Let $f : Xrightarrow Y$ and $C, D subseteq Y$. Fill and prove formulas.
Definition:
Let $f: Arightarrow B$ be function and $Xsubseteq A$.
The image of set X is defined as a set $f[X] = {bin B :|: exists ain X: f(a) = b}$. Inverse image of set $Ysubseteq B$ is defined as $f^{-1} = {ain A:|:f(a)in Y}$.
Let $f : Xrightarrow Y$ and $C, D subseteq Y$. Fill and prove formulas:
$$f[f^{-1}[C]] : ? : C$$
$$f[f^{-1}[C]] = : ?$$
I have read few articles about functions, images and inverse images and found that these two formulas should probably be written the following way:
$$f[f^{-1}[C]] subseteq C text{ for all functions}$$
$$f[f^{-1}[C]] = C text{ for surjective (onto) functions}$$
I could not find any proofs of them and it is what I am looking for.
elementary-set-theory
asked Dec 8 at 23:41
whiskeyo
536
add a comment |
Definition:
Let $f: Arightarrow B$ be function and $Xsubseteq A$.
The image of set X is defined as a set $f[X] = {bin B :|: exists ain X: f(a) = b}$. Inverse image of set $Ysubseteq B$ is defined as $f^{-1} = {ain A:|:f(a)in Y}$.
Let $f : Xrightarrow Y$ and $C, D subseteq Y$. Fill and prove formulas:
$$f[f^{-1}[C]] : ? : C$$
$$f[f^{-1}[C]] = : ?$$
I have read few articles about functions, images and inverse images and found that these two formulas should probably be written the following way:
$$f[f^{-1}[C]] subseteq C text{ for all functions}$$
$$f[f^{-1}[C]] = C text{ for surjective (onto) functions}$$
I could not find any proofs of them and it is what I am looking for.
elementary-set-theory
asked Dec 8 at 23:41
whiskeyo
536
add a comment |
Definition:
Let $f: Arightarrow B$ be function and $Xsubseteq A$.
The image of set X is defined as a set $f[X] = {bin B :|: exists ain X: f(a) = b}$. Inverse image of set $Ysubseteq B$ is defined as $f^{-1} = {ain A:|:f(a)in Y}$.
Let $f : Xrightarrow Y$ and $C, D subseteq Y$. Fill and prove formulas:
$$f[f^{-1}[C]] : ? : C$$
$$f[f^{-1}[C]] = : ?$$
I have read few articles about functions, images and inverse images and found that these two formulas should probably be written the following way:
$$f[f^{-1}[C]] subseteq C text{ for all functions}$$
$$f[f^{-1}[C]] = C text{ for surjective (onto) functions}$$
I could not find any proofs of them and it is what I am looking for.
elementary-set-theory
asked Dec 8 at 23:41
whiskeyo
536
Definition:
Let $f: Arightarrow B$ be function and $Xsubseteq A$.
The image of set X is defined as a set $f[X] = {bin B :|: exists ain X: f(a) = b}$. Inverse image of set $Ysubseteq B$ is defined as $f^{-1} = {ain A:|:f(a)in Y}$.
Let $f : Xrightarrow Y$ and $C, D subseteq Y$. Fill and prove formulas:
$$f[f^{-1}[C]] : ? : C$$
$$f[f^{-1}[C]] = : ?$$
I have read few articles about functions, images and inverse images and found that these two formulas should probably be written the following way:
$$f[f^{-1}[C]] subseteq C text{ for all functions}$$
$$f[f^{-1}[C]] = C text{ for surjective (onto) functions}$$
I could not find any proofs of them and it is what I am looking for.
elementary-set-theory
elementary-set-theory
asked Dec 8 at 23:41
whiskeyo
536
asked Dec 8 at 23:41
whiskeyo
536
edited Dec 9 at 1:33
asked Dec 8 at 23:41
whiskeyo
536
asked Dec 8 at 23:41
whiskeyo
536
asked Dec 8 at 23:41
whiskeyo
536
536
add a comment |
add a comment |
1 Answer
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By definition $xin f^{-1}(C)implies f(x)in C$. Thus $f(f^{-1}(C))subseteq C$.
Now if $f$ is surjective, then for each $cin C$, $f^{-1}(c)not=emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(bigcup_{cin C}c)=f(bigcup_{cin C}f^{-1}(c))=bigcup_{cin C} f(f^{-1})(c)=bigcup_{cin C}c=C$.
answered Dec 9 at 1:11
Chris Custer
10.8k3724
add a comment |
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1 Answer
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1 Answer
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By definition $xin f^{-1}(C)implies f(x)in C$. Thus $f(f^{-1}(C))subseteq C$.
Now if $f$ is surjective, then for each $cin C$, $f^{-1}(c)not=emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(bigcup_{cin C}c)=f(bigcup_{cin C}f^{-1}(c))=bigcup_{cin C} f(f^{-1})(c)=bigcup_{cin C}c=C$.
answered Dec 9 at 1:11
Chris Custer
10.8k3724
add a comment |
By definition $xin f^{-1}(C)implies f(x)in C$. Thus $f(f^{-1}(C))subseteq C$.
Now if $f$ is surjective, then for each $cin C$, $f^{-1}(c)not=emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(bigcup_{cin C}c)=f(bigcup_{cin C}f^{-1}(c))=bigcup_{cin C} f(f^{-1})(c)=bigcup_{cin C}c=C$.
answered Dec 9 at 1:11
Chris Custer
10.8k3724
add a comment |
By definition $xin f^{-1}(C)implies f(x)in C$. Thus $f(f^{-1}(C))subseteq C$.
Now if $f$ is surjective, then for each $cin C$, $f^{-1}(c)not=emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(bigcup_{cin C}c)=f(bigcup_{cin C}f^{-1}(c))=bigcup_{cin C} f(f^{-1})(c)=bigcup_{cin C}c=C$.
answered Dec 9 at 1:11
Chris Custer
10.8k3724
By definition $xin f^{-1}(C)implies f(x)in C$. Thus $f(f^{-1}(C))subseteq C$.
Now if $f$ is surjective, then for each $cin C$, $f^{-1}(c)not=emptyset$. Hence $f(f^{-1})(c)=c$. So $f(f^{-1})(C)=f(f^{-1})(bigcup_{cin C}c)=f(bigcup_{cin C}f^{-1}(c))=bigcup_{cin C} f(f^{-1})(c)=bigcup_{cin C}c=C$.
answered Dec 9 at 1:11
Chris Custer
10.8k3724
answered Dec 9 at 1:11
Chris Custer
10.8k3724
answered Dec 9 at 1:11
Chris Custer
10.8k3724
answered Dec 9 at 1:11
Chris Custer
10.8k3724
10.8k3724
add a comment |
add a comment |
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