Mixing
Structure of Proofs of Invertibility?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
When trying to prove the invertibility of an objects (say, the composition of two functions), do we always structure the proofs by showing, not necessarily in this order, (1) that the inverse of the function composed with the function is equal to the identity, and (2) that the function composed with the inverse is equal to the identity? This would show that the function is equal to its inverse, which is our goal in the proof, right?
I ask because I'm trying to remember how to prove such theorems when asked on exams, so if they always follow this structure/format, then I'll always know how to proceed, regardless of the specifics of the theorem I'm being asked to prove.
The following is an example of one such proof that follows this structure.
Theorem: If $f: X to Y$ and $g: Y to Z$ are invertible, then $g circ f: X to Z$ is also invertible, with inverse $f^-1 circ g^-1$.
Proof: Suppose that $f: X to Y$ and $g: Y to Z$ are invertible.
We first show that the inverse of the function composed with the function is equal to the identity.
$(f^-1 circ g^-1) circ (g circ f) = f^-1 circ ( g^-1 circ g ) circ f$ (By associativity of composition.)
$= f^-1 circ (i_Y) circ f$ (Since the composition of a function with its inverse is equal to the identity on the domain of the function.)
$= f^-1 circ f$ (Since the identity on $Y$ just maps from $Y$ back to $Y$.)
$= i_X$ (Since $f^-1 circ f$ just maps from $X$ back to $X$.)
And we now prove that the function composed with the inverse is equal to the identity.
$(g circ f) circ (f^-1 circ g^-1) = g circ (f circ f^-1) circ g^-1$
$= g circ (i_Y) circ g^-1$
$= g circ g^-1$
$= i_Z$
Therefore, $(g circ f)^-1 = f^-1 circ g^-1$.
$Q.E.D.$
functions proof-writing inverse-function function-and-relation-composition
asked Jul 29 at 19:28
The Pointer
2,4482829
 |Â
show 2 more comments
up vote
0
down vote
favorite
When trying to prove the invertibility of an objects (say, the composition of two functions), do we always structure the proofs by showing, not necessarily in this order, (1) that the inverse of the function composed with the function is equal to the identity, and (2) that the function composed with the inverse is equal to the identity? This would show that the function is equal to its inverse, which is our goal in the proof, right?
I ask because I'm trying to remember how to prove such theorems when asked on exams, so if they always follow this structure/format, then I'll always know how to proceed, regardless of the specifics of the theorem I'm being asked to prove.
The following is an example of one such proof that follows this structure.
Theorem: If $f: X to Y$ and $g: Y to Z$ are invertible, then $g circ f: X to Z$ is also invertible, with inverse $f^-1 circ g^-1$.
Proof: Suppose that $f: X to Y$ and $g: Y to Z$ are invertible.
We first show that the inverse of the function composed with the function is equal to the identity.
$(f^-1 circ g^-1) circ (g circ f) = f^-1 circ ( g^-1 circ g ) circ f$ (By associativity of composition.)
$= f^-1 circ (i_Y) circ f$ (Since the composition of a function with its inverse is equal to the identity on the domain of the function.)
$= f^-1 circ f$ (Since the identity on $Y$ just maps from $Y$ back to $Y$.)
$= i_X$ (Since $f^-1 circ f$ just maps from $X$ back to $X$.)
And we now prove that the function composed with the inverse is equal to the identity.
$(g circ f) circ (f^-1 circ g^-1) = g circ (f circ f^-1) circ g^-1$
$= g circ (i_Y) circ g^-1$
$= g circ g^-1$
$= i_Z$
Therefore, $(g circ f)^-1 = f^-1 circ g^-1$.
$Q.E.D.$
functions proof-writing inverse-function function-and-relation-composition
asked Jul 29 at 19:28
The Pointer
2,4482829
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
1
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
1
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When trying to prove the invertibility of an objects (say, the composition of two functions), do we always structure the proofs by showing, not necessarily in this order, (1) that the inverse of the function composed with the function is equal to the identity, and (2) that the function composed with the inverse is equal to the identity? This would show that the function is equal to its inverse, which is our goal in the proof, right?
I ask because I'm trying to remember how to prove such theorems when asked on exams, so if they always follow this structure/format, then I'll always know how to proceed, regardless of the specifics of the theorem I'm being asked to prove.
The following is an example of one such proof that follows this structure.
Theorem: If $f: X to Y$ and $g: Y to Z$ are invertible, then $g circ f: X to Z$ is also invertible, with inverse $f^-1 circ g^-1$.
Proof: Suppose that $f: X to Y$ and $g: Y to Z$ are invertible.
We first show that the inverse of the function composed with the function is equal to the identity.
$(f^-1 circ g^-1) circ (g circ f) = f^-1 circ ( g^-1 circ g ) circ f$ (By associativity of composition.)
