Mixing
Prove that $(((f_{*})^*)^*)_{*}: P(P(P(P(A))))rightarrow P(P(P(P(B))))$ is injective
$begingroup$
Let $f^*$ be the image of $f$ and $f_{*}$ be the preimage of $f$. How can I prove (step by step, using definitions of images and preimages) that if $f: A rightarrow B$ is injective, then $$(((f_{*})^*)^*)_{*}: P(P(P(P(A))))rightarrow P(P(P(P(B))))$$ is injective?
functions logic
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
$endgroup$
add a comment |
$begingroup$
Let $f^*$ be the image of $f$ and $f_{*}$ be the preimage of $f$. How can I prove (step by step, using definitions of images and preimages) that if $f: A rightarrow B$ is injective, then $$(((f_{*})^*)^*)_{*}: P(P(P(P(A))))rightarrow P(P(P(P(B))))$$ is injective?
functions logic
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
$endgroup$
1
$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
1
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34
add a comment |
$begingroup$
Let $f^*$ be the image of $f$ and $f_{*}$ be the preimage of $f$. How can I prove (step by step, using definitions of images and preimages) that if $f: A rightarrow B$ is injective, then $$(((f_{*})^*)^*)_{*}: P(P(P(P(A))))rightarrow P(P(P(P(B))))$$ is injective?
functions logic
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
$endgroup$
Let $f^*$ be the image of $f$ and $f_{*}$ be the preimage of $f$. How can I prove (step by step, using definitions of images and preimages) that if $f: A rightarrow B$ is injective, then $$(((f_{*})^*)^*)_{*}: P(P(P(P(A))))rightarrow P(P(P(P(B))))$$ is injective?
functions logic
functions logic
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
edited Jan 10 at 22:09
Alex Kruckman
27.2k22557
27.2k22557
asked Jan 10 at 21:06
nene123nene123
284
asked Jan 10 at 21:06
nene123nene123
284
asked Jan 10 at 21:06
nene123nene123
284
284
1
$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
1
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34
add a comment |
1
$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
1
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34
1
1
$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
1
1
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Prove the following claims:
- If $f$ is injective, then $f_*$ is injective.
- If $f$ is injective, then $f^*$ is surjective.
- If $f$ is surjective, then $f_*$ is surjective.
- If $f$ is surjective, then $f^*$ is injective.
Now $f$ injective $Rightarrow$ $f_*$ injective $Rightarrow$ $(f_*)^*$ surjective $Rightarrow$ $((f_*)^*)^*$ injective $Rightarrow$ $(((f_*)^*)^*)_*$ injective.
By 1, then 2, then 4, then 1. (Ok, so you don't have to prove 3 if you don't want to).
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
add a comment |
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$begingroup$
Hint: Prove the following claims:
- If $f$ is injective, then $f_*$ is injective.
- If $f$ is injective, then $f^*$ is surjective.
- If $f$ is surjective, then $f_*$ is surjective.
- If $f$ is surjective, then $f^*$ is injective.
Now $f$ injective $Rightarrow$ $f_*$ injective $Rightarrow$ $(f_*)^*$ surjective $Rightarrow$ $((f_*)^*)^*$ injective $Rightarrow$ $(((f_*)^*)^*)_*$ injective.
By 1, then 2, then 4, then 1. (Ok, so you don't have to prove 3 if you don't want to).
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
add a comment |
$begingroup$
Hint: Prove the following claims:
- If $f$ is injective, then $f_*$ is injective.
- If $f$ is injective, then $f^*$ is surjective.
- If $f$ is surjective, then $f_*$ is surjective.
- If $f$ is surjective, then $f^*$ is injective.
Now $f$ injective $Rightarrow$ $f_*$ injective $Rightarrow$ $(f_*)^*$ surjective $Rightarrow$ $((f_*)^*)^*$ injective $Rightarrow$ $(((f_*)^*)^*)_*$ injective.
By 1, then 2, then 4, then 1. (Ok, so you don't have to prove 3 if you don't want to).
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
add a comment |
$begingroup$
Hint: Prove the following claims:
- If $f$ is injective, then $f_*$ is injective.
- If $f$ is injective, then $f^*$ is surjective.
- If $f$ is surjective, then $f_*$ is surjective.
- If $f$ is surjective, then $f^*$ is injective.
Now $f$ injective $Rightarrow$ $f_*$ injective $Rightarrow$ $(f_*)^*$ surjective $Rightarrow$ $((f_*)^*)^*$ injective $Rightarrow$ $(((f_*)^*)^*)_*$ injective.
By 1, then 2, then 4, then 1. (Ok, so you don't have to prove 3 if you don't want to).
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
$endgroup$
Hint: Prove the following claims:
- If $f$ is injective, then $f_*$ is injective.
- If $f$ is injective, then $f^*$ is surjective.
- If $f$ is surjective, then $f_*$ is surjective.
- If $f$ is surjective, then $f^*$ is injective.
Now $f$ injective $Rightarrow$ $f_*$ injective $Rightarrow$ $(f_*)^*$ surjective $Rightarrow$ $((f_*)^*)^*$ injective $Rightarrow$ $(((f_*)^*)^*)_*$ injective.
By 1, then 2, then 4, then 1. (Ok, so you don't have to prove 3 if you don't want to).
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
answered Jan 10 at 21:53
Alex KruckmanAlex Kruckman
27.2k22557
27.2k22557
add a comment |
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$begingroup$
Guess it's easier to prove that $f$ injective $Rightarrow$ $f^*$ injective and $f_*$ injective. The result will follow.
$endgroup$
– lisyarus
Jan 10 at 21:10
1
$begingroup$
@lisyarus I think it would be harder to prove that claim, because it's not true ($f^*$ need not be injective).
$endgroup$
– Alex Kruckman
Jan 10 at 21:46
$begingroup$
@AlexKruckman Oh, of course, thank you. I meant what's in your excellent answer.
$endgroup$
– lisyarus
Jan 10 at 23:34