Is there any problem to my solution for Problem 2.3.9 in pinter's set theory.
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Problem 9.
Let $f:Arightarrow C$ and $g:Arightarrow B$ be functions. Prove that there exists a function $h:Brightarrow C$ such that $f=hcirc g $ if and only if $forall x,yin A$
$$g(x)=g(y) Rightarrow f(x)=f(y)$$
Prove that $h$ is unique.
$(Rightarrow)$
$g(x)=g(y)Rightarrow h(g(x))=h(g(y))Rightarrow f(x)=f(y)$
$(Leftarrow)$
Let
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_1in C; if; yin B-g(A)$$
Actually this function satisfy the suggestion in problem
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_2in C; if; yin B-g(A)$$
is also satisfy above, so $h$ is not unique, I don't know what is problem.
elementary-set-theory
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
asked Aug 24 at 1:36
백주ìƒÂ
1089
add a comment |Â
up vote
0
down vote
favorite
Problem 9.
Let $f:Arightarrow C$ and $g:Arightarrow B$ be functions. Prove that there exists a function $h:Brightarrow C$ such that $f=hcirc g $ if and only if $forall x,yin A$
$$g(x)=g(y) Rightarrow f(x)=f(y)$$
Prove that $h$ is unique.
$(Rightarrow)$
$g(x)=g(y)Rightarrow h(g(x))=h(g(y))Rightarrow f(x)=f(y)$
$(Leftarrow)$
Let
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_1in C; if; yin B-g(A)$$
Actually this function satisfy the suggestion in problem
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_2in C; if; yin B-g(A)$$
is also satisfy above, so $h$ is not unique, I don't know what is problem.
elementary-set-theory
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
asked Aug 24 at 1:36
백주ìƒÂ
1089
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem 9.
Let $f:Arightarrow C$ and $g:Arightarrow B$ be functions. Prove that there exists a function $h:Brightarrow C$ such that $f=hcirc g $ if and only if $forall x,yin A$
$$g(x)=g(y) Rightarrow f(x)=f(y)$$
Prove that $h$ is unique.
$(Rightarrow)$
$g(x)=g(y)Rightarrow h(g(x))=h(g(y))Rightarrow f(x)=f(y)$
$(Leftarrow)$
Let
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_1in C; if; yin B-g(A)$$
Actually this function satisfy the suggestion in problem
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_2in C; if; yin B-g(A)$$
is also satisfy above, so $h$ is not unique, I don't know what is problem.
elementary-set-theory
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
asked Aug 24 at 1:36
백주ìƒÂ
1089
Problem 9.
Let $f:Arightarrow C$ and $g:Arightarrow B$ be functions. Prove that there exists a function $h:Brightarrow C$ such that $f=hcirc g $ if and only if $forall x,yin A$
$$g(x)=g(y) Rightarrow f(x)=f(y)$$
Prove that $h$ is unique.
$(Rightarrow)$
$g(x)=g(y)Rightarrow h(g(x))=h(g(y))Rightarrow f(x)=f(y)$
$(Leftarrow)$
Let
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_1in C; if; yin B-g(A)$$
Actually this function satisfy the suggestion in problem
$$h(y)=Big{^f(x); if ; yin g(A);And; g(x)=y_z_2in C; if; yin B-g(A)$$
is also satisfy above, so $h$ is not unique, I don't know what is problem.
elementary-set-theory
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
asked Aug 24 at 1:36
백주ìƒÂ
1089
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
edited Aug 24 at 1:47
Andrés E. Caicedo
63.3k7153238
63.3k7153238
asked Aug 24 at 1:36
백주ìƒÂ
1089
asked Aug 24 at 1:36
백주ìƒÂ
1089
asked Aug 24 at 1:36
백주ìƒÂ
1089
1089
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
Unless there's extra information, there's no reason to think $h$ will be unique. Assume $xin B$ and $yin C$, and let $f(a)=y$ and $g(a)=x$ for all $ain A$. Then if $B$ and $C$ both have more than one element, it's easy to see there will be more than one such $h$.
TL;DR - The question as presented here is flawed; $h$ needn't be unique.
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Unless there's extra information, there's no reason to think $h$ will be unique. Assume $xin B$ and $yin C$, and let $f(a)=y$ and $g(a)=x$ for all $ain A$. Then if $B$ and $C$ both have more than one element, it's easy to see there will be more than one such $h$.
TL;DR - The question as presented here is flawed; $h$ needn't be unique.
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
add a comment |Â
up vote
1
down vote
accepted
Unless there's extra information, there's no reason to think $h$ will be unique. Assume $xin B$ and $yin C$, and let $f(a)=y$ and $g(a)=x$ for all $ain A$. Then if $B$ and $C$ both have more than one element, it's easy to see there will be more than one such $h$.
TL;DR - The question as presented here is flawed; $h$ needn't be unique.
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Unless there's extra information, there's no reason to think $h$ will be unique. Assume $xin B$ and $yin C$, and let $f(a)=y$ and $g(a)=x$ for all $ain A$. Then if $B$ and $C$ both have more than one element, it's easy to see there will be more than one such $h$.
TL;DR - The question as presented here is flawed; $h$ needn't be unique.
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
Unless there's extra information, there's no reason to think $h$ will be unique. Assume $xin B$ and $yin C$, and let $f(a)=y$ and $g(a)=x$ for all $ain A$. Then if $B$ and $C$ both have more than one element, it's easy to see there will be more than one such $h$.
TL;DR - The question as presented here is flawed; $h$ needn't be unique.
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
answered Aug 24 at 1:55
Malice Vidrine
5,56921019
5,56921019
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
add a comment |Â
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
The matter is quite different if $g$ is required to be surjective, and perhaps this is what the exercise was intended to be. But I think the exercise as it appears in the text is ill posed.
– Malice Vidrine
Aug 24 at 2:02
add a comment |Â
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