it's inverse is the following relation:
$$R^{-1}:={(y,x)|(x,y)in R}$$
And $R$ is called a function if $forall x,y_1,y_2$, $(x,y_1)in R land (x,y_2)in R implies y_1=y_2$. In the case of functions, we can always define a formal inverse, and we call a function $f$ injective if it's formal inverse is also a function. And not all of the surjections are injections, for example let
$$f:={(a,c),(b,d),(a,e)}$$
Now, $f$ is a surjection from $A$ to $B$, but it's not an injection, because it's inverse is not a function.
answered Nov 19 at 15:25
Botond
5,1462732
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A relation is a set of ordered pairs, and it's inverse is the following relation:
$$R^{-1}:={(y,x)|(x,y)in R}$$
And $R$ is called a function if $forall x,y_1,y_2$, $(x,y_1)in R land (x,y_2)in R implies y_1=y_2$. In the case of functions, we can always define a formal inverse, and we call a function $f$ injective if it's formal inverse is also a function. And not all of the surjections are injections, for example let
$$f:={(a,c),(b,d),(a,e)}$$
Now, $f$ is a surjection from $A$ to $B$, but it's not an injection, because it's inverse is not a function.
answered Nov 19 at 15:25
Botond
5,1462732
add a comment |
up vote
0
down vote
A relation is a set of ordered pairs, and it's inverse is the following relation:
$$R^{-1}:={(y,x)|(x,y)in R}$$
And $R$ is called a function if $forall x,y_1,y_2$, $(x,y_1)in R land (x,y_2)in R implies y_1=y_2$. In the case of functions, we can always define a formal inverse, and we call a function $f$ injective if it's formal inverse is also a function. And not all of the surjections are injections, for example let
$$f:={(a,c),(b,d),(a,e)}$$
Now, $f$ is a surjection from $A$ to $B$, but it's not an injection, because it's inverse is not a function.
answered Nov 19 at 15:25
Botond
5,1462732
add a comment |
up vote
0
down vote
up vote
0
down vote
A relation is a set of ordered pairs, and it's inverse is the following relation:
$$R^{-1}:={(y,x)|(x,y)in R}$$
And $R$ is called a function if $forall x,y_1,y_2$, $(x,y_1)in R land (x,y_2)in R implies y_1=y_2$. In the case of functions, we can always define a formal inverse, and we call a function $f$ injective if it's formal inverse is also a function. And not all of the surjections are injections, for example let
$$f:={(a,c),(b,d),(a,e)}$$
Now, $f$ is a surjection from $A$ to $B$, but it's not an injection, because it's inverse is not a function.
answered Nov 19 at 15:25
Botond
5,1462732
A relation is a set of ordered pairs, and it's inverse is the following relation:
$$R^{-1}:={(y,x)|(x,y)in R}$$
And $R$ is called a function if $forall x,y_1,y_2$, $(x,y_1)in R land (x,y_2)in R implies y_1=y_2$. In the case of functions, we can always define a formal inverse, and we call a function $f$ injective if it's formal inverse is also a function. And not all of the surjections are injections, for example let
$$f:={(a,c),(b,d),(a,e)}$$
Now, $f$ is a surjection from $A$ to $B$, but it's not an injection, because it's inverse is not a function.
answered Nov 19 at 15:25
Botond
5,1462732
answered Nov 19 at 15:25
Botond
5,1462732
answered Nov 19 at 15:25
Botond
5,1462732
answered Nov 19 at 15:25
Botond
5,1462732
5,1462732
add a comment |
add a comment |
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Inverse defined from the range of the given function is a function.
– Thomas Shelby
Nov 19 at 15:11