Characterization of invertible functions
$begingroup$
Lemma
I have managed to prove first part that f is bijection.
Attempt
I'm stuck on proving that $g=h=f^{-1}$
I tried the set theoretic approach, trying to show that the $2$ functions $g$ and $f^{-1}$ are subsets of each other.
Now I just need to show that $b=b'$ so that $(a,b)$ belongs to $f^{-1}$
Any help would be appreciated.
elementary-set-theory inverse
edited Jan 7 at 9:28
drhab
103k545136
asked Jan 7 at 8:58
799169799169
102
$endgroup$
add a comment |
$begingroup$
Lemma
I have managed to prove first part that f is bijection.
Attempt
I'm stuck on proving that $g=h=f^{-1}$
I tried the set theoretic approach, trying to show that the $2$ functions $g$ and $f^{-1}$ are subsets of each other.
Now I just need to show that $b=b'$ so that $(a,b)$ belongs to $f^{-1}$
Any help would be appreciated.
elementary-set-theory inverse
edited Jan 7 at 9:28
drhab
103k545136
asked Jan 7 at 8:58
799169799169
102
$endgroup$
add a comment |
$begingroup$
Lemma
I have managed to prove first part that f is bijection.
Attempt
I'm stuck on proving that $g=h=f^{-1}$
I tried the set theoretic approach, trying to show that the $2$ functions $g$ and $f^{-1}$ are subsets of each other.
Now I just need to show that $b=b'$ so that $(a,b)$ belongs to $f^{-1}$
Any help would be appreciated.
elementary-set-theory inverse
edited Jan 7 at 9:28
drhab
103k545136
asked Jan 7 at 8:58
799169799169
102
$endgroup$
Lemma
I have managed to prove first part that f is bijection.
Attempt
I'm stuck on proving that $g=h=f^{-1}$
I tried the set theoretic approach, trying to show that the $2$ functions $g$ and $f^{-1}$ are subsets of each other.
Now I just need to show that $b=b'$ so that $(a,b)$ belongs to $f^{-1}$
Any help would be appreciated.
elementary-set-theory inverse
elementary-set-theory inverse
edited Jan 7 at 9:28
drhab
103k545136
asked Jan 7 at 8:58
799169799169
102
edited Jan 7 at 9:28
drhab
103k545136
asked Jan 7 at 8:58
799169799169
102
edited Jan 7 at 9:28
drhab
103k545136
edited Jan 7 at 9:28
drhab
103k545136
edited Jan 7 at 9:28
drhab
103k545136
103k545136
asked Jan 7 at 8:58
799169799169
102
asked Jan 7 at 8:58
799169799169
102
asked Jan 7 at 8:58
799169799169
102
102
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
Let it be that $gcirc f=mathsf{id}_A$ and $fcirc h=mathsf{id}_B$.
Then: $$g=gcircmathsf{id}_B=gcirc(fcirc h)=(gcirc f)circ h=mathsf{id}_Acirc h=h$$
The first and fifth equalities are based on neutrality of identities.
The third equality is based on associativity of composition.
answered Jan 7 at 9:26
drhabdrhab
103k545136
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let it be that $gcirc f=mathsf{id}_A$ and $fcirc h=mathsf{id}_B$.
Then: $$g=gcircmathsf{id}_B=gcirc(fcirc h)=(gcirc f)circ h=mathsf{id}_Acirc h=h$$
The first and fifth equalities are based on neutrality of identities.
The third equality is based on associativity of composition.
answered Jan 7 at 9:26
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Let it be that $gcirc f=mathsf{id}_A$ and $fcirc h=mathsf{id}_B$.
Then: $$g=gcircmathsf{id}_B=gcirc(fcirc h)=(gcirc f)circ h=mathsf{id}_Acirc h=h$$
The first and fifth equalities are based on neutrality of identities.
The third equality is based on associativity of composition.
answered Jan 7 at 9:26
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Let it be that $gcirc f=mathsf{id}_A$ and $fcirc h=mathsf{id}_B$.
Then: $$g=gcircmathsf{id}_B=gcirc(fcirc h)=(gcirc f)circ h=mathsf{id}_Acirc h=h$$
The first and fifth equalities are based on neutrality of identities.
The third equality is based on associativity of composition.
answered Jan 7 at 9:26
drhabdrhab
103k545136
$endgroup$
Let it be that $gcirc f=mathsf{id}_A$ and $fcirc h=mathsf{id}_B$.
Then: $$g=gcircmathsf{id}_B=gcirc(fcirc h)=(gcirc f)circ h=mathsf{id}_Acirc h=h$$
The first and fifth equalities are based on neutrality of identities.
The third equality is based on associativity of composition.
answered Jan 7 at 9:26
drhabdrhab
103k545136
edited Jan 8 at 6:43
answered Jan 7 at 9:26
drhabdrhab
103k545136
answered Jan 7 at 9:26
drhabdrhab
103k545136
answered Jan 7 at 9:26
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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