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Identity function
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I was told that identity functions have the definition when followed.
If f ∘ e = f, e = idx. If e ∘ f = f, e = idy.
The textbook had a question below the definition asking, for any functions f and g, does either of g ∘ f = idx, f ∘ g = idy implies the other.
My answer was,
f ∘ g = idy
f ∘ g ∘ f = idy ∘ f = f
f ∘ (g ∘ f) = f
(g ∘ f) = idx (from the definition of f ∘ e = f, e = idx)
and similarly
g ∘ f = idx
f ∘ g ∘ f = f ∘ idx = f
(f ∘ g) ∘ f = f
(f ∘ g) = idy (from the definition of e ∘ f = f, e = idy)
But the answer was, no they do not imply the other and it didn't give me the proof, but showed a counter example.
counter example
Okay, I understand the counter example, but where in my proof has gone wrong? I didn't assume anything, just followed definition in the proof.
Edit: The text book says, (translated)
Let's figure out whether identity and inverse elements exist for composites. The identity element is a function e that satisfies, f ∘ e = e ∘ f = f for all f: X->Y. However, the function that satisfies f ∘ e = f is e: X->X, and the function that satisfies e ∘ f = f is e: Y->Y and hence they are different. Therefore, identity element of composites does not exist nor the inverse element. Despite this, we define the function e that satisfies f ∘ e = f as the identity function of X and write as id_X, and the function e that satisfies e ∘ f = f as the identity function of Y and write as id_Y. For each set, there exists one identity function.
Edit2: updated some lines for readability
functions function-and-relation-composition
asked Jul 14 at 22:03
Seung
63
 |Â
show 11 more comments
up vote
0
down vote
favorite
I was told that identity functions have the definition when followed.
If f ∘ e = f, e = idx. If e ∘ f = f, e = idy.
The textbook had a question below the definition asking, for any functions f and g, does either of g ∘ f = idx, f ∘ g = idy implies the other.
My answer was,
f ∘ g = idy
f ∘ g ∘ f = idy ∘ f = f
f ∘ (g ∘ f) = f
(g ∘ f) = idx (from the definition of f ∘ e = f, e = idx)
and similarly
g ∘ f = idx
f ∘ g ∘ f = f ∘ idx = f
(f ∘ g) ∘ f = f
(f ∘ g) = idy (from the definition of e ∘ f = f, e = idy)
But the answer was, no they do not imply the other and it didn't give me the proof, but showed a counter example.
counter example
Okay, I understand the counter example, but where in my proof has gone wrong? I didn't assume anything, just followed definition in the proof.
Edit: The text book says, (translated)
Let's figure out whether identity and inverse elements exist for composites. The identity element is a function e that satisfies, f ∘ e = e ∘ f = f for all f: X->Y. However, the function that satisfies f ∘ e = f is e: X->X, and the function that satisfies e ∘ f = f is e: Y->Y and hence they are different. Therefore, identity element of composites does not exist nor the inverse element. Despite this, we define the function e that satisfies f ∘ e = f as the identity function of X and write as id_X, and the function e that satisfies e ∘ f = f as the identity function of Y and write as id_Y. For each set, there exists one identity function.
Edit2: updated some lines for readability
functions function-and-relation-composition
asked Jul 14 at 22:03
Seung
63
Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
1
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
1
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36
 |Â
show 11 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was told that identity functions have the definition when followed.
If f ∘ e = f, e = idx. If e ∘ f = f, e = idy.
The textbook had a question below the definition asking, for any functions f and g, does either of g ∘ f = idx, f ∘ g = idy implies the other.
My answer was,
f ∘ g = idy
f ∘ g ∘ f = idy ∘ f = f
f ∘ (g ∘ f) = f
(g ∘ f) = idx (from the definition of f ∘ e = f, e = idx)
and similarly
g ∘ f = idx
f ∘ g ∘ f = f ∘ idx = f
(f ∘ g) ∘ f = f
(f ∘ g) = idy (from the definition of e ∘ f = f, e = idy)
But the answer was, no they do not imply the other and it didn't give me the proof, but showed a counter example.
counter example
Okay, I understand the counter example, but where in my proof has gone wrong? I didn't assume anything, just followed definition in the proof.
