Mixing
Images and Inverses
$begingroup$
Definition 0.2.10 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $S subseteq B$. Then the set $f^{-1}(S)={ xin A : f(x) in S}$ is called the inverse image of the set $S$ under the function $f$.
Definition 0.2.14 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $T subseteq A$. Then the set $f(T)={xin B : x=f(t) text{ for some }t in T }$
is called the image of the set $T$ under the function $f$.
How do these two definitions differ from traditional usage of $f$ and $f^{-1}$ as in a calculus course?
My attempt to make any sense of it:
They are similar in many ways. In the definitions, for a function $f$ that maps $A$ to $B$ is similar to plugging a value $x$ into $f(x)$ and getting an output $y$. But, in the definitions given above, $f$ is invertible if it is one-to-one and onto, but how does this differ from the traditional $f$ used in calculus? I cannot seem to find the ties or correlations.
elementary-set-theory
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
$endgroup$
add a comment |
$begingroup$
Definition 0.2.10 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $S subseteq B$. Then the set $f^{-1}(S)={ xin A : f(x) in S}$ is called the inverse image of the set $S$ under the function $f$.
Definition 0.2.14 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $T subseteq A$. Then the set $f(T)={xin B : x=f(t) text{ for some }t in T }$
is called the image of the set $T$ under the function $f$.
How do these two definitions differ from traditional usage of $f$ and $f^{-1}$ as in a calculus course?
My attempt to make any sense of it:
They are similar in many ways. In the definitions, for a function $f$ that maps $A$ to $B$ is similar to plugging a value $x$ into $f(x)$ and getting an output $y$. But, in the definitions given above, $f$ is invertible if it is one-to-one and onto, but how does this differ from the traditional $f$ used in calculus? I cannot seem to find the ties or correlations.
elementary-set-theory
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
$endgroup$
1
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
1
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36
add a comment |
$begingroup$
Definition 0.2.10 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $S subseteq B$. Then the set $f^{-1}(S)={ xin A : f(x) in S}$ is called the inverse image of the set $S$ under the function $f$.
Definition 0.2.14 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $T subseteq A$. Then the set $f(T)={xin B : x=f(t) text{ for some }t in T }$
is called the image of the set $T$ under the function $f$.
How do these two definitions differ from traditional usage of $f$ and $f^{-1}$ as in a calculus course?
My attempt to make any sense of it:
They are similar in many ways. In the definitions, for a function $f$ that maps $A$ to $B$ is similar to plugging a value $x$ into $f(x)$ and getting an output $y$. But, in the definitions given above, $f$ is invertible if it is one-to-one and onto, but how does this differ from the traditional $f$ used in calculus? I cannot seem to find the ties or correlations.
elementary-set-theory
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
$endgroup$
Definition 0.2.10 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $S subseteq B$. Then the set $f^{-1}(S)={ xin A : f(x) in S}$ is called the inverse image of the set $S$ under the function $f$.
Definition 0.2.14 Let $A$ and $B$ be sets, and $f: A rightarrow B$ be a function. Let $T subseteq A$. Then the set $f(T)={xin B : x=f(t) text{ for some }t in T }$
is called the image of the set $T$ under the function $f$.
How do these two definitions differ from traditional usage of $f$ and $f^{-1}$ as in a calculus course?
My attempt to make any sense of it:
They are similar in many ways. In the definitions, for a function $f$ that maps $A$ to $B$ is similar to plugging a value $x$ into $f(x)$ and getting an output $y$. But, in the definitions given above, $f$ is invertible if it is one-to-one and onto, but how does this differ from the traditional $f$ used in calculus? I cannot seem to find the ties or correlations.
elementary-set-theory
elementary-set-theory
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
edited Jan 9 at 5:30
Xander Henderson
14.3k103554
14.3k103554
asked Jan 9 at 5:22
RyanRyan
1457
asked Jan 9 at 5:22
RyanRyan
1457
asked Jan 9 at 5:22
RyanRyan
1457
1457
1
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
1
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36
add a comment |
1
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
1
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36
1
1
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
1
1
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A., $ E.g. we usually assume that a subset of $Bbb R$ is never a member of $Bbb R,$ so that if $f:Bbb R to Bbb R$ and $Tsubset Bbb R$ then $f(T)={f(x):xin T}$ and $f^{-1}(T)=f^{-1}T={x:f(x)in T}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:Ato B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.
Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $emptyset$ and ${emptyset}$ both belong to the domain of a function $f$, then it is unclear what $f({emptyset })$ "should" mean. In Set Theory a function $f:Ato B$ $is$ its graph, so $f$ is a certain type of subset of $Atimes B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote ${f(t): tin Tcap dom(f)},$ the image of $T$ under $f.$
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
add a comment |
$begingroup$
Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[Bbb R] = Bbb R_{geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
$endgroup$
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
add a comment |
$begingroup$
The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.
That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:{1,2,3}to{1,2,3},f={(1,1),(2,1),(3,2)}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,
$f^{-1}(phi)=f^{-1}({3})=phi$, since no element maps to $3$.
$f^{-1}({1})={1,2}$, since both $1,2$ map to $1$.
$f^{-1}({1,3})={1,2}$, since both $1,2$ map to $1$ while no element maps to $3$.
$f^{-1}({2})={3}$, since only $3$ maps to $2$.
The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have
$f(phi)=phi$
$f({1})=f({2})={1}$, since both $1,2$ map to $1$.
$f({3})={2}$, since $3$ maps to $2$.
$f({1,3})={1,2}$, since $1$ maps to $1$ and $3$ maps to $2$.
Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as ${1,2}$. Similarly for the inverse of the function, if it exists, and the inverse image.
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
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add a comment |
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why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A., $ E.g. we usually assume that a subset of $Bbb R$ is never a member of $Bbb R,$ so that if $f:Bbb R to Bbb R$ and $Tsubset Bbb R$ then $f(T)={f(x):xin T}$ and $f^{-1}(T)=f^{-1}T={x:f(x)in T}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:Ato B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.
Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $emptyset$ and ${emptyset}$ both belong to the domain of a function $f$, then it is unclear what $f({emptyset })$ "should" mean. In Set Theory a function $f:Ato B$ $is$ its graph, so $f$ is a certain type of subset of $Atimes B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote ${f(t): tin Tcap dom(f)},$ the image of $T$ under $f.$
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
add a comment |
$begingroup$
It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A., $ E.g. we usually assume that a subset of $Bbb R$ is never a member of $Bbb R,$ so that if $f:Bbb R to Bbb R$ and $Tsubset Bbb R$ then $f(T)={f(x):xin T}$ and $f^{-1}(T)=f^{-1}T={x:f(x)in T}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:Ato B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.
Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $emptyset$ and ${emptyset}$ both belong to the domain of a function $f$, then it is unclear what $f({emptyset })$ "should" mean. In Set Theory a function $f:Ato B$ $is$ its graph, so $f$ is a certain type of subset of $Atimes B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote ${f(t): tin Tcap dom(f)},$ the image of $T$ under $f.$
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
add a comment |
$begingroup$
It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A., $ E.g. we usually assume that a subset of $Bbb R$ is never a member of $Bbb R,$ so that if $f:Bbb R to Bbb R$ and $Tsubset Bbb R$ then $f(T)={f(x):xin T}$ and $f^{-1}(T)=f^{-1}T={x:f(x)in T}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:Ato B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.
Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $emptyset$ and ${emptyset}$ both belong to the domain of a function $f$, then it is unclear what $f({emptyset })$ "should" mean. In Set Theory a function $f:Ato B$ $is$ its graph, so $f$ is a certain type of subset of $Atimes B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote ${f(t): tin Tcap dom(f)},$ the image of $T$ under $f.$
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
$endgroup$
It is common and convenient in many subjects to assume that some set $A$ that we are interested in is an anti-transitive set: That no member of $A$ is a subset of $A., $ E.g. we usually assume that a subset of $Bbb R$ is never a member of $Bbb R,$ so that if $f:Bbb R to Bbb R$ and $Tsubset Bbb R$ then $f(T)={f(x):xin T}$ and $f^{-1}(T)=f^{-1}T={x:f(x)in T}$ are unambiguous definitions. And if $A,B$ are anti-transitive sets and $f:Ato B$ is one-to-one (injective) and if $b$ belongs to the (unambiguous) set $f(A)$, then defining $a=f^{-1}(b)iff f(a)=b$ is also unambiguous. The anti-transitive assumption occurs, often tacitly, in many books and papers.
