surjectivity equivalent to injectivity $fcirc f circ f=f$
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Let $f:Emapsto E$ be a map such that $fcirc fcirc f = f$.
Show that: $$f text{ injective} Leftrightarrow f text{ surjective}$$
Let $xin E$, then $f(f(f(x)))=f(x)$.
We have:
$$f text{ injective} Leftrightarrow forall xin E,;f(f(x))=x$$
$$ Leftrightarrow forall xin E,;x=f(z) text{ and } z=f(x)$$
$$ Leftrightarrow f text{ surjective}$$.
Do you validate this answer ?
real-analysis proof-verification applications
edited Nov 12 at 21:15
rtybase
10k21433
asked Nov 12 at 20:47
Smilia
576516
add a comment |
up vote
0
down vote
favorite
Let $f:Emapsto E$ be a map such that $fcirc fcirc f = f$.
Show that: $$f text{ injective} Leftrightarrow f text{ surjective}$$
Let $xin E$, then $f(f(f(x)))=f(x)$.
We have:
$$f text{ injective} Leftrightarrow forall xin E,;f(f(x))=x$$
$$ Leftrightarrow forall xin E,;x=f(z) text{ and } z=f(x)$$
$$ Leftrightarrow f text{ surjective}$$.
Do you validate this answer ?
real-analysis proof-verification applications
edited Nov 12 at 21:15
rtybase
10k21433
asked Nov 12 at 20:47
Smilia
576516
1
I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:Emapsto E$ be a map such that $fcirc fcirc f = f$.
Show that: $$f text{ injective} Leftrightarrow f text{ surjective}$$
Let $xin E$, then $f(f(f(x)))=f(x)$.
We have:
$$f text{ injective} Leftrightarrow forall xin E,;f(f(x))=x$$
$$ Leftrightarrow forall xin E,;x=f(z) text{ and } z=f(x)$$
$$ Leftrightarrow f text{ surjective}$$.
Do you validate this answer ?
real-analysis proof-verification applications
edited Nov 12 at 21:15
rtybase
10k21433
asked Nov 12 at 20:47
Smilia
576516
Let $f:Emapsto E$ be a map such that $fcirc fcirc f = f$.
Show that: $$f text{ injective} Leftrightarrow f text{ surjective}$$
Let $xin E$, then $f(f(f(x)))=f(x)$.
We have:
$$f text{ injective} Leftrightarrow forall xin E,;f(f(x))=x$$
$$ Leftrightarrow forall xin E,;x=f(z) text{ and } z=f(x)$$
$$ Leftrightarrow f text{ surjective}$$.
Do you validate this answer ?
real-analysis proof-verification applications
real-analysis proof-verification applications
edited Nov 12 at 21:15
rtybase
10k21433
asked Nov 12 at 20:47
Smilia
576516
edited Nov 12 at 21:15
rtybase
10k21433
asked Nov 12 at 20:47
Smilia
576516
edited Nov 12 at 21:15
rtybase
10k21433
edited Nov 12 at 21:15
rtybase
10k21433
edited Nov 12 at 21:15
rtybase
10k21433
10k21433
asked Nov 12 at 20:47
Smilia
576516
asked Nov 12 at 20:47
Smilia
576516
asked Nov 12 at 20:47
Smilia
576516
576516
1
I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01
add a comment |
1
I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01
1
1
I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01
add a comment |
1 Answer
1
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up vote
1
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Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $xin E$, there is a $yin E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.
answered Nov 12 at 21:04
user3482749
72939
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $xin E$, there is a $yin E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.
answered Nov 12 at 21:04
user3482749
72939
add a comment |
up vote
1
down vote
accepted
Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $xin E$, there is a $yin E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.
answered Nov 12 at 21:04
user3482749
72939
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $xin E$, there is a $yin E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.
answered Nov 12 at 21:04
user3482749
72939
Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $xin E$, there is a $yin E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.
answered Nov 12 at 21:04
user3482749
72939
answered Nov 12 at 21:04
user3482749
72939
answered Nov 12 at 21:04
user3482749
72939
answered Nov 12 at 21:04
user3482749
72939
72939
add a comment |
add a comment |
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I think the last equivalence does not hold either. But if you replace the equivalences with implications, then I think it is correct. Thus you proved that $f$ injective implies $f$ surjective.
– araomis
Nov 12 at 20:56
oh right I agree but then I reflect about surjectivity implies injectivity
– Smilia
Nov 12 at 21:01