Edit: The text book says, (translated)
Let's figure out whether identity and inverse elements exist for composites. The identity element is a function e that satisfies, f ∘ e = e ∘ f = f for all f: X->Y. However, the function that satisfies f ∘ e = f is e: X->X, and the function that satisfies e ∘ f = f is e: Y->Y and hence they are different. Therefore, identity element of composites does not exist nor the inverse element. Despite this, we define the function e that satisfies f ∘ e = f as the identity function of X and write as id_X, and the function e that satisfies e ∘ f = f as the identity function of Y and write as id_Y. For each set, there exists one identity function.
Edit2: updated some lines for readability
functions function-and-relation-composition
asked Jul 14 at 22:03
Seung
63
I was told that identity functions have the definition when followed.
If f ∘ e = f, e = idx. If e ∘ f = f, e = idy.
The textbook had a question below the definition asking, for any functions f and g, does either of g ∘ f = idx, f ∘ g = idy implies the other.
My answer was,
f ∘ g = idy
f ∘ g ∘ f = idy ∘ f = f
f ∘ (g ∘ f) = f
(g ∘ f) = idx (from the definition of f ∘ e = f, e = idx)
and similarly
g ∘ f = idx
f ∘ g ∘ f = f ∘ idx = f
(f ∘ g) ∘ f = f
(f ∘ g) = idy (from the definition of e ∘ f = f, e = idy)
But the answer was, no they do not imply the other and it didn't give me the proof, but showed a counter example.
counter example
Okay, I understand the counter example, but where in my proof has gone wrong? I didn't assume anything, just followed definition in the proof.
Edit: The text book says, (translated)
Let's figure out whether identity and inverse elements exist for composites. The identity element is a function e that satisfies, f ∘ e = e ∘ f = f for all f: X->Y. However, the function that satisfies f ∘ e = f is e: X->X, and the function that satisfies e ∘ f = f is e: Y->Y and hence they are different. Therefore, identity element of composites does not exist nor the inverse element. Despite this, we define the function e that satisfies f ∘ e = f as the identity function of X and write as id_X, and the function e that satisfies e ∘ f = f as the identity function of Y and write as id_Y. For each set, there exists one identity function.
Edit2: updated some lines for readability
functions function-and-relation-composition
asked Jul 14 at 22:03
Seung
63
edited Jul 14 at 23:26
asked Jul 14 at 22:03
Seung
63
asked Jul 14 at 22:03
Seung
63
asked Jul 14 at 22:03
Seung
63
63
Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
1
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
1
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36
 |Â
show 11 more comments
Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
1
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
1
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36
Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
1
1
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
1
1
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36
 |Â
show 11 more comments
2 Answers
2
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up vote
1
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The problem is that you have no justification for jumping from$$fcirc(gcirc f)=fimplies gcirc f=operatornameid_X,tag1$$to $gcirc f=operatornameid_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $fcirc g=operatornameid_Y$ is that $f$ is surjective, not that it is injective.
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
 |Â
show 2 more comments
up vote
0
down vote
Okay.
You have two sets, $X$ and $Y$ and a class of functions that map $X to Y$. They want to know if there is a function $id$ so that $idcirc f = fcirc id = f$ for any $f: Xto Y$.
I'm actually not sure why they even bothered to ask as the $idcirc f = f$ would require that $id: Yto Y$ and $fcirc id = f$ would require $id:Xto X$ which is impossible as $X$ and $Y$ are different sets.
However then it asks if there are $id_X: Xto X$ that can act as a right identity with $fcirc id_X = f$ for all $f:Xto Y$. The answer is, yes, the identity element $i:Xto X$ where $i(x) = x$ for all $x in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X to X$.
(If $h(x) = w ne x$ then there is some function $f$ where $f(w) ne f(x)$ and so $f(h(x)) = f(w) ne f(x) $. So the only function that works is $i(x) = x$.)