Difficulties arise, especially in Set Theory, when we cannot safely assume anti-transitivity. E.g. if $emptyset$ and ${emptyset}$ both belong to the domain of a function $f$, then it is unclear what $f({emptyset })$ "should" mean. In Set Theory a function $f:Ato B$ $is$ its graph, so $f$ is a certain type of subset of $Atimes B.$ And in Set Theory it is preferable to write $b=f(a)$ to only mean that $(a,b)in f.$ And then the notation $f''T$ (read $f$-double-prime-$T$) is used to denote ${f(t): tin Tcap dom(f)},$ the image of $T$ under $f.$
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
answered Jan 9 at 7:31
DanielWainfleetDanielWainfleet
34.9k31648
34.9k31648
add a comment |
add a comment |
$begingroup$
Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[Bbb R] = Bbb R_{geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
$endgroup$
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
add a comment |
$begingroup$
Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[Bbb R] = Bbb R_{geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
$endgroup$
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
add a comment |
$begingroup$
Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[Bbb R] = Bbb R_{geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
$endgroup$
Usually when we think about the image of a function, we imagine the whole function. What the author defined here is a generalization: let $f(x) = x^2$, for example; then $f[Bbb R] = Bbb R_{geq 0}$ as expected, but also we can look specifically at some part of the graph, like e.g. $f[[-1,1]] = [0;1]$. Try to sketch those in your mind: it's like "cropping" the function to those specific interesting parts to see what they look like. The same for the inverse image: you're looking for all the values that make your function output something in your set.
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
edited Jan 9 at 8:42
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
answered Jan 9 at 5:31
Lucas HenriqueLucas Henrique
1,037414
1,037414
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
add a comment |
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
$begingroup$
I'm trying to be very intuitive and even informal. If you want a more formal description, just warn me.
$endgroup$
– Lucas Henrique
Jan 9 at 5:32
1
1
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
Thank you for the response. Your response is fine, as I am really trying to digest most of the material in this chapter. Thanks again.
$endgroup$
– Ryan
Jan 9 at 5:33
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
$begingroup$
$f(Bbb R)=Bbb R^+cup{0}$
$endgroup$
– Shubham Johri
Jan 9 at 6:15
add a comment |
$begingroup$
The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.
That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:{1,2,3}to{1,2,3},f={(1,1),(2,1),(3,2)}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,
$f^{-1}(phi)=f^{-1}({3})=phi$, since no element maps to $3$.
$f^{-1}({1})={1,2}$, since both $1,2$ map to $1$.
$f^{-1}({1,3})={1,2}$, since both $1,2$ map to $1$ while no element maps to $3$.
$f^{-1}({2})={3}$, since only $3$ maps to $2$.
The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have
$f(phi)=phi$
$f({1})=f({2})={1}$, since both $1,2$ map to $1$.
$f({3})={2}$, since $3$ maps to $2$.
$f({1,3})={1,2}$, since $1$ maps to $1$ and $3$ maps to $2$.
Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as ${1,2}$. Similarly for the inverse of the function, if it exists, and the inverse image.
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.
That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:{1,2,3}to{1,2,3},f={(1,1),(2,1),(3,2)}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,
$f^{-1}(phi)=f^{-1}({3})=phi$, since no element maps to $3$.
$f^{-1}({1})={1,2}$, since both $1,2$ map to $1$.
$f^{-1}({1,3})={1,2}$, since both $1,2$ map to $1$ while no element maps to $3$.
$f^{-1}({2})={3}$, since only $3$ maps to $2$.
The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have
$f(phi)=phi$
$f({1})=f({2})={1}$, since both $1,2$ map to $1$.
$f({3})={2}$, since $3$ maps to $2$.
