Mixing
Prove cancellation law for inverse function
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Trying to do an exercise from Analysis I, Tao.
edit: The paragraph introducting/defining inverse:
If $f$ is bijective, then for every $y in Y$, there is exactly one $x$ such that $f(x)=y$ (there is at least one because of surjectivity, and at most one because of injectivity). This value of $x$ is denoted $f^-1(y)$; thus $f^-1$ is a function from $Y$ to $X$. We call $f^-1$ the inverse of $f$.
(The paragraph stands alone, but I assume $f:X rightarrow Y$)
Exercise 3.3.6.
Let $f:Xrightarrow Y$ be a bijective function, and let $f^-1:Y rightarrow X$ be its inverse. Verify the cancellation laws $f^-1(f(x))=x$ for all $x in X$ and $f(f^-1(y))=y$ for all $y in Y$. Conclude that $f^-1$ is also invertible, and has $f$ as its inverse (thus $(f^-1)^-1=f$).
My attempt at a proof:
(1) Let $y in Y$. Since $f$ is bijective, we have exactly one $x in X$, denoted $f^-1(y)$, such that $y=f(x)=f(f^-1(y))$.
(2) Now let $x in X$. Then $f(x)=y$ for exactly one $y in Y$. Furthermore, since $f$ is bijective we have exactly one $x_0 in X$, denoted $f^-1(y)$, such that $y=f(x_0)$. Thus $x_0=x$, and so $x=f^-1(y)=f^-1(f(x))$. Observe that we have shown that for all $x in X$, there is exactly one $ y in Y$, namely $y=f(x)$, such that $f^-1(y)=x$, thus $f^-1$ is invertible and has $f$ as its inverse.
$blacksquare$
I'm not sure if the proof is correct. I believe that (1) proved the cancellation law $y=f(f^-1(y))$.
I believe that (2) proved the cancellation law $x=f^-1(f(x))$ and also that $f^-1$ is onto and one to one, thus invertible.
I'm not sure if the last claim (that $f$ is the inverse of $f^-1$) was properly justified.
real-analysis proof-verification proof-writing
asked Jul 28 at 14:30
Ken Tjhia
307
add a comment |Â
up vote
1
down vote
favorite
Trying to do an exercise from Analysis I, Tao.
edit: The paragraph introducting/defining inverse:
If $f$ is bijective, then for every $y in Y$, there is exactly one $x$ such that $f(x)=y$ (there is at least one because of surjectivity, and at most one because of injectivity). This value of $x$ is denoted $f^-1(y)$; thus $f^-1$ is a function from $Y$ to $X$. We call $f^-1$ the inverse of $f$.
(The paragraph stands alone, but I assume $f:X rightarrow Y$)
Exercise 3.3.6.
Let $f:Xrightarrow Y$ be a bijective function, and let $f^-1:Y rightarrow X$ be its inverse. Verify the cancellation laws $f^-1(f(x))=x$ for all $x in X$ and $f(f^-1(y))=y$ for all $y in Y$. Conclude that $f^-1$ is also invertible, and has $f$ as its inverse (thus $(f^-1)^-1=f$).
My attempt at a proof:
(1) Let $y in Y$. Since $f$ is bijective, we have exactly one $x in X$, denoted $f^-1(y)$, such that $y=f(x)=f(f^-1(y))$.
(2) Now let $x in X$. Then $f(x)=y$ for exactly one $y in Y$. Furthermore, since $f$ is bijective we have exactly one $x_0 in X$, denoted $f^-1(y)$, such that $y=f(x_0)$. Thus $x_0=x$, and so $x=f^-1(y)=f^-1(f(x))$. Observe that we have shown that for all $x in X$, there is exactly one $ y in Y$, namely $y=f(x)$, such that $f^-1(y)=x$, thus $f^-1$ is invertible and has $f$ as its inverse.
$blacksquare$
I'm not sure if the proof is correct. I believe that (1) proved the cancellation law $y=f(f^-1(y))$.
I believe that (2) proved the cancellation law $x=f^-1(f(x))$ and also that $f^-1$ is onto and one to one, thus invertible.
