Mixing
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 yesterday
K. Meyer
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up vote
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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 yesterday
K. Meyer
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday
add a comment |
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up vote
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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 yesterday
K. Meyer
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 yesterday
K. Meyer
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
K. Meyer
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
K. Meyer
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
K. Meyer
11
asked yesterday
K. Meyer
11
11
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
K. Meyer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday
add a comment |
Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday
Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday
Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday
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 yesterday
Botond
4,9632732
add a comment |
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 yesterday
Botond
4,9632732
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 yesterday
Botond
4,9632732
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 yesterday
Botond
4,9632732
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 yesterday
Botond
4,9632732
answered yesterday
Botond
4,9632732
answered yesterday
Botond
4,9632732
answered yesterday
Botond
4,9632732
4,9632732
add a comment |
add a comment |
K. Meyer is a new contributor. Be nice, and check out our Code of Conduct.
K. Meyer is a new contributor. Be nice, and check out our Code of Conduct.
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Inverse defined from the range of the given function is a function.
– Thomas Shelby
yesterday