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
952618
$endgroup$
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
952618
$endgroup$
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
952618
$endgroup$
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
952618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
952618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
952618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
952618
asked Dec 1 '18 at 19:51
Jxt921Jxt921
952618
952618
add a comment |
add a comment |
1 Answer
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$begingroup$
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
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add a comment |
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$begingroup$
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 |
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