Mixing
Uniqueness in the Universal Property of Quotient Maps
$begingroup$
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
asked Jan 2 at 15:32
user193319user193319
2,3292924
$endgroup$
add a comment |
$begingroup$
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
asked Jan 2 at 15:32
user193319user193319
2,3292924
$endgroup$
1
$begingroup$
The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37
add a comment |
$begingroup$
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
asked Jan 2 at 15:32
user193319user193319
2,3292924
$endgroup$
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
asked Jan 2 at 15:32
user193319user193319
2,3292924
asked Jan 2 at 15:32
user193319user193319
2,3292924
asked Jan 2 at 15:32
user193319user193319
2,3292924
asked Jan 2 at 15:32
user193319user193319
2,3292924
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
add a comment |
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
add a comment |
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.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
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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.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
$p$ 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.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
$p$ 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.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$p$ 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.
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 18:01
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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The uniqueness of $f$ follows from the surjectivity of $p$. No continuity is involved.
$endgroup$
– Andreas Blass
Jan 2 at 15:37