a problem on composition of functions
$begingroup$
Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.
I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.
functions
asked 4 hours ago
user649511user649511
234
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user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.
I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.
functions
asked 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago
add a comment |
$begingroup$
Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.
I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.
functions
asked 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $f colon A to A$ be a function such that $f circ f=f$. If $f$ is one-to-one then prove that $f$ is also onto.
I know in my head that the func. $f$ is $f(x)=x$, but I can't develop a proof for the above statement.
functions
functions
asked 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user649511user649511
234
New contributor
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edited 15 mins ago
user649511
asked 4 hours ago
user649511user649511
234
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user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
user649511user649511
234
asked 4 hours ago
user649511user649511
234
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago
add a comment |
$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago
$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.
Suppose $f$ is not surjective, and let
$B = f(A) tag 1$
be the image of $A$ under $f$; since
$f circ f = f, tag 2$
it is clear that every element of $B$ is fixed under $f$, for
$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$
furthermore, for $c in A$,
$f(c) = c Longrightarrow c in B; tag 4$
thus $B$ is precisely the set of fixed points of $f$.
We have assumed $f$ not surjective; then by the above we have
$B subsetneq A, tag 5$
which implies
$exists a in A setminus B; tag 6$
if
$b = f(a) in B, tag 7$
then
$f(b) = f(f(a)) = f(a) = b; tag 8$
we note
$B ni b ne a in A setminus B, tag 9$
which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
$endgroup$
add a comment |
$begingroup$
We know
$fcirc f=f$ so $f([f(x)]) =f (x)$.
Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...
i see someone downvoted the question
if you feel its a dumb question please know I'm still in school learning relations and functions ! :)
edited 3 hours ago
Graham Kemp
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
add a comment |
$begingroup$
Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.
QED
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.
Suppose $f$ is not surjective, and let
$B = f(A) tag 1$
be the image of $A$ under $f$; since
$f circ f = f, tag 2$
it is clear that every element of $B$ is fixed under $f$, for
$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$
furthermore, for $c in A$,
$f(c) = c Longrightarrow c in B; tag 4$
thus $B$ is precisely the set of fixed points of $f$.
We have assumed $f$ not surjective; then by the above we have
$B subsetneq A, tag 5$
which implies
$exists a in A setminus B; tag 6$
if
$b = f(a) in B, tag 7$
then
$f(b) = f(f(a)) = f(a) = b; tag 8$
we note
$B ni b ne a in A setminus B, tag 9$
which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
$endgroup$
add a comment |
$begingroup$
What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.
Suppose $f$ is not surjective, and let
$B = f(A) tag 1$
be the image of $A$ under $f$; since
$f circ f = f, tag 2$
it is clear that every element of $B$ is fixed under $f$, for
$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$
furthermore, for $c in A$,
$f(c) = c Longrightarrow c in B; tag 4$
thus $B$ is precisely the set of fixed points of $f$.
We have assumed $f$ not surjective; then by the above we have
$B subsetneq A, tag 5$
which implies
$exists a in A setminus B; tag 6$
if
$b = f(a) in B, tag 7$
then
$f(b) = f(f(a)) = f(a) = b; tag 8$
we note
$B ni b ne a in A setminus B, tag 9$
which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
$endgroup$
add a comment |
$begingroup$
What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.
Suppose $f$ is not surjective, and let
$B = f(A) tag 1$
be the image of $A$ under $f$; since
$f circ f = f, tag 2$
it is clear that every element of $B$ is fixed under $f$, for
$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$
furthermore, for $c in A$,
$f(c) = c Longrightarrow c in B; tag 4$
thus $B$ is precisely the set of fixed points of $f$.
We have assumed $f$ not surjective; then by the above we have
$B subsetneq A, tag 5$
which implies
$exists a in A setminus B; tag 6$
if
$b = f(a) in B, tag 7$
then
$f(b) = f(f(a)) = f(a) = b; tag 8$
we note
$B ni b ne a in A setminus B, tag 9$
which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
$endgroup$
What I like about this problem is that it places no restriction on the carinality $A$; its conclusion binds for some very large sets indeed.