Likewise $i:Yto Y$ $i(y) = y$ can act as $id_Y: Yto Y$ that can act as a left identity with $id_Ycirc f = f$ for all $f: Xto Y$.
So they are asking if you have a $g:Yto X$ so that $gcirc f: Xto X = id_X$ or $fcirc g: Yto Y= id_Y$ does one imply the other.
And the answer is no.
The thing is if $|X| > |Y|$ then $f: X to Y$ can not be injective so there must be $win X,x in X$ and $wne x$ so that $f(x)=f(w)$. $gcirc f:Xto X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $gcirc f(w) = g(f(w)) = g(f(x)) = gcirc f(x)$.
answered Jul 14 at 23:27
fleablood
60.5k22575
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The problem is that you have no justification for jumping from$$fcirc(gcirc f)=fimplies gcirc f=operatornameid_X,tag1$$to $gcirc f=operatornameid_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $fcirc g=operatornameid_Y$ is that $f$ is surjective, not that it is injective.
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
 |Â
show 2 more comments
up vote
1
down vote
The problem is that you have no justification for jumping from$$fcirc(gcirc f)=fimplies gcirc f=operatornameid_X,tag1$$to $gcirc f=operatornameid_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $fcirc g=operatornameid_Y$ is that $f$ is surjective, not that it is injective.
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
The problem is that you have no justification for jumping from$$fcirc(gcirc f)=fimplies gcirc f=operatornameid_X,tag1$$to $gcirc f=operatornameid_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $fcirc g=operatornameid_Y$ is that $f$ is surjective, not that it is injective.
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
The problem is that you have no justification for jumping from$$fcirc(gcirc f)=fimplies gcirc f=operatornameid_X,tag1$$to $gcirc f=operatornameid_X$. In fact, if $f$ was injective then, yes, you could deduce that $(1)$ hods, but all that you can deduce from $fcirc g=operatornameid_Y$ is that $f$ is surjective, not that it is injective.
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
edited Jul 14 at 22:14
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
answered Jul 14 at 22:08
José Carlos Santos
114k1698177
114k1698177
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
 |Â
show 2 more comments
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
f∘(g∘f)=f is in a from of f ∘ e = f isn't it? So from the definition I am given, g∘f = e = idx.
– Seung
Jul 14 at 22:25
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
I have no idea about what you mean with “is in a form ofâ€Â. Yes, $fcircoperatornameid_X=f$, but what you asserted was that $fcirc(gcirc f)=fimplies gcirc f=operatornameid_X$, without justification.
– José Carlos Santos
Jul 14 at 22:30
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
Okay if I say h = g o f, f o (g o f) = f becomes f o h = f. What I am saying is that this is the definition of idx I am given. f o e = f, e = idx. So f o (g o f) = f, (g o f) = idx. Sorry for the unbearable readability here.
– Seung
Jul 14 at 22:35
1
1
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
Yes, $operatornameid_X$ is the only function from $X$ into itself such that for every $fcolon Xlongrightarrow Y$, $fcircoperatornameid_X=f$. But for a specific $fcolon Xlongrightarrow Y$ there may be other functions from $X$ into itself (besides $operatornameid_X$) with the same property.
– José Carlos Santos
Jul 14 at 22:41
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
For the first line of your comment: Are you saying the id_X is universal? For all functions, they have the same id_X? I thought id_X is a property of a function. Second line: That means there may exist multiple functions e that satisfies f o e = f. Are those other functions not identity functions then? Only the universal id_X is the identity function?
– Seung
Jul 14 at 23:04
 |Â
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up vote
0
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Okay.
You have two sets, $X$ and $Y$ and a class of functions that map $X to Y$. They want to know if there is a function $id$ so that $idcirc f = fcirc id = f$ for any $f: Xto Y$.
I'm actually not sure why they even bothered to ask as the $idcirc f = f$ would require that $id: Yto Y$ and $fcirc id = f$ would require $id:Xto X$ which is impossible as $X$ and $Y$ are different sets.
