Proof that for any function $f:Ato B$ there exists a set $C$ and two functions $g:Ato C,h:Cto B$ not equal to...
Proof that for any function $f:Ato B$ there exists a set $C$ and two functions $g:Ato C,h:Cto B$ not equal to $f$ such that $f=hcirc g$?
I really have no clue how to tackle this problem. I have strong evidence to conclude this is true, but I don't know how to prove it.
I think this may be solved using category theory, knowing if in the category Set, for any morphism $f:Alongrightarrow B$, there are two morphisms such that their composition equals $f$. The axioms for category tells the opposite, that for any two morphism there exists their composition morphism, but is it true the other way around in this context? And if this is not true, what condition does $f$ need to have in order to not have this property?
functions category-theory
asked Jan 2 at 23:36
Garmekain
1,304720
add a comment |
Proof that for any function $f:Ato B$ there exists a set $C$ and two functions $g:Ato C,h:Cto B$ not equal to $f$ such that $f=hcirc g$?
I really have no clue how to tackle this problem. I have strong evidence to conclude this is true, but I don't know how to prove it.
I think this may be solved using category theory, knowing if in the category Set, for any morphism $f:Alongrightarrow B$, there are two morphisms such that their composition equals $f$. The axioms for category tells the opposite, that for any two morphism there exists their composition morphism, but is it true the other way around in this context? And if this is not true, what condition does $f$ need to have in order to not have this property?
functions category-theory
asked Jan 2 at 23:36
Garmekain
1,304720
2
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57
add a comment |
Proof that for any function $f:Ato B$ there exists a set $C$ and two functions $g:Ato C,h:Cto B$ not equal to $f$ such that $f=hcirc g$?
I really have no clue how to tackle this problem. I have strong evidence to conclude this is true, but I don't know how to prove it.
I think this may be solved using category theory, knowing if in the category Set, for any morphism $f:Alongrightarrow B$, there are two morphisms such that their composition equals $f$. The axioms for category tells the opposite, that for any two morphism there exists their composition morphism, but is it true the other way around in this context? And if this is not true, what condition does $f$ need to have in order to not have this property?
functions category-theory
asked Jan 2 at 23:36
Garmekain
1,304720
Proof that for any function $f:Ato B$ there exists a set $C$ and two functions $g:Ato C,h:Cto B$ not equal to $f$ such that $f=hcirc g$?
I really have no clue how to tackle this problem. I have strong evidence to conclude this is true, but I don't know how to prove it.
I think this may be solved using category theory, knowing if in the category Set, for any morphism $f:Alongrightarrow B$, there are two morphisms such that their composition equals $f$. The axioms for category tells the opposite, that for any two morphism there exists their composition morphism, but is it true the other way around in this context? And if this is not true, what condition does $f$ need to have in order to not have this property?
functions category-theory
functions category-theory
asked Jan 2 at 23:36
Garmekain
1,304720
asked Jan 2 at 23:36
Garmekain
1,304720
asked Jan 2 at 23:36
Garmekain
1,304720
asked Jan 2 at 23:36
Garmekain
1,304720
asked Jan 2 at 23:36
Garmekain
1,304720
1,304720
2
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57
add a comment |
2
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57
2
2
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57
add a comment |
3 Answers
3
active
oldest
votes
Assuming that $Anevarnothing$, you can let
$$ C = {; {{a},{A,B}} mid a in A } $$
Then as a matter of (mainstream) set theory no element of $C$ can equal an element of $A$ or $B$.
Let $g$ be the natural bijection $Ato C$ and $h = f circ g^{-1}$.
answered Jan 3 at 0:03
Henning Makholm
238k16303540
add a comment |
Pick $x in A^c $ and define $C= A cup {x} $ and $g (y) = y$ for all $ y in A$. and define $h: A cup {x} to B$ with $h = f$ on $A$, and $h(x) = f(a)$ for some $a in A$. Then observe that $f = hog$.