Suppose $f$ is not surjective, and let
$B = f(A) tag 1$
be the image of $A$ under $f$; since
$f circ f = f, tag 2$
it is clear that every element of $B$ is fixed under $f$, for
$b in B Longrightarrow exists c in A, ; b = f(c) Longrightarrow f(b) = f(f(c)) = f(c) = b; tag 3$
furthermore, for $c in A$,
$f(c) = c Longrightarrow c in B; tag 4$
thus $B$ is precisely the set of fixed points of $f$.
We have assumed $f$ not surjective; then by the above we have
$B subsetneq A, tag 5$
which implies
$exists a in A setminus B; tag 6$
if
$b = f(a) in B, tag 7$
then
$f(b) = f(f(a)) = f(a) = b; tag 8$
we note
$B ni b ne a in A setminus B, tag 9$
which contradicts the given hypothesis that $f$ is an injective map. Thus $f$ is in fact surjective.
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
edited 1 hour ago
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
answered 1 hour ago
Robert LewisRobert Lewis
47.4k23067
47.4k23067
add a comment |
add a comment |
$begingroup$
We know
$fcirc f=f$ so $f([f(x)]) =f (x)$.
Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...
i see someone downvoted the question
if you feel its a dumb question please know I'm still in school learning relations and functions ! :)
edited 3 hours ago
Graham Kemp
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
add a comment |
$begingroup$
We know
$fcirc f=f$ so $f([f(x)]) =f (x)$.
Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...
i see someone downvoted the question
if you feel its a dumb question please know I'm still in school learning relations and functions ! :)
edited 3 hours ago
Graham Kemp
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
add a comment |
$begingroup$
We know
$fcirc f=f$ so $f([f(x)]) =f (x)$.
Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...
i see someone downvoted the question
if you feel its a dumb question please know I'm still in school learning relations and functions ! :)
edited 3 hours ago
Graham Kemp
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know
$fcirc f=f$ so $f([f(x)]) =f (x)$.
Comparing both sides, which shows $[f (x)] = x$ (since $f$ is one-to-one)
so for all $x$ in codomain of $f$ there is an $x$ in domain such that $f(x) = x$ .. so it is onto.. yay...
i see someone downvoted the question
if you feel its a dumb question please know I'm still in school learning relations and functions ! :)
edited 3 hours ago
Graham Kemp
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Graham Kemp
86.2k43478
edited 3 hours ago
Graham Kemp
86.2k43478
edited 3 hours ago
Graham Kemp
86.2k43478
86.2k43478
answered 4 hours ago
user649511user649511
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
user649511user649511
234
answered 4 hours ago
user649511user649511
234
234
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user649511 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
add a comment |
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
$begingroup$
Questions get marked down or voted to close when they don't show enough initiative. To do that, this attempt should be included as part of the question.
$endgroup$
– Graham Kemp
3 hours ago
add a comment |
$begingroup$
Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.
QED
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
$endgroup$
add a comment |
$begingroup$
Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.
QED
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
$endgroup$
add a comment |
$begingroup$
Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.
QED
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
$endgroup$
Let $x in A$. Let $y=f(x).$ Then $f(y) = f(f(x)) = (f circ f) (x) = f(x)$ $implies y=x,$ if f was assumed to be injective. So if $f$ was injective then $f(x)=x,$ for all $x in A.$ So $f$ is the identity map on $A.$ But then $f$ is clearly surjective, as claimed.
QED
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
answered 1 hour ago
Dbchatto67Dbchatto67
1,301219
1,301219
add a comment |
add a comment |
user649511 is a new contributor. Be nice, and check out our Code of Conduct.
user649511 is a new contributor. Be nice, and check out our Code of Conduct.
user649511 is a new contributor. Be nice, and check out our Code of Conduct.
user649511 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome! What have you tried?
$endgroup$
– user458276
4 hours ago
$begingroup$
Don't try to prove what is the f function. I think that would be much harder than going through the definitions of one-to-one and onto. You can assume you have the one-to-one property and you use the "fof=f" equality somehow (possibly with another thoerem?) to result to the definition of onto.
$endgroup$
– frabala
4 hours ago
$begingroup$
thank you.. got it..
$endgroup$
– user649511
4 hours ago
$begingroup$
What are the definitions? What does it mean to be onto?
$endgroup$
– JavaMan
4 hours ago
$begingroup$
onto means that for a function from set A to set B every element in set B has a pre image in A . i.e. f(x) = y for every y in B
$endgroup$
– user649511
4 hours ago