Pushout in the category of Sets: proof
$begingroup$
Let $fcolon Zto X$ and $gcolon Zto Y$ be functions. What is a pushout of $f$ and $g$ in $mathsf{Set}$? Different sources and questions on this site claim that it's the quotient of the disjoint union $Xsqcup Y$ by the equivalence relation $sim_R$ generated by the relation $R = { ((0,x),(1,y)) in (Xsqcup Y)times (Xsqcup Y) mid exists z in Z, x = f(z)$ and $y = g(z) }$.
Define functions $i_1colon Xto Xsqcup Y/{sim_R}$ and $i_2colon Yto Xsqcup Y/{sim_R}$ by setting $i_1$ to map each $x in X$ to the equivalence class $[(0,x)]$ and $i_2$ to map each $y in Y$ to the equivalence class $[(1,y)]$.
For $Xsqcup Y/{sim_R}$ together with $i_1$ and $i_2$ to be a pushout in $mathsf{Set}$, we first must have $i_2circ f = i_2circ g$. It's easy to check that it's so.
Now, let $j_1colon Xto Q$ and $j_2colon Yto Q$ be functions so that $j_1circ f = j_2circ g$. We seek a function $ucolon Xsqcup Y/{sim_R} to Q$. $ucirc i_1 = j_1$ and $ucirc i_2 = j_2$. Of course, for these identities to hold, this hypothetical function $u$ must satisfy $u([(0,x)]) = j_1(x)$ and $u([(1,y)]) = j_2(y)$ for any $x in X$ and $y in Y$. But I'm not sure how to prove that this data indeed defines a function. We need to prove that it is independent of the choice of $x$ and $y$. Of course, this needs to use the fact that $j_1circ f = j_2circ g$.
elementary-set-theory category-theory
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
$endgroup$
add a comment |
$begingroup$
Let $fcolon Zto X$ and $gcolon Zto Y$ be functions. What is a pushout of $f$ and $g$ in $mathsf{Set}$? Different sources and questions on this site claim that it's the quotient of the disjoint union $Xsqcup Y$ by the equivalence relation $sim_R$ generated by the relation $R = { ((0,x),(1,y)) in (Xsqcup Y)times (Xsqcup Y) mid exists z in Z, x = f(z)$ and $y = g(z) }$.
Define functions $i_1colon Xto Xsqcup Y/{sim_R}$ and $i_2colon Yto Xsqcup Y/{sim_R}$ by setting $i_1$ to map each $x in X$ to the equivalence class $[(0,x)]$ and $i_2$ to map each $y in Y$ to the equivalence class $[(1,y)]$.
For $Xsqcup Y/{sim_R}$ together with $i_1$ and $i_2$ to be a pushout in $mathsf{Set}$, we first must have $i_2circ f = i_2circ g$. It's easy to check that it's so.
Now, let $j_1colon Xto Q$ and $j_2colon Yto Q$ be functions so that $j_1circ f = j_2circ g$. We seek a function $ucolon Xsqcup Y/{sim_R} to Q$. $ucirc i_1 = j_1$ and $ucirc i_2 = j_2$. Of course, for these identities to hold, this hypothetical function $u$ must satisfy $u([(0,x)]) = j_1(x)$ and $u([(1,y)]) = j_2(y)$ for any $x in X$ and $y in Y$. But I'm not sure how to prove that this data indeed defines a function. We need to prove that it is independent of the choice of $x$ and $y$. Of course, this needs to use the fact that $j_1circ f = j_2circ g$.
elementary-set-theory category-theory
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
$endgroup$
add a comment |
$begingroup$
Let $fcolon Zto X$ and $gcolon Zto Y$ be functions. What is a pushout of $f$ and $g$ in $mathsf{Set}$? Different sources and questions on this site claim that it's the quotient of the disjoint union $Xsqcup Y$ by the equivalence relation $sim_R$ generated by the relation $R = { ((0,x),(1,y)) in (Xsqcup Y)times (Xsqcup Y) mid exists z in Z, x = f(z)$ and $y = g(z) }$.