However then it asks if there are $id_X: Xto X$ that can act as a right identity with $fcirc id_X = f$ for all $f:Xto Y$. The answer is, yes, the identity element $i:Xto X$ where $i(x) = x$ for all $x in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X to X$.
(If $h(x) = w ne x$ then there is some function $f$ where $f(w) ne f(x)$ and so $f(h(x)) = f(w) ne f(x) $. So the only function that works is $i(x) = x$.)
Likewise $i:Yto Y$ $i(y) = y$ can act as $id_Y: Yto Y$ that can act as a left identity with $id_Ycirc f = f$ for all $f: Xto Y$.
So they are asking if you have a $g:Yto X$ so that $gcirc f: Xto X = id_X$ or $fcirc g: Yto Y= id_Y$ does one imply the other.
And the answer is no.
The thing is if $|X| > |Y|$ then $f: X to Y$ can not be injective so there must be $win X,x in X$ and $wne x$ so that $f(x)=f(w)$. $gcirc f:Xto X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $gcirc f(w) = g(f(w)) = g(f(x)) = gcirc f(x)$.
answered Jul 14 at 23:27
fleablood
60.5k22575
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
add a comment |Â
up vote
0
down vote
Okay.
You have two sets, $X$ and $Y$ and a class of functions that map $X to Y$. They want to know if there is a function $id$ so that $idcirc f = fcirc id = f$ for any $f: Xto Y$.
I'm actually not sure why they even bothered to ask as the $idcirc f = f$ would require that $id: Yto Y$ and $fcirc id = f$ would require $id:Xto X$ which is impossible as $X$ and $Y$ are different sets.
However then it asks if there are $id_X: Xto X$ that can act as a right identity with $fcirc id_X = f$ for all $f:Xto Y$. The answer is, yes, the identity element $i:Xto X$ where $i(x) = x$ for all $x in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X to X$.
(If $h(x) = w ne x$ then there is some function $f$ where $f(w) ne f(x)$ and so $f(h(x)) = f(w) ne f(x) $. So the only function that works is $i(x) = x$.)
Likewise $i:Yto Y$ $i(y) = y$ can act as $id_Y: Yto Y$ that can act as a left identity with $id_Ycirc f = f$ for all $f: Xto Y$.
So they are asking if you have a $g:Yto X$ so that $gcirc f: Xto X = id_X$ or $fcirc g: Yto Y= id_Y$ does one imply the other.
And the answer is no.
The thing is if $|X| > |Y|$ then $f: X to Y$ can not be injective so there must be $win X,x in X$ and $wne x$ so that $f(x)=f(w)$. $gcirc f:Xto X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $gcirc f(w) = g(f(w)) = g(f(x)) = gcirc f(x)$.
answered Jul 14 at 23:27
fleablood
60.5k22575
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Okay.
You have two sets, $X$ and $Y$ and a class of functions that map $X to Y$. They want to know if there is a function $id$ so that $idcirc f = fcirc id = f$ for any $f: Xto Y$.
I'm actually not sure why they even bothered to ask as the $idcirc f = f$ would require that $id: Yto Y$ and $fcirc id = f$ would require $id:Xto X$ which is impossible as $X$ and $Y$ are different sets.
However then it asks if there are $id_X: Xto X$ that can act as a right identity with $fcirc id_X = f$ for all $f:Xto Y$. The answer is, yes, the identity element $i:Xto X$ where $i(x) = x$ for all $x in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X to X$.
(If $h(x) = w ne x$ then there is some function $f$ where $f(w) ne f(x)$ and so $f(h(x)) = f(w) ne f(x) $. So the only function that works is $i(x) = x$.)
Likewise $i:Yto Y$ $i(y) = y$ can act as $id_Y: Yto Y$ that can act as a left identity with $id_Ycirc f = f$ for all $f: Xto Y$.
So they are asking if you have a $g:Yto X$ so that $gcirc f: Xto X = id_X$ or $fcirc g: Yto Y= id_Y$ does one imply the other.