$= f^-1 circ (i_Y) circ f$ (Since the composition of a function with its inverse is equal to the identity on the domain of the function.)
$= f^-1 circ f$ (Since the identity on $Y$ just maps from $Y$ back to $Y$.)
$= i_X$ (Since $f^-1 circ f$ just maps from $X$ back to $X$.)
And we now prove that the function composed with the inverse is equal to the identity.
$(g circ f) circ (f^-1 circ g^-1) = g circ (f circ f^-1) circ g^-1$
$= g circ (i_Y) circ g^-1$
$= g circ g^-1$
$= i_Z$
Therefore, $(g circ f)^-1 = f^-1 circ g^-1$.
$Q.E.D.$
functions proof-writing inverse-function function-and-relation-composition
asked Jul 29 at 19:28
The Pointer
2,4482829
When trying to prove the invertibility of an objects (say, the composition of two functions), do we always structure the proofs by showing, not necessarily in this order, (1) that the inverse of the function composed with the function is equal to the identity, and (2) that the function composed with the inverse is equal to the identity? This would show that the function is equal to its inverse, which is our goal in the proof, right?
I ask because I'm trying to remember how to prove such theorems when asked on exams, so if they always follow this structure/format, then I'll always know how to proceed, regardless of the specifics of the theorem I'm being asked to prove.
The following is an example of one such proof that follows this structure.
Theorem: If $f: X to Y$ and $g: Y to Z$ are invertible, then $g circ f: X to Z$ is also invertible, with inverse $f^-1 circ g^-1$.
Proof: Suppose that $f: X to Y$ and $g: Y to Z$ are invertible.
We first show that the inverse of the function composed with the function is equal to the identity.
$(f^-1 circ g^-1) circ (g circ f) = f^-1 circ ( g^-1 circ g ) circ f$ (By associativity of composition.)
$= f^-1 circ (i_Y) circ f$ (Since the composition of a function with its inverse is equal to the identity on the domain of the function.)
$= f^-1 circ f$ (Since the identity on $Y$ just maps from $Y$ back to $Y$.)
$= i_X$ (Since $f^-1 circ f$ just maps from $X$ back to $X$.)
And we now prove that the function composed with the inverse is equal to the identity.
$(g circ f) circ (f^-1 circ g^-1) = g circ (f circ f^-1) circ g^-1$
$= g circ (i_Y) circ g^-1$
$= g circ g^-1$
$= i_Z$
Therefore, $(g circ f)^-1 = f^-1 circ g^-1$.
$Q.E.D.$
functions proof-writing inverse-function function-and-relation-composition
asked Jul 29 at 19:28
The Pointer
2,4482829
asked Jul 29 at 19:28
The Pointer
2,4482829
asked Jul 29 at 19:28
The Pointer
2,4482829
asked Jul 29 at 19:28
The Pointer
2,4482829
2,4482829
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
1
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
1
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38
 |Â
show 2 more comments
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
1
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
1
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
1
1
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
1
1
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, the proof is correct and your description of the inverse function is correct and clear. Note that not every function is its own inverse.
If $gof$ and $fog$ are both identity functions, then $f$ and $g$ are inverses of each other.
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, the proof is correct and your description of the inverse function is correct and clear. Note that not every function is its own inverse.
If $gof$ and $fog$ are both identity functions, then $f$ and $g$ are inverses of each other.
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
add a comment |Â
up vote
2
down vote
accepted
Yes, the proof is correct and your description of the inverse function is correct and clear. Note that not every function is its own inverse.
If $gof$ and $fog$ are both identity functions, then $f$ and $g$ are inverses of each other.
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, the proof is correct and your description of the inverse function is correct and clear. Note that not every function is its own inverse.
If $gof$ and $fog$ are both identity functions, then $f$ and $g$ are inverses of each other.
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
Yes, the proof is correct and your description of the inverse function is correct and clear. Note that not every function is its own inverse.
If $gof$ and $fog$ are both identity functions, then $f$ and $g$ are inverses of each other.
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 19:34
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
add a comment |Â
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
Very good. Thank you for the clarification. :)
– The Pointer
Jul 29 at 19:37
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%2f2866344%2fstructure-of-proofs-of-invertibility%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
There are different kinds of inverses : inverses in a group, inverses of a matrix , inverses of a map and so on. Not sure they can all be handled in the same way.
– Peter
Jul 29 at 19:34
@Peter Oh, ok. But what I wrote would be true for function inverses?
– The Pointer
Jul 29 at 19:34
1
Yes (see answer below). In the case of elements of a group $ab=e$ implies $ba=e$, also in the case of multiplication of square matrices $AB=I$ implies $BA=I$
– Peter
Jul 29 at 19:36
@Peter Ok, thank you for the clarification.
– The Pointer
Jul 29 at 19:37
1
In the case of maps you should always verify $f o g$ AND $g o f$
– Peter
Jul 29 at 19:38