Define functions $i_1colon Xto Xsqcup Y/{sim_R}$ and $i_2colon Yto Xsqcup Y/{sim_R}$ by setting $i_1$ to map each $x in X$ to the equivalence class $[(0,x)]$ and $i_2$ to map each $y in Y$ to the equivalence class $[(1,y)]$.
For $Xsqcup Y/{sim_R}$ together with $i_1$ and $i_2$ to be a pushout in $mathsf{Set}$, we first must have $i_2circ f = i_2circ g$. It's easy to check that it's so.
Now, let $j_1colon Xto Q$ and $j_2colon Yto Q$ be functions so that $j_1circ f = j_2circ g$. We seek a function $ucolon Xsqcup Y/{sim_R} to Q$. $ucirc i_1 = j_1$ and $ucirc i_2 = j_2$. Of course, for these identities to hold, this hypothetical function $u$ must satisfy $u([(0,x)]) = j_1(x)$ and $u([(1,y)]) = j_2(y)$ for any $x in X$ and $y in Y$. But I'm not sure how to prove that this data indeed defines a function. We need to prove that it is independent of the choice of $x$ and $y$. Of course, this needs to use the fact that $j_1circ f = j_2circ g$.
elementary-set-theory category-theory
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
$endgroup$
Let $fcolon Zto X$ and $gcolon Zto Y$ be functions. What is a pushout of $f$ and $g$ in $mathsf{Set}$? Different sources and questions on this site claim that it's the quotient of the disjoint union $Xsqcup Y$ by the equivalence relation $sim_R$ generated by the relation $R = { ((0,x),(1,y)) in (Xsqcup Y)times (Xsqcup Y) mid exists z in Z, x = f(z)$ and $y = g(z) }$.
Define functions $i_1colon Xto Xsqcup Y/{sim_R}$ and $i_2colon Yto Xsqcup Y/{sim_R}$ by setting $i_1$ to map each $x in X$ to the equivalence class $[(0,x)]$ and $i_2$ to map each $y in Y$ to the equivalence class $[(1,y)]$.
For $Xsqcup Y/{sim_R}$ together with $i_1$ and $i_2$ to be a pushout in $mathsf{Set}$, we first must have $i_2circ f = i_2circ g$. It's easy to check that it's so.
Now, let $j_1colon Xto Q$ and $j_2colon Yto Q$ be functions so that $j_1circ f = j_2circ g$. We seek a function $ucolon Xsqcup Y/{sim_R} to Q$. $ucirc i_1 = j_1$ and $ucirc i_2 = j_2$. Of course, for these identities to hold, this hypothetical function $u$ must satisfy $u([(0,x)]) = j_1(x)$ and $u([(1,y)]) = j_2(y)$ for any $x in X$ and $y in Y$. But I'm not sure how to prove that this data indeed defines a function. We need to prove that it is independent of the choice of $x$ and $y$. Of course, this needs to use the fact that $j_1circ f = j_2circ g$.
elementary-set-theory category-theory
elementary-set-theory category-theory
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
986618
986618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is clear that $u$ thus defined is really a function on all of the quotient space, because the quotient map is surjective. Whenever you define anything on a quotient by means of representatives, you check that it is well defined by considering different representatives.
If I denote by $pi$ the quotient map, we want to show that if $pi((0,x)) = pi((1,y))$, then $u(pi((0,x))$ = $u(pi((1,y))$. The condition implies that there is a (possibly non-unique) $z in Z$ with $f(z) = x$ and $g(z) = y$. Applying $j_1$ to the first and $j_2$ to the second equality yields $j_1 (x) = j_2(y)$ by the condition. But that implies precisely $u(pi((0,x))$ = $u(pi((1,y))$. Hence, $u$ is well defined. Can you explain why $u$ is also unique?
As a continuation, you may want to consider the following: this diagrammatic property of the pushout characterizes the pushout. If any other set $P$ has the pushout property, then there is a canonical bijection between the 'abstract' pushout you defined and the new pushout $P$. The proof of course uses the property of the pushout.
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
$endgroup$
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%2f3021738%2fpushout-in-the-category-of-sets-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is clear that $u$ thus defined is really a function on all of the quotient space, because the quotient map is surjective. Whenever you define anything on a quotient by means of representatives, you check that it is well defined by considering different representatives.
