Mixing
Simplifying a proof that $f$ and $g$ are mutually inverse functions knowing that $f$ is an injection
To prove that $f$ and $g$ are mutually inverse bijections $Ato B$ and $Bto A$, it is necessary to prove:
- that really $f:Ato B$ and $g:Bto A$;
- for every $x_0in A$ and $y=f(x_0)$ we have $x_1=x_0$ for $x_1=g(y)$;
- for every $y_0in B$ and $x=g(y_0)$ we have $y_1=y_0$ for $y_1=f(x)$.
Now suppose we already know that $f$ is an injection. Knowing this which items of the above list may be skipped without breaking the proof that they are mutually inverse?
functions inverse-function
asked Nov 20 '18 at 23:01
porton
1,90911227
add a comment |
To prove that $f$ and $g$ are mutually inverse bijections $Ato B$ and $Bto A$, it is necessary to prove:
- that really $f:Ato B$ and $g:Bto A$;
- for every $x_0in A$ and $y=f(x_0)$ we have $x_1=x_0$ for $x_1=g(y)$;
- for every $y_0in B$ and $x=g(y_0)$ we have $y_1=y_0$ for $y_1=f(x)$.
Now suppose we already know that $f$ is an injection. Knowing this which items of the above list may be skipped without breaking the proof that they are mutually inverse?
functions inverse-function
asked Nov 20 '18 at 23:01
porton
1,90911227
Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07
add a comment |
To prove that $f$ and $g$ are mutually inverse bijections $Ato B$ and $Bto A$, it is necessary to prove:
- that really $f:Ato B$ and $g:Bto A$;
- for every $x_0in A$ and $y=f(x_0)$ we have $x_1=x_0$ for $x_1=g(y)$;
- for every $y_0in B$ and $x=g(y_0)$ we have $y_1=y_0$ for $y_1=f(x)$.
Now suppose we already know that $f$ is an injection. Knowing this which items of the above list may be skipped without breaking the proof that they are mutually inverse?
functions inverse-function
asked Nov 20 '18 at 23:01
porton
1,90911227
To prove that $f$ and $g$ are mutually inverse bijections $Ato B$ and $Bto A$, it is necessary to prove:
- that really $f:Ato B$ and $g:Bto A$;
- for every $x_0in A$ and $y=f(x_0)$ we have $x_1=x_0$ for $x_1=g(y)$;
- for every $y_0in B$ and $x=g(y_0)$ we have $y_1=y_0$ for $y_1=f(x)$.
Now suppose we already know that $f$ is an injection. Knowing this which items of the above list may be skipped without breaking the proof that they are mutually inverse?
functions inverse-function
functions inverse-function
asked Nov 20 '18 at 23:01
porton
1,90911227
asked Nov 20 '18 at 23:01
porton
1,90911227
asked Nov 20 '18 at 23:01
porton
1,90911227
asked Nov 20 '18 at 23:01
porton
1,90911227
asked Nov 20 '18 at 23:01
porton
1,90911227
1,90911227
Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07
add a comment |
Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07
Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07
Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07
add a comment |
1 Answer
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It's a bit easier if we rephrase your conditions thus:
$f in A to B$ and $g in B to A$
- for all $a in A$, $g(f(a)) = a$
- for all $b in B$, $f(g(b)) = b$
If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.
So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
It's a bit easier if we rephrase your conditions thus:
$f in A to B$ and $g in B to A$
- for all $a in A$, $g(f(a)) = a$
- for all $b in B$, $f(g(b)) = b$
If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.
So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
add a comment |
It's a bit easier if we rephrase your conditions thus:
$f in A to B$ and $g in B to A$
- for all $a in A$, $g(f(a)) = a$
- for all $b in B$, $f(g(b)) = b$
If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.
So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
add a comment |
It's a bit easier if we rephrase your conditions thus:
$f in A to B$ and $g in B to A$
- for all $a in A$, $g(f(a)) = a$
- for all $b in B$, $f(g(b)) = b$
If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.
So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
It's a bit easier if we rephrase your conditions thus:
$f in A to B$ and $g in B to A$
- for all $a in A$, $g(f(a)) = a$
- for all $b in B$, $f(g(b)) = b$
If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.
So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
edited Nov 20 '18 at 23:36
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
answered Nov 20 '18 at 23:29
Rob Arthan
29.1k42866
29.1k42866
add a comment |
add a comment |
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Hm, it seems that all items involve $g$. So just removing an item would break the proof, even if I add that $f$ is an injection. Well, maybe there is any way to simplify the proof under assumption of injectivity of $f$ anyway?
– porton
Nov 20 '18 at 23:07