Proving that a bijection exists
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Let $f$ be a bijective function from $A$ to $B$. Let $x in A$ and $y in B$ . Prove that there exists a bijection $g$ defined from $A$ to $B$ such that $g(x)=y$ .
Here is my solution but i don't know if it is correct.
Assume that $f(x) neq y$ , because if they were equal taking $g$ the same bijection as $f$ would work.
Since $f$ is a bijection we have $|A|=|B|$ and there exists $x_1 in A$ and $y_1 in B $ such that $f(x_1)=y$ and $f(x)=y_1$
Now taking $g(a)=f(a)$ if $a neq x_1$ or$ x$ , and $g(x)=y $ and
$g(x_1)=y_1$ completes the proof.
analysis functions discrete-mathematics
asked Nov 18 at 10:33
Arben_Ajredini
184
add a comment |
up vote
2
down vote
favorite
Let $f$ be a bijective function from $A$ to $B$. Let $x in A$ and $y in B$ . Prove that there exists a bijection $g$ defined from $A$ to $B$ such that $g(x)=y$ .
Here is my solution but i don't know if it is correct.
Assume that $f(x) neq y$ , because if they were equal taking $g$ the same bijection as $f$ would work.
Since $f$ is a bijection we have $|A|=|B|$ and there exists $x_1 in A$ and $y_1 in B $ such that $f(x_1)=y$ and $f(x)=y_1$
Now taking $g(a)=f(a)$ if $a neq x_1$ or$ x$ , and $g(x)=y $ and
$g(x_1)=y_1$ completes the proof.
analysis functions discrete-mathematics
asked Nov 18 at 10:33
Arben_Ajredini
184
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f$ be a bijective function from $A$ to $B$. Let $x in A$ and $y in B$ . Prove that there exists a bijection $g$ defined from $A$ to $B$ such that $g(x)=y$ .
Here is my solution but i don't know if it is correct.
Assume that $f(x) neq y$ , because if they were equal taking $g$ the same bijection as $f$ would work.
Since $f$ is a bijection we have $|A|=|B|$ and there exists $x_1 in A$ and $y_1 in B $ such that $f(x_1)=y$ and $f(x)=y_1$
Now taking $g(a)=f(a)$ if $a neq x_1$ or$ x$ , and $g(x)=y $ and
$g(x_1)=y_1$ completes the proof.
analysis functions discrete-mathematics
asked Nov 18 at 10:33
Arben_Ajredini
184
Let $f$ be a bijective function from $A$ to $B$. Let $x in A$ and $y in B$ . Prove that there exists a bijection $g$ defined from $A$ to $B$ such that $g(x)=y$ .
Here is my solution but i don't know if it is correct.
Assume that $f(x) neq y$ , because if they were equal taking $g$ the same bijection as $f$ would work.
Since $f$ is a bijection we have $|A|=|B|$ and there exists $x_1 in A$ and $y_1 in B $ such that $f(x_1)=y$ and $f(x)=y_1$
Now taking $g(a)=f(a)$ if $a neq x_1$ or$ x$ , and $g(x)=y $ and
$g(x_1)=y_1$ completes the proof.
analysis functions discrete-mathematics
analysis functions discrete-mathematics
asked Nov 18 at 10:33
Arben_Ajredini
184
asked Nov 18 at 10:33
Arben_Ajredini
184
edited Nov 18 at 18:55
asked Nov 18 at 10:33
Arben_Ajredini
184
asked Nov 18 at 10:33
Arben_Ajredini
184
asked Nov 18 at 10:33
Arben_Ajredini
184
184
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05
add a comment |
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05
1
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05
add a comment |
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 18 at 10:36
Your proof is correct. A similar proof shows that if $f$ is any function, one-to-one, or onto, then $g$ is any function, one-to-one, or onto respectively, with $g(x)=y$.
– palmpo
Nov 18 at 16:05