Prove the inverse relation $f^{-1}$ of a function $f : Arightarrow B$ is a function from B to A if and only...
This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
proof-explanation relations inverse-function
asked Nov 26 at 1:45
Nicholas Cousar
302211
add a comment |
This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
proof-explanation relations inverse-function
asked Nov 26 at 1:45
Nicholas Cousar
302211
"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50
add a comment |
This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
proof-explanation relations inverse-function
asked Nov 26 at 1:45
Nicholas Cousar
302211
This proof is from "Mathematical Proofs: A Transition to Advanced Mathematics"(4th Ed.) on page 268. I understand how $f^{-1} : Bto A$ being a well-defined function implies that $f$ must be injective, but I do not understand how this condition implies that $f$ must also be surjective. Furthermore, how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
proof-explanation relations inverse-function
proof-explanation relations inverse-function
asked Nov 26 at 1:45
Nicholas Cousar
302211
asked Nov 26 at 1:45
Nicholas Cousar
302211
asked Nov 26 at 1:45
Nicholas Cousar
302211
asked Nov 26 at 1:45
Nicholas Cousar
302211
asked Nov 26 at 1:45
Nicholas Cousar
302211
302211
"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50
add a comment |
"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50
"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50
"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50
add a comment |
1 Answer
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It's actually trivial.
The definition of $f^{-1}:B to A$ being surjective is:
$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.
And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.
So $f^{-1}(f(a)) = a$.
Hence $f^{-1}$ is surjective.
====
how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.
So by fact 1) $f(a) = b$.
And by fact 2) we have $(b,a) in f^{-1}$.
ANd by fact 1) again, that means $f^{-1}(b) =a$.
So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.
And that is exactly the definition of surjective.
answered Nov 26 at 1:53
fleablood
68.1k22684
add a comment |
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It's actually trivial.
The definition of $f^{-1}:B to A$ being surjective is:
$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.
And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.
So $f^{-1}(f(a)) = a$.
Hence $f^{-1}$ is surjective.
====
how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.
So by fact 1) $f(a) = b$.
And by fact 2) we have $(b,a) in f^{-1}$.
ANd by fact 1) again, that means $f^{-1}(b) =a$.
So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.
And that is exactly the definition of surjective.
answered Nov 26 at 1:53
fleablood
68.1k22684
add a comment |
It's actually trivial.
The definition of $f^{-1}:B to A$ being surjective is:
$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.
And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.
So $f^{-1}(f(a)) = a$.
Hence $f^{-1}$ is surjective.
====
how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.
So by fact 1) $f(a) = b$.
And by fact 2) we have $(b,a) in f^{-1}$.
ANd by fact 1) again, that means $f^{-1}(b) =a$.
So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.
And that is exactly the definition of surjective.
answered Nov 26 at 1:53
fleablood
68.1k22684
add a comment |
It's actually trivial.
The definition of $f^{-1}:B to A$ being surjective is:
$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.
And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.
So $f^{-1}(f(a)) = a$.
Hence $f^{-1}$ is surjective.
====
how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.
So by fact 1) $f(a) = b$.
And by fact 2) we have $(b,a) in f^{-1}$.
ANd by fact 1) again, that means $f^{-1}(b) =a$.
So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.
And that is exactly the definition of surjective.
answered Nov 26 at 1:53
fleablood
68.1k22684
It's actually trivial.
The definition of $f^{-1}:B to A$ being surjective is:
$f^{-1}:B to A$ is surjective if for every $a in A$ there exists a $bin B$ so that $f^{-1}(b) = a$.
And as $f:A to B$ is a function for every $a in A$ there is an $f(a) in B$.
So $f^{-1}(f(a)) = a$.
Hence $f^{-1}$ is surjective.
====
how exactly does the proof of surjectivety make use of the following two facts presented at the beginning of the proof?
$f:A to B$ is a function so for every $a in A$ there is exactly one distinct $(a,b) in f$ for precisely one distinct $bin B$. That's the definition of function.
So by fact 1) $f(a) = b$.
And by fact 2) we have $(b,a) in f^{-1}$.
ANd by fact 1) again, that means $f^{-1}(b) =a$.
So for every $a in A$ we have a $b in B$ so that $f^{-1}(b) = a$.
And that is exactly the definition of surjective.
answered Nov 26 at 1:53
fleablood
68.1k22684
edited Nov 26 at 2:06
answered Nov 26 at 1:53
fleablood
68.1k22684
answered Nov 26 at 1:53
fleablood
68.1k22684
answered Nov 26 at 1:53
fleablood
68.1k22684
68.1k22684
add a comment |
add a comment |
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"how this condition implies that f must also be surjective." Because $f$ is a function. It's domain is $A$ so for all $a in A$ then $f(a)in B$ exists. And there is an $b=f(a)in B$ so that $f^{-1}(b) =a$. And that, by definition, means $f^{-1}:Bto A$ is surjective.
– fleablood
Nov 26 at 1:50