$f({1,3})={1,2}$, since $1$ maps to $1$ and $3$ maps to $2$.
Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as ${1,2}$. Similarly for the inverse of the function, if it exists, and the inverse image.
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.
That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:{1,2,3}to{1,2,3},f={(1,1),(2,1),(3,2)}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,
$f^{-1}(phi)=f^{-1}({3})=phi$, since no element maps to $3$.
$f^{-1}({1})={1,2}$, since both $1,2$ map to $1$.
$f^{-1}({1,3})={1,2}$, since both $1,2$ map to $1$ while no element maps to $3$.
$f^{-1}({2})={3}$, since only $3$ maps to $2$.
The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have
$f(phi)=phi$
$f({1})=f({2})={1}$, since both $1,2$ map to $1$.
$f({3})={2}$, since $3$ maps to $2$.
$f({1,3})={1,2}$, since $1$ maps to $1$ and $3$ maps to $2$.
Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as ${1,2}$. Similarly for the inverse of the function, if it exists, and the inverse image.
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
$endgroup$
The inverse image is a function from the power set of the codomain to the power set of the domain. That is,$$f^{-1}:P(B)to P(A)$$that operates on a subset of the codomain, say $S$, to return a subset of the domain, say $M$, containing values that map to some member of $S$ under $f$. It is not required for each member of $S$ to have a pre-image. At the same time, some members of $S$ can have more than one pre-image, in case $f$ is not injective.
That is to say, $f$ needn't be invertible for defining the inverse image of a subset of its codomain. For example, consider the function $f:{1,2,3}to{1,2,3},f={(1,1),(2,1),(3,2)}$. Clearly, $f$ is not invertible since it is neither injective nor surjective. But,
$f^{-1}(phi)=f^{-1}({3})=phi$, since no element maps to $3$.
$f^{-1}({1})={1,2}$, since both $1,2$ map to $1$.
$f^{-1}({1,3})={1,2}$, since both $1,2$ map to $1$ while no element maps to $3$.
$f^{-1}({2})={3}$, since only $3$ maps to $2$.
The direct image, or just the image, is a function from the power set of the domain to the power set of the range, or codomain, of $f$. That is,$$f:P(A)to P(B)$$is a function that takes in a subset of the domain, say $T$, and returns the set $R$ containing the images of all the elements of $T$. In the example given above, we have
$f(phi)=phi$
$f({1})=f({2})={1}$, since both $1,2$ map to $1$.
$f({3})={2}$, since $3$ maps to $2$.
$f({1,3})={1,2}$, since $1$ maps to $1$ and $3$ maps to $2$.
Remark. Although both the function and the direct image use the notation $f$, you can detect if it is the latter that is being talked about by the argument of $f$. In case of direct image, the argument will be a set, such as ${1,2}$. Similarly for the inverse of the function, if it exists, and the inverse image.
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 6:49
Shubham JohriShubham Johri
5,122717
5,122717
add a comment |
add a comment |
$begingroup$
why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
$endgroup$
add a comment |
$begingroup$
why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
$endgroup$
add a comment |
$begingroup$
why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
$endgroup$
why will there be any difference it seems the $f$ and $ f^{-1} $ defined here is exactly as that of calculus . one thing you may say as a difference is in calculus most generally we restrict ourselves to metric spaces as the two sets.
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
answered Jan 9 at 5:33
Bijayan RayBijayan Ray
25110
25110
add a comment |
add a comment |
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1
$begingroup$
The first definition doesn't look quite right.
$endgroup$
– Lucas Henrique
Jan 9 at 5:23
1
$begingroup$
You're right! I made a typo error. I've updated it.
$endgroup$
– Ryan
Jan 9 at 5:26
$begingroup$
You may define $f(T)={f(t):tin T}$
$endgroup$
– Shubham Johri
Jan 9 at 6:55
$begingroup$
@ShubhamJohri this is not good, especially at the start, because in first glance there is no reason for this to be a set
$endgroup$
– Holo
Jan 9 at 8:45
$begingroup$
Beg your pardon?
$endgroup$
– Shubham Johri
Jan 9 at 9:36