Uniqueness in the Universal Property of Quotient Maps












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Here is Munkres' way of phrasing the universal property of quotient maps:




Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.




Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?



I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).






general-topology quotient-spaces universal-property







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  • 1




    $begingroup$
    The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
    $endgroup$
    – Andreas Blass
    Jan 2 at 15:37
















0












$begingroup$


Here is Munkres' way of phrasing the universal property of quotient maps:




Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.




Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?



I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).






general-topology quotient-spaces universal-property







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
    $endgroup$
    – Andreas Blass
    Jan 2 at 15:37














0












0








0





$begingroup$


Here is Munkres' way of phrasing the universal property of quotient maps:




Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.




Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?



I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).






general-topology quotient-spaces universal-property







share|cite|improve this question









$endgroup$




Here is Munkres' way of phrasing the universal property of quotient maps:




Let $p : X to Y$ be a quotient map. Let $Z$ be a space and let $g : X to Z$ be a map that is constant on each set $p^{-1}({y})$, for $y in Y$. Then $g$ induces a map $f : Y to Z$ such that $f circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.




Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f circ p = g$, so that if $h : Y to Z$ is some other map such that $h circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?



I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ {0,1}$ is homeomorphic to $S^1$).






general-topology quotient-spaces universal-property




general-topology quotient-spaces universal-property






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asked Jan 2 at 15:32









user193319user193319

2,3292924




2,3292924








  • 1




    $begingroup$
    The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
    $endgroup$
    – Andreas Blass
    Jan 2 at 15:37














  • 1




    $begingroup$
    The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
    $endgroup$
    – Andreas Blass
    Jan 2 at 15:37








1




1




$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37




$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37










1 Answer
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$begingroup$

$p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.






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    $begingroup$

    $p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.






        share|cite|improve this answer









        $endgroup$



        $p$ is onto in Munkres' definition. It follows from this and $f circ p = g$ that $f$ is unique: take $y in Y$. $y = p(x)$ for some $x in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 18:01









        Henno BrandsmaHenno Brandsma

        106k347114




        106k347114






























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