I'm not sure if the last claim (that $f$ is the inverse of $f^-1$) was properly justified.
real-analysis proof-verification proof-writing
asked Jul 28 at 14:30
Ken Tjhia
307
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Trying to do an exercise from Analysis I, Tao.
edit: The paragraph introducting/defining inverse:
If $f$ is bijective, then for every $y in Y$, there is exactly one $x$ such that $f(x)=y$ (there is at least one because of surjectivity, and at most one because of injectivity). This value of $x$ is denoted $f^-1(y)$; thus $f^-1$ is a function from $Y$ to $X$. We call $f^-1$ the inverse of $f$.
(The paragraph stands alone, but I assume $f:X rightarrow Y$)
Exercise 3.3.6.
Let $f:Xrightarrow Y$ be a bijective function, and let $f^-1:Y rightarrow X$ be its inverse. Verify the cancellation laws $f^-1(f(x))=x$ for all $x in X$ and $f(f^-1(y))=y$ for all $y in Y$. Conclude that $f^-1$ is also invertible, and has $f$ as its inverse (thus $(f^-1)^-1=f$).
My attempt at a proof:
(1) Let $y in Y$. Since $f$ is bijective, we have exactly one $x in X$, denoted $f^-1(y)$, such that $y=f(x)=f(f^-1(y))$.
(2) Now let $x in X$. Then $f(x)=y$ for exactly one $y in Y$. Furthermore, since $f$ is bijective we have exactly one $x_0 in X$, denoted $f^-1(y)$, such that $y=f(x_0)$. Thus $x_0=x$, and so $x=f^-1(y)=f^-1(f(x))$. Observe that we have shown that for all $x in X$, there is exactly one $ y in Y$, namely $y=f(x)$, such that $f^-1(y)=x$, thus $f^-1$ is invertible and has $f$ as its inverse.
$blacksquare$
I'm not sure if the proof is correct. I believe that (1) proved the cancellation law $y=f(f^-1(y))$.
I believe that (2) proved the cancellation law $x=f^-1(f(x))$ and also that $f^-1$ is onto and one to one, thus invertible.
I'm not sure if the last claim (that $f$ is the inverse of $f^-1$) was properly justified.
real-analysis proof-verification proof-writing
asked Jul 28 at 14:30
Ken Tjhia
307
Trying to do an exercise from Analysis I, Tao.
edit: The paragraph introducting/defining inverse:
If $f$ is bijective, then for every $y in Y$, there is exactly one $x$ such that $f(x)=y$ (there is at least one because of surjectivity, and at most one because of injectivity). This value of $x$ is denoted $f^-1(y)$; thus $f^-1$ is a function from $Y$ to $X$. We call $f^-1$ the inverse of $f$.
(The paragraph stands alone, but I assume $f:X rightarrow Y$)
Exercise 3.3.6.
Let $f:Xrightarrow Y$ be a bijective function, and let $f^-1:Y rightarrow X$ be its inverse. Verify the cancellation laws $f^-1(f(x))=x$ for all $x in X$ and $f(f^-1(y))=y$ for all $y in Y$. Conclude that $f^-1$ is also invertible, and has $f$ as its inverse (thus $(f^-1)^-1=f$).
My attempt at a proof:
(1) Let $y in Y$. Since $f$ is bijective, we have exactly one $x in X$, denoted $f^-1(y)$, such that $y=f(x)=f(f^-1(y))$.
(2) Now let $x in X$. Then $f(x)=y$ for exactly one $y in Y$. Furthermore, since $f$ is bijective we have exactly one $x_0 in X$, denoted $f^-1(y)$, such that $y=f(x_0)$. Thus $x_0=x$, and so $x=f^-1(y)=f^-1(f(x))$. Observe that we have shown that for all $x in X$, there is exactly one $ y in Y$, namely $y=f(x)$, such that $f^-1(y)=x$, thus $f^-1$ is invertible and has $f$ as its inverse.
$blacksquare$
I'm not sure if the proof is correct. I believe that (1) proved the cancellation law $y=f(f^-1(y))$.
I believe that (2) proved the cancellation law $x=f^-1(f(x))$ and also that $f^-1$ is onto and one to one, thus invertible.
I'm not sure if the last claim (that $f$ is the inverse of $f^-1$) was properly justified.
real-analysis proof-verification proof-writing
asked Jul 28 at 14:30
Ken Tjhia
307
edited Jul 28 at 14:49
asked Jul 28 at 14:30
Ken Tjhia
307
asked Jul 28 at 14:30
Ken Tjhia
307
asked Jul 28 at 14:30
Ken Tjhia
307
307
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53
add a comment |Â
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The definition of $f^-1$ in the first paragraph is $f(f^-1(y)) = y$.
