Prove $ g circ f $ is not one to one The Next CEO of Stack OverflowShowing two functions are bijectionsProof strategy for $(=>)$: If $g circ f = id_A$, then f onto $iff$ g 1-1. [Chartrand 3Ed P239 9.72]A multivariate function with bounded partial derivatives is LipschitzProving a function $F$ is surjective if and only if $f$ is injectiveShow that if $g(f(x)) $ is one-to-one and $f$ is onto, then $g$ is one-to-oneIs this example being bijective?What is a one to one relationship that never changes?Consider the function $f: mathbbC to mathbbR^2$ given by $f(a+bi) = (a,b)$. Show that $f$ is one-to-one and onto.how is my proof on composition functionsIf $g circ f$ is injective and $f$ is surjective, then $g$ is injective.
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Prove $ g circ f $ is not one to one
The Next CEO of Stack OverflowShowing two functions are bijectionsProof strategy for $(=>)$: If $g circ f = id_A$, then f onto $iff$ g 1-1. [Chartrand 3Ed P239 9.72]A multivariate function with bounded partial derivatives is LipschitzProving a function $F$ is surjective if and only if $f$ is injectiveShow that if $g(f(x)) $ is one-to-one and $f$ is onto, then $g$ is one-to-oneIs this example being bijective?What is a one to one relationship that never changes?Consider the function $f: mathbbC to mathbbR^2$ given by $f(a+bi) = (a,b)$. Show that $f$ is one-to-one and onto.how is my proof on composition functionsIf $g circ f$ is injective and $f$ is surjective, then $g$ is injective.
$begingroup$
Given f is Onto and g is not one to one
Now $exists b_1, b_2 in B $ such that $(b_1,c)$ and $(b_2,c) in g$. Now since f is onto so that
$exists a_1, a_2 in A $ such that $(a_1,b_1)$ and $(a_2,b_2) in f$ and so $(a_1,c)$ and $(a_2,c) in g circ f$ . So its not one to one .
Is this correct ? Thanks
functions proof-verification
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
$endgroup$
add a comment |
$begingroup$
Given f is Onto and g is not one to one
Now $exists b_1, b_2 in B $ such that $(b_1,c)$ and $(b_2,c) in g$. Now since f is onto so that
$exists a_1, a_2 in A $ such that $(a_1,b_1)$ and $(a_2,b_2) in f$ and so $(a_1,c)$ and $(a_2,c) in g circ f$ . So its not one to one .
Is this correct ? Thanks
functions proof-verification
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
$endgroup$
$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
2
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41
add a comment |
$begingroup$
Given f is Onto and g is not one to one
Now $exists b_1, b_2 in B $ such that $(b_1,c)$ and $(b_2,c) in g$. Now since f is onto so that
$exists a_1, a_2 in A $ such that $(a_1,b_1)$ and $(a_2,b_2) in f$ and so $(a_1,c)$ and $(a_2,c) in g circ f$ . So its not one to one .
Is this correct ? Thanks
functions proof-verification
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
$endgroup$
Given f is Onto and g is not one to one
Now $exists b_1, b_2 in B $ such that $(b_1,c)$ and $(b_2,c) in g$. Now since f is onto so that
$exists a_1, a_2 in A $ such that $(a_1,b_1)$ and $(a_2,b_2) in f$ and so $(a_1,c)$ and $(a_2,c) in g circ f$ . So its not one to one .
Is this correct ? Thanks
functions proof-verification
functions proof-verification
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
edited Mar 20 at 9:50
J. Deff
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
asked Mar 20 at 9:39
J. DeffJ. Deff
653517
653517
$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
2
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41
add a comment |
$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
2
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41
$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
2
2
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The idea is correct, but
- when you wrote $exists a_1,a_2in A$, you should have added that $a_1neq a_2$;
- the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
The idea is correct, but
- when you wrote $exists a_1,a_2in A$, you should have added that $a_1neq a_2$;
- the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
The idea is correct, but
- when you wrote $exists a_1,a_2in A$, you should have added that $a_1neq a_2$;
- the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
The idea is correct, but
- when you wrote $exists a_1,a_2in A$, you should have added that $a_1neq a_2$;
- the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
The idea is correct, but
- when you wrote $exists a_1,a_2in A$, you should have added that $a_1neq a_2$;
- the final conclusion should be “So it's not one-to-one”, instead of “So its not Onto”.
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 20 at 9:49
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
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$begingroup$
Seems correct. First f, then g.
$endgroup$
– Wuestenfux
Mar 20 at 9:41
2
$begingroup$
$1 - 1 = 0$, use one-to-one or injective
$endgroup$
– Jens Renders
Mar 20 at 9:41