Mixing
Let f:$A to B$ and $g:B to A$. Prove that…
$begingroup$
Let $f:Ato B$ and $g:B to A$. Prove that $f$ is one-to-one and onto if $f∘g$ is one-to-one and $g∘f$ is onto.
I realize the logic behind it, I realize $g∘f:B to B$ being surjective (onto) implies that everything in $B$ is being hit. Now $f∘g:A to A$ being injective means that there exists only one path from $A$ to $A$. I am just not sure how to combine the information to show that $f$ is bijective.
elementary-set-theory
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
$endgroup$
add a comment |
$begingroup$
Let $f:Ato B$ and $g:B to A$. Prove that $f$ is one-to-one and onto if $f∘g$ is one-to-one and $g∘f$ is onto.
I realize the logic behind it, I realize $g∘f:B to B$ being surjective (onto) implies that everything in $B$ is being hit. Now $f∘g:A to A$ being injective means that there exists only one path from $A$ to $A$. I am just not sure how to combine the information to show that $f$ is bijective.
elementary-set-theory
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
$endgroup$
add a comment |
$begingroup$
Let $f:Ato B$ and $g:B to A$. Prove that $f$ is one-to-one and onto if $f∘g$ is one-to-one and $g∘f$ is onto.
I realize the logic behind it, I realize $g∘f:B to B$ being surjective (onto) implies that everything in $B$ is being hit. Now $f∘g:A to A$ being injective means that there exists only one path from $A$ to $A$. I am just not sure how to combine the information to show that $f$ is bijective.
elementary-set-theory
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
$endgroup$
Let $f:Ato B$ and $g:B to A$. Prove that $f$ is one-to-one and onto if $f∘g$ is one-to-one and $g∘f$ is onto.
I realize the logic behind it, I realize $g∘f:B to B$ being surjective (onto) implies that everything in $B$ is being hit. Now $f∘g:A to A$ being injective means that there exists only one path from $A$ to $A$. I am just not sure how to combine the information to show that $f$ is bijective.
elementary-set-theory
elementary-set-theory
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 1:12
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
asked Jan 27 at 0:34
ragnvaldr.jsragnvaldr.js
85
85
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
If $f circ g$ is injective, $g$ has to be injective. Indeed, let $g(a) = g(b)$. Then $(f circ g)(a) = (f circ g)(b)$ and by the injectivity of $f circ g$, $a = b$. Similarly, surjectivity of $g circ f$ implies surjectivity of $g$. Hence $g$ is bijective. Now apply $g^{-1}$ to see that $f$ is also bijective.
answered Jan 27 at 0:42
KlausKlaus
2,732113
$endgroup$
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
If $f circ g$ is injective, $g$ has to be injective. Indeed, let $g(a) = g(b)$. Then $(f circ g)(a) = (f circ g)(b)$ and by the injectivity of $f circ g$, $a = b$. Similarly, surjectivity of $g circ f$ implies surjectivity of $g$. Hence $g$ is bijective. Now apply $g^{-1}$ to see that $f$ is also bijective.
answered Jan 27 at 0:42
KlausKlaus
2,732113
$endgroup$
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
add a comment |
$begingroup$
If $f circ g$ is injective, $g$ has to be injective. Indeed, let $g(a) = g(b)$. Then $(f circ g)(a) = (f circ g)(b)$ and by the injectivity of $f circ g$, $a = b$. Similarly, surjectivity of $g circ f$ implies surjectivity of $g$. Hence $g$ is bijective. Now apply $g^{-1}$ to see that $f$ is also bijective.
answered Jan 27 at 0:42
KlausKlaus
2,732113
$endgroup$
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
add a comment |
$begingroup$
If $f circ g$ is injective, $g$ has to be injective. Indeed, let $g(a) = g(b)$. Then $(f circ g)(a) = (f circ g)(b)$ and by the injectivity of $f circ g$, $a = b$. Similarly, surjectivity of $g circ f$ implies surjectivity of $g$. Hence $g$ is bijective. Now apply $g^{-1}$ to see that $f$ is also bijective.
answered Jan 27 at 0:42
KlausKlaus
2,732113
$endgroup$
If $f circ g$ is injective, $g$ has to be injective. Indeed, let $g(a) = g(b)$. Then $(f circ g)(a) = (f circ g)(b)$ and by the injectivity of $f circ g$, $a = b$. Similarly, surjectivity of $g circ f$ implies surjectivity of $g$. Hence $g$ is bijective. Now apply $g^{-1}$ to see that $f$ is also bijective.
answered Jan 27 at 0:42
KlausKlaus
2,732113
answered Jan 27 at 0:42
KlausKlaus
2,732113
answered Jan 27 at 0:42
KlausKlaus
2,732113
answered Jan 27 at 0:42
KlausKlaus
2,732113
2,732113
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
add a comment |
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
What do you mean by "apply g inverse"?
$endgroup$
– ragnvaldr.js
Jan 27 at 0:47
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
Consider $f = f circ g circ g^{-1} = g^{-1} circ g circ f$.
$endgroup$
– Klaus
Jan 27 at 0:49
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
$begingroup$
I haven't seen that before. I knew how to prove it for g, but that last step was what I was missing. Thanks.
$endgroup$
– ragnvaldr.js
Jan 27 at 0:52
add a comment |
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