If we let $y = f(x)$ we get $f(f^-1(f(x))) = f(x)$. Since $f$ is injective we
have $f^-1(f(x)) = x$.
To show that $f^-1$ is bijective, note that if $f^-1(a)= f^-1(b)$, then applying $f$ to both sides yields $a=b$, hence $f^-1$ is injective. Furthermore,
$f^-1(f(x)) = x$, hence $f^-1$ is surjective.
If $g$ is the inverse of $f^-1$, then, as in the first paragraph of the question, we have
$f^-1(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and
so $(f^-1)^-1 = f$.
answered Jul 28 at 15:12
copper.hat
122k557156
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The definition of $f^-1$ in the first paragraph is $f(f^-1(y)) = y$.
If we let $y = f(x)$ we get $f(f^-1(f(x))) = f(x)$. Since $f$ is injective we
have $f^-1(f(x)) = x$.
To show that $f^-1$ is bijective, note that if $f^-1(a)= f^-1(b)$, then applying $f$ to both sides yields $a=b$, hence $f^-1$ is injective. Furthermore,
$f^-1(f(x)) = x$, hence $f^-1$ is surjective.
If $g$ is the inverse of $f^-1$, then, as in the first paragraph of the question, we have
$f^-1(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and
so $(f^-1)^-1 = f$.
answered Jul 28 at 15:12
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
accepted
The definition of $f^-1$ in the first paragraph is $f(f^-1(y)) = y$.
If we let $y = f(x)$ we get $f(f^-1(f(x))) = f(x)$. Since $f$ is injective we
have $f^-1(f(x)) = x$.
To show that $f^-1$ is bijective, note that if $f^-1(a)= f^-1(b)$, then applying $f$ to both sides yields $a=b$, hence $f^-1$ is injective. Furthermore,
$f^-1(f(x)) = x$, hence $f^-1$ is surjective.
If $g$ is the inverse of $f^-1$, then, as in the first paragraph of the question, we have
$f^-1(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and
so $(f^-1)^-1 = f$.
answered Jul 28 at 15:12
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The definition of $f^-1$ in the first paragraph is $f(f^-1(y)) = y$.
If we let $y = f(x)$ we get $f(f^-1(f(x))) = f(x)$. Since $f$ is injective we
have $f^-1(f(x)) = x$.
To show that $f^-1$ is bijective, note that if $f^-1(a)= f^-1(b)$, then applying $f$ to both sides yields $a=b$, hence $f^-1$ is injective. Furthermore,
$f^-1(f(x)) = x$, hence $f^-1$ is surjective.
If $g$ is the inverse of $f^-1$, then, as in the first paragraph of the question, we have
$f^-1(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and
so $(f^-1)^-1 = f$.
answered Jul 28 at 15:12
copper.hat
122k557156
The definition of $f^-1$ in the first paragraph is $f(f^-1(y)) = y$.
If we let $y = f(x)$ we get $f(f^-1(f(x))) = f(x)$. Since $f$ is injective we
have $f^-1(f(x)) = x$.
To show that $f^-1$ is bijective, note that if $f^-1(a)= f^-1(b)$, then applying $f$ to both sides yields $a=b$, hence $f^-1$ is injective. Furthermore,
$f^-1(f(x)) = x$, hence $f^-1$ is surjective.
If $g$ is the inverse of $f^-1$, then, as in the first paragraph of the question, we have
$f^-1(g(y)) = y$, and applying $f$ to both sides yields $g(y) = f(y)$ and
so $(f^-1)^-1 = f$.
answered Jul 28 at 15:12
copper.hat
122k557156
answered Jul 28 at 15:12
copper.hat
122k557156
answered Jul 28 at 15:12
copper.hat
122k557156
answered Jul 28 at 15:12
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865294%2fprove-cancellation-law-for-inverse-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I must say that actually the exercise makes no sense to me. This because in my optics - if $f^-1$ is defined to be the inverse of $f$ then by definition we have $f^-1(f(x))=x$ and $f(f^-1(y))=y$. How is "inverse function" defined in your material?
– Vera
Jul 28 at 14:44
@Vera I added the paragraph where function inverse was discussed.
– Ken Tjhia
Jul 28 at 14:53