If I denote by $pi$ the quotient map, we want to show that if $pi((0,x)) = pi((1,y))$, then $u(pi((0,x))$ = $u(pi((1,y))$. The condition implies that there is a (possibly non-unique) $z in Z$ with $f(z) = x$ and $g(z) = y$. Applying $j_1$ to the first and $j_2$ to the second equality yields $j_1 (x) = j_2(y)$ by the condition. But that implies precisely $u(pi((0,x))$ = $u(pi((1,y))$. Hence, $u$ is well defined. Can you explain why $u$ is also unique?
As a continuation, you may want to consider the following: this diagrammatic property of the pushout characterizes the pushout. If any other set $P$ has the pushout property, then there is a canonical bijection between the 'abstract' pushout you defined and the new pushout $P$. The proof of course uses the property of the pushout.
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
$endgroup$
add a comment |
$begingroup$
It is clear that $u$ thus defined is really a function on all of the quotient space, because the quotient map is surjective. Whenever you define anything on a quotient by means of representatives, you check that it is well defined by considering different representatives.
If I denote by $pi$ the quotient map, we want to show that if $pi((0,x)) = pi((1,y))$, then $u(pi((0,x))$ = $u(pi((1,y))$. The condition implies that there is a (possibly non-unique) $z in Z$ with $f(z) = x$ and $g(z) = y$. Applying $j_1$ to the first and $j_2$ to the second equality yields $j_1 (x) = j_2(y)$ by the condition. But that implies precisely $u(pi((0,x))$ = $u(pi((1,y))$. Hence, $u$ is well defined. Can you explain why $u$ is also unique?
As a continuation, you may want to consider the following: this diagrammatic property of the pushout characterizes the pushout. If any other set $P$ has the pushout property, then there is a canonical bijection between the 'abstract' pushout you defined and the new pushout $P$. The proof of course uses the property of the pushout.
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
$endgroup$
add a comment |
$begingroup$
It is clear that $u$ thus defined is really a function on all of the quotient space, because the quotient map is surjective. Whenever you define anything on a quotient by means of representatives, you check that it is well defined by considering different representatives.
If I denote by $pi$ the quotient map, we want to show that if $pi((0,x)) = pi((1,y))$, then $u(pi((0,x))$ = $u(pi((1,y))$. The condition implies that there is a (possibly non-unique) $z in Z$ with $f(z) = x$ and $g(z) = y$. Applying $j_1$ to the first and $j_2$ to the second equality yields $j_1 (x) = j_2(y)$ by the condition. But that implies precisely $u(pi((0,x))$ = $u(pi((1,y))$. Hence, $u$ is well defined. Can you explain why $u$ is also unique?
As a continuation, you may want to consider the following: this diagrammatic property of the pushout characterizes the pushout. If any other set $P$ has the pushout property, then there is a canonical bijection between the 'abstract' pushout you defined and the new pushout $P$. The proof of course uses the property of the pushout.
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
$endgroup$
It is clear that $u$ thus defined is really a function on all of the quotient space, because the quotient map is surjective. Whenever you define anything on a quotient by means of representatives, you check that it is well defined by considering different representatives.
If I denote by $pi$ the quotient map, we want to show that if $pi((0,x)) = pi((1,y))$, then $u(pi((0,x))$ = $u(pi((1,y))$. The condition implies that there is a (possibly non-unique) $z in Z$ with $f(z) = x$ and $g(z) = y$. Applying $j_1$ to the first and $j_2$ to the second equality yields $j_1 (x) = j_2(y)$ by the condition. But that implies precisely $u(pi((0,x))$ = $u(pi((1,y))$. Hence, $u$ is well defined. Can you explain why $u$ is also unique?
As a continuation, you may want to consider the following: this diagrammatic property of the pushout characterizes the pushout. If any other set $P$ has the pushout property, then there is a canonical bijection between the 'abstract' pushout you defined and the new pushout $P$. The proof of course uses the property of the pushout.
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
answered Dec 1 '18 at 20:13
Thomas BakxThomas Bakx
3309
3309
add a comment |
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.
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%2f3021738%2fpushout-in-the-category-of-sets-proof%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