"Note that you always can assume $A$ is proper subset of a bigger set, this guarantees that $ A^c neq emptyset $ "
answered Jan 2 at 23:46
Red shoes
4,726621
add a comment |
Since this is tagged category-theory and neither of the answers provided
work in an arbitrary category, let's try once more with the constraint
that the category has a terminal object and products --which is essentially
what the other solutions utilise.
Given arrow $f : A → B$, we seek a new pair $g : A → C, h : C → B$.
- Let $C = A × 1$ --note that in 𝒮ℯ𝓉, 1 is any singleton set such as
1 = {*}. Then we naturally have $A → C = A × 1$ by $Id × !$ where $! : X → 1$ is the unique
map to the terminal object.
So take $g : A → A×1$ to be $Id × !$, which in 𝒮ℯ𝓉 acts $a ↦ (a, *)$.Now for $C = A × 1 → B$ we simply ignore the terminal object and apply $f$,
that is $h = f ∘ proj_1$, which in 𝒮ℯ𝓉 acts $(x, y) ↦ f,x$.
Hence we have produced new items $C, g, h$ and it remains to check that
$f = h ∘ g$.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{=}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
Indeed,
begin{calc}
h ∘ g
step{Definitions}
(f ∘ proj₁) ∘ (Id × !)
step{ Composition is associtive }
f ∘ (proj₁ ∘ (Id × !))
step{ Projection on products }
f ∘ Id
step{ Identity maps }
f
end{calc}
Neato! :-)
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060101%2fproof-that-for-any-function-fa-to-b-there-exists-a-set-c-and-two-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assuming that $Anevarnothing$, you can let
$$ C = {; {{a},{A,B}} mid a in A } $$
Then as a matter of (mainstream) set theory no element of $C$ can equal an element of $A$ or $B$.
Let $g$ be the natural bijection $Ato C$ and $h = f circ g^{-1}$.
answered Jan 3 at 0:03
Henning Makholm
238k16303540
add a comment |
Assuming that $Anevarnothing$, you can let
$$ C = {; {{a},{A,B}} mid a in A } $$
Then as a matter of (mainstream) set theory no element of $C$ can equal an element of $A$ or $B$.
Let $g$ be the natural bijection $Ato C$ and $h = f circ g^{-1}$.
answered Jan 3 at 0:03
Henning Makholm
238k16303540
add a comment |
Assuming that $Anevarnothing$, you can let
$$ C = {; {{a},{A,B}} mid a in A } $$
Then as a matter of (mainstream) set theory no element of $C$ can equal an element of $A$ or $B$.
Let $g$ be the natural bijection $Ato C$ and $h = f circ g^{-1}$.
answered Jan 3 at 0:03
Henning Makholm
238k16303540
Assuming that $Anevarnothing$, you can let
$$ C = {; {{a},{A,B}} mid a in A } $$
Then as a matter of (mainstream) set theory no element of $C$ can equal an element of $A$ or $B$.
Let $g$ be the natural bijection $Ato C$ and $h = f circ g^{-1}$.
answered Jan 3 at 0:03
Henning Makholm
238k16303540
answered Jan 3 at 0:03
Henning Makholm
238k16303540
answered Jan 3 at 0:03
Henning Makholm
238k16303540
answered Jan 3 at 0:03
Henning Makholm
238k16303540
238k16303540
add a comment |
add a comment |
Pick $x in A^c $ and define $C= A cup {x} $ and $g (y) = y$ for all $ y in A$. and define $h: A cup {x} to B$ with $h = f$ on $A$, and $h(x) = f(a)$ for some $a in A$. Then observe that $f = hog$.
"Note that you always can assume $A$ is proper subset of a bigger set, this guarantees that $ A^c neq emptyset $ "
answered Jan 2 at 23:46
Red shoes
4,726621
add a comment |
Pick $x in A^c $ and define $C= A cup {x} $ and $g (y) = y$ for all $ y in A$. and define $h: A cup {x} to B$ with $h = f$ on $A$, and $h(x) = f(a)$ for some $a in A$. Then observe that $f = hog$.
