For $f:Xto Y$ and $A$ a subset of $X$, why are $A$ and $f^{-1}(f(A))$ not necessarily equal?
I need help with this question:
Let $f: X to Y$ be a function, and let $A$ be a subset of $X$. Why are the sets $A$ and $f^{-1}(f(A))$ not necessarily equal? Under what conditions are they equal?
Thanks!
functions
edited Nov 29 '18 at 19:57
Blue
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
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I need help with this question:
Let $f: X to Y$ be a function, and let $A$ be a subset of $X$. Why are the sets $A$ and $f^{-1}(f(A))$ not necessarily equal? Under what conditions are they equal?
Thanks!
functions
edited Nov 29 '18 at 19:57
Blue
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
add a comment |
I need help with this question:
Let $f: X to Y$ be a function, and let $A$ be a subset of $X$. Why are the sets $A$ and $f^{-1}(f(A))$ not necessarily equal? Under what conditions are they equal?
Thanks!
functions
edited Nov 29 '18 at 19:57
Blue
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
I need help with this question:
Let $f: X to Y$ be a function, and let $A$ be a subset of $X$. Why are the sets $A$ and $f^{-1}(f(A))$ not necessarily equal? Under what conditions are they equal?
Thanks!
functions
functions
edited Nov 29 '18 at 19:57
Blue
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
edited Nov 29 '18 at 19:57
Blue
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
edited Nov 29 '18 at 19:57
Blue
47.7k870151
edited Nov 29 '18 at 19:57
Blue
47.7k870151
edited Nov 29 '18 at 19:57
Blue
47.7k870151
47.7k870151
asked Nov 29 '18 at 14:25
FridaFrida
1
asked Nov 29 '18 at 14:25
FridaFrida
1
asked Nov 29 '18 at 14:25
FridaFrida
1
1
add a comment |
add a comment |
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Consider the function $f:{1,2,3}rightarrow{2,3,4}, f(1)=2, f(2)=f(3)=3$ and let $A={1,2}subseteq{1,2,3}$.
$f(A)=f({1,2})={2,3}implies f^{-1}(f(A))=f^{-1}({2,3})={1,2,3}neq{1,2}=A$
We define the direct image of $Asubseteq X$ as $f(A)={f(x): xin A}$ and the inverse image of $Bsubseteq Y$ as $f^{-1}(B)={x: xin X, f(x)in B }$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)in f(A), forall xin A$. This means $Asubseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $exists yin X-A$ such that $f(y)in f(A)$, then $yin f^{-1}(f(A))$ but $ynotin A$.
The sufficient and necessary condition for $f^{-1}(f(A))=A, forall Asubseteq X$, is that $f$ is injective. You already know from above that $Asubseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))subseteq A,forall Asubseteq X$ given $f$ is injective?
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
add a comment |
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Consider the function $f:{1,2,3}rightarrow{2,3,4}, f(1)=2, f(2)=f(3)=3$ and let $A={1,2}subseteq{1,2,3}$.
$f(A)=f({1,2})={2,3}implies f^{-1}(f(A))=f^{-1}({2,3})={1,2,3}neq{1,2}=A$
We define the direct image of $Asubseteq X$ as $f(A)={f(x): xin A}$ and the inverse image of $Bsubseteq Y$ as $f^{-1}(B)={x: xin X, f(x)in B }$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)in f(A), forall xin A$. This means $Asubseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $exists yin X-A$ such that $f(y)in f(A)$, then $yin f^{-1}(f(A))$ but $ynotin A$.
The sufficient and necessary condition for $f^{-1}(f(A))=A, forall Asubseteq X$, is that $f$ is injective. You already know from above that $Asubseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))subseteq A,forall Asubseteq X$ given $f$ is injective?
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
add a comment |
Consider the function $f:{1,2,3}rightarrow{2,3,4}, f(1)=2, f(2)=f(3)=3$ and let $A={1,2}subseteq{1,2,3}$.
$f(A)=f({1,2})={2,3}implies f^{-1}(f(A))=f^{-1}({2,3})={1,2,3}neq{1,2}=A$
We define the direct image of $Asubseteq X$ as $f(A)={f(x): xin A}$ and the inverse image of $Bsubseteq Y$ as $f^{-1}(B)={x: xin X, f(x)in B }$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)in f(A), forall xin A$. This means $Asubseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $exists yin X-A$ such that $f(y)in f(A)$, then $yin f^{-1}(f(A))$ but $ynotin A$.
The sufficient and necessary condition for $f^{-1}(f(A))=A, forall Asubseteq X$, is that $f$ is injective. You already know from above that $Asubseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))subseteq A,forall Asubseteq X$ given $f$ is injective?
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
add a comment |
Consider the function $f:{1,2,3}rightarrow{2,3,4}, f(1)=2, f(2)=f(3)=3$ and let $A={1,2}subseteq{1,2,3}$.
$f(A)=f({1,2})={2,3}implies f^{-1}(f(A))=f^{-1}({2,3})={1,2,3}neq{1,2}=A$
We define the direct image of $Asubseteq X$ as $f(A)={f(x): xin A}$ and the inverse image of $Bsubseteq Y$ as $f^{-1}(B)={x: xin X, f(x)in B }$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)in f(A), forall xin A$. This means $Asubseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $exists yin X-A$ such that $f(y)in f(A)$, then $yin f^{-1}(f(A))$ but $ynotin A$.
The sufficient and necessary condition for $f^{-1}(f(A))=A, forall Asubseteq X$, is that $f$ is injective. You already know from above that $Asubseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))subseteq A,forall Asubseteq X$ given $f$ is injective?
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
Consider the function $f:{1,2,3}rightarrow{2,3,4}, f(1)=2, f(2)=f(3)=3$ and let $A={1,2}subseteq{1,2,3}$.
$f(A)=f({1,2})={2,3}implies f^{-1}(f(A))=f^{-1}({2,3})={1,2,3}neq{1,2}=A$
We define the direct image of $Asubseteq X$ as $f(A)={f(x): xin A}$ and the inverse image of $Bsubseteq Y$ as $f^{-1}(B)={x: xin X, f(x)in B }$. Clearly, the inverse image of $f(A)$, that is $f^{-1}(f(A)),$ contains all the elements of $A$ as $f(x)in f(A), forall xin A$. This means $Asubseteq f^{-1}(f(A))$. But this is not to say that $f^{-1}(f(A))$ contains only the elements of $A$. If $exists yin X-A$ such that $f(y)in f(A)$, then $yin f^{-1}(f(A))$ but $ynotin A$.
The sufficient and necessary condition for $f^{-1}(f(A))=A, forall Asubseteq X$, is that $f$ is injective. You already know from above that $Asubseteq f^{-1}(f(A))$. Can you prove the converse: $f^{-1}(f(A))subseteq A,forall Asubseteq X$ given $f$ is injective?
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
answered Nov 29 '18 at 15:13
Shubham JohriShubham Johri
4,684717
4,684717
add a comment |
add a comment |
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