And the answer is no.
The thing is if $|X| > |Y|$ then $f: X to Y$ can not be injective so there must be $win X,x in X$ and $wne x$ so that $f(x)=f(w)$. $gcirc f:Xto X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $gcirc f(w) = g(f(w)) = g(f(x)) = gcirc f(x)$.
answered Jul 14 at 23:27
fleablood
60.5k22575
Okay.
You have two sets, $X$ and $Y$ and a class of functions that map $X to Y$. They want to know if there is a function $id$ so that $idcirc f = fcirc id = f$ for any $f: Xto Y$.
I'm actually not sure why they even bothered to ask as the $idcirc f = f$ would require that $id: Yto Y$ and $fcirc id = f$ would require $id:Xto X$ which is impossible as $X$ and $Y$ are different sets.
However then it asks if there are $id_X: Xto X$ that can act as a right identity with $fcirc id_X = f$ for all $f:Xto Y$. The answer is, yes, the identity element $i:Xto X$ where $i(x) = x$ for all $x in X$ is such an element and with a little bit of playing it's clear that that is the is the only such function from $X to X$.
(If $h(x) = w ne x$ then there is some function $f$ where $f(w) ne f(x)$ and so $f(h(x)) = f(w) ne f(x) $. So the only function that works is $i(x) = x$.)
Likewise $i:Yto Y$ $i(y) = y$ can act as $id_Y: Yto Y$ that can act as a left identity with $id_Ycirc f = f$ for all $f: Xto Y$.
So they are asking if you have a $g:Yto X$ so that $gcirc f: Xto X = id_X$ or $fcirc g: Yto Y= id_Y$ does one imply the other.
And the answer is no.
The thing is if $|X| > |Y|$ then $f: X to Y$ can not be injective so there must be $win X,x in X$ and $wne x$ so that $f(x)=f(w)$. $gcirc f:Xto X$ can't be $id_x$. $id_X(w) = w$ and $id_X(x) = x$ but $gcirc f(w) = g(f(w)) = g(f(x)) = gcirc f(x)$.
answered Jul 14 at 23:27
fleablood
60.5k22575
edited Jul 15 at 2:32
answered Jul 14 at 23:27
fleablood
60.5k22575
answered Jul 14 at 23:27
fleablood
60.5k22575
answered Jul 14 at 23:27
fleablood
60.5k22575
60.5k22575
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
add a comment |Â
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
The question does not specify anything about function f or g. The answer, however, is given as a counterexample that uses f: X->Y, g: Y->X. With regards to your answer, I am trying to understand.
– Seung
Jul 14 at 23:37
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
I think this link has something to say with regards to identity element and identity function. en.wikipedia.org/wiki/Identity_function#Algebraic_property.
– Seung
Jul 14 at 23:41
add a comment |Â
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Who told you that definition? It's totally wrong...
– Eric Wofsey
Jul 14 at 22:11
@EricWofsey in a textbook I am using. What is correct then?
– Seung
Jul 14 at 22:23
If $f(x) = x^2$ and if $e(x) = -x$ then $fcirc e(x)= f(-x) = (-x)^2 = x^2 = f(x)$. So does that mean $e$ is the identity function? Are different functions supposed to have different function because if $g(x) = e^x$ then $gcirc e(x) = e^-x ne g(x)$. Are some functions supposed to have more than one identity function. .... (In short I think either the book or you screwed up on the definition. That is not the definition but it is a consequence.
– fleablood
Jul 14 at 22:24
1
Usually the identy function is simply defined as $e: xmapsto x$ for all $x$ in the domain.
– fleablood
Jul 14 at 22:25
1
Is this an abstract algebra text book? Are there any restrictions on the functions? Are they all 1-1 and onto. Do they all have the same domains and codomains. You define $Id_X=e$ as the function where $fcirc e = f$. Is that supposed to be true for all $f$. Has it been proven there is such a function? That it is unique? This is just too.... strange.... without context.
– fleablood
Jul 14 at 22:36