Mixing
When is a space homeomorphic to a quotient space?
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Is the following theorem true? It seems straightforward but I haven't seen it published anywhere, not even as a corollary, so I'm concerned I've missed something. Discussions that introduce quotient spaces all seem to dance around this very simple and useful fact. Why don't they just come right out and say it?
Let $X$ and $Y$ be a topological spaces. Let $sim$ be an equivalence relation on $X$. Then $Y$ is homeomorphic to the quotient space $X/{sim}$ iff there exists a quotient map $f:X to Y$ that induces the same partition as $sim$.
general-topology equivalence-relations quotient-spaces
asked Jan 6 at 16:05
TJCrowTJCrow
665
$endgroup$
|
show 1 more comment
$begingroup$
Is the following theorem true? It seems straightforward but I haven't seen it published anywhere, not even as a corollary, so I'm concerned I've missed something. Discussions that introduce quotient spaces all seem to dance around this very simple and useful fact. Why don't they just come right out and say it?
Let $X$ and $Y$ be a topological spaces. Let $sim$ be an equivalence relation on $X$. Then $Y$ is homeomorphic to the quotient space $X/{sim}$ iff there exists a quotient map $f:X to Y$ that induces the same partition as $sim$.
general-topology equivalence-relations quotient-spaces
asked Jan 6 at 16:05
TJCrowTJCrow
665
$endgroup$
$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
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Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
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@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
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– TJCrow
Jan 6 at 17:34
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Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37
|
show 1 more comment
$begingroup$
Is the following theorem true? It seems straightforward but I haven't seen it published anywhere, not even as a corollary, so I'm concerned I've missed something. Discussions that introduce quotient spaces all seem to dance around this very simple and useful fact. Why don't they just come right out and say it?
Let $X$ and $Y$ be a topological spaces. Let $sim$ be an equivalence relation on $X$. Then $Y$ is homeomorphic to the quotient space $X/{sim}$ iff there exists a quotient map $f:X to Y$ that induces the