"Note that you always can assume $A$ is proper subset of a bigger set, this guarantees that $ A^c neq emptyset $ "
answered Jan 2 at 23:46
Red shoes
4,726621
add a comment |
Pick $x in A^c $ and define $C= A cup {x} $ and $g (y) = y$ for all $ y in A$. and define $h: A cup {x} to B$ with $h = f$ on $A$, and $h(x) = f(a)$ for some $a in A$. Then observe that $f = hog$.
"Note that you always can assume $A$ is proper subset of a bigger set, this guarantees that $ A^c neq emptyset $ "
answered Jan 2 at 23:46
Red shoes
4,726621
Pick $x in A^c $ and define $C= A cup {x} $ and $g (y) = y$ for all $ y in A$. and define $h: A cup {x} to B$ with $h = f$ on $A$, and $h(x) = f(a)$ for some $a in A$. Then observe that $f = hog$.
"Note that you always can assume $A$ is proper subset of a bigger set, this guarantees that $ A^c neq emptyset $ "
answered Jan 2 at 23:46
Red shoes
4,726621
edited Jan 2 at 23:51
answered Jan 2 at 23:46
Red shoes
4,726621
answered Jan 2 at 23:46
Red shoes
4,726621
answered Jan 2 at 23:46
Red shoes
4,726621
4,726621
add a comment |
add a comment |
Since this is tagged category-theory and neither of the answers provided
work in an arbitrary category, let's try once more with the constraint
that the category has a terminal object and products --which is essentially
what the other solutions utilise.
Given arrow $f : A → B$, we seek a new pair $g : A → C, h : C → B$.
- Let $C = A × 1$ --note that in 𝒮ℯ𝓉, 1 is any singleton set such as
1 = {*}. Then we naturally have $A → C = A × 1$ by $Id × !$ where $! : X → 1$ is the unique
map to the terminal object.
So take $g : A → A×1$ to be $Id × !$, which in 𝒮ℯ𝓉 acts $a ↦ (a, *)$.Now for $C = A × 1 → B$ we simply ignore the terminal object and apply $f$,
that is $h = f ∘ proj_1$, which in 𝒮ℯ𝓉 acts $(x, y) ↦ f,x$.
Hence we have produced new items $C, g, h$ and it remains to check that
$f = h ∘ g$.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{=}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
Indeed,
begin{calc}
h ∘ g
step{Definitions}
(f ∘ proj₁) ∘ (Id × !)
step{ Composition is associtive }
f ∘ (proj₁ ∘ (Id × !))
step{ Projection on products }
f ∘ Id
step{ Identity maps }
f
end{calc}
Neato! :-)
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
add a comment |
Since this is tagged category-theory and neither of the answers provided
work in an arbitrary category, let's try once more with the constraint
that the category has a terminal object and products --which is essentially
what the other solutions utilise.
Given arrow $f : A → B$, we seek a new pair $g : A → C, h : C → B$.
- Let $C = A × 1$ --note that in 𝒮ℯ𝓉, 1 is any singleton set such as
1 = {*}. Then we naturally have $A → C = A × 1$ by $Id × !$ where $! : X → 1$ is the unique
map to the terminal object.
So take $g : A → A×1$ to be $Id × !$, which in 𝒮ℯ𝓉 acts $a ↦ (a, *)$.Now for $C = A × 1 → B$ we simply ignore the terminal object and apply $f$,
that is $h = f ∘ proj_1$, which in 𝒮ℯ𝓉 acts $(x, y) ↦ f,x$.
Hence we have produced new items $C, g, h$ and it remains to check that
$f = h ∘ g$.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{=}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
Indeed,
begin{calc}
h ∘ g
step{Definitions}
(f ∘ proj₁) ∘ (Id × !)
step{ Composition is associtive }
f ∘ (proj₁ ∘ (Id × !))
step{ Projection on products }
f ∘ Id
step{ Identity maps }
f
end{calc}
Neato! :-)
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
add a comment |
Since this is tagged category-theory and neither of the answers provided
work in an arbitrary category, let's try once more with the constraint
that the category has a terminal object and products --which is essentially
what the other solutions utilise.
Given arrow $f : A → B$, we seek a new pair $g : A → C, h : C → B$.
- Let $C = A × 1$ --note that in 𝒮ℯ𝓉, 1 is any singleton set such as
1 = {*}. Then we naturally have $A → C = A × 1$ by $Id × !$ where $! : X → 1$ is the unique
map to the terminal object.
So take $g : A → A×1$ to be $Id × !$, which in 𝒮ℯ𝓉 acts $a ↦ (a, *)$.Now for $C = A × 1 → B$ we simply ignore the terminal object and apply $f$,
that is $h = f ∘ proj_1$, which in 𝒮ℯ𝓉 acts $(x, y) ↦ f,x$.
Hence we have produced new items $C, g, h$ and it remains to check that
$f = h ∘ g$.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{=}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
Indeed,
begin{calc}
h ∘ g
step{Definitions}
(f ∘ proj₁) ∘ (Id × !)
step{ Composition is associtive }
f ∘ (proj₁ ∘ (Id × !))
step{ Projection on products }
f ∘ Id
step{ Identity maps }
f
end{calc}
Neato! :-)
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
Since this is tagged category-theory and neither of the answers provided
work in an arbitrary category, let's try once more with the constraint
that the category has a terminal object and products --which is essentially
what the other solutions utilise.
Given arrow $f : A → B$, we seek a new pair $g : A → C, h : C → B$.
- Let $C = A × 1$ --note that in 𝒮ℯ𝓉, 1 is any singleton set such as
1 = {*}. Then we naturally have $A → C = A × 1$ by $Id × !$ where $! : X → 1$ is the unique
map to the terminal object.
So take $g : A → A×1$ to be $Id × !$, which in 𝒮ℯ𝓉 acts $a ↦ (a, *)$.Now for $C = A × 1 → B$ we simply ignore the terminal object and apply $f$,
that is $h = f ∘ proj_1$, which in 𝒮ℯ𝓉 acts $(x, y) ↦ f,x$.
Hence we have produced new items $C, g, h$ and it remains to check that
$f = h ∘ g$.
$defstepWith#1#2{ \ #1 & quad color{green}{{;text{#2};}} \ & }defstep#1{ stepWith{=}{#1} }newenvironment{calc}{begin{align*} & }{end{align*}}$
Indeed,
begin{calc}
h ∘ g
step{Definitions}
(f ∘ proj₁) ∘ (Id × !)
step{ Composition is associtive }
f ∘ (proj₁ ∘ (Id × !))
step{ Projection on products }
f ∘ Id
step{ Identity maps }
f
end{calc}
Neato! :-)
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
answered Jan 3 at 21:49
Musa Al-hassy
1,3291711
1,3291711
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
add a comment |
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
1
1
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
However this does not necessary make $g$ and $h$ different from $f$. For example, a lattice has products and a terminal object, but $Atimes 1$ is the same object as $A$, so your $g$ would be an identity and $h=f$.
– Henning Makholm
Jan 3 at 23:56
Excellent observation!
– Musa Al-hassy
2 days ago
Excellent observation!
– Musa Al-hassy
2 days ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060101%2fproof-that-for-any-function-fa-to-b-there-exists-a-set-c-and-two-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
If "not equal to $f$" is interpreted set-theoretically, then it is not true if $A=varnothing$, since then $f$ and $g$ are necessarily both the empty function.
– Henning Makholm
Jan 2 at 23:42
Or can $g$ count as "not equal to $f$" for you if only $f$ and $g$ are assigned different codomains even though $f(a)=g(a)$ for all $ain A$?
– Henning Makholm
Jan 2 at 23:46
@HenningMakholm "not equal to $f$" as in set theory. I thought about the empty set as an exception, are there any other functions?
– Garmekain
Jan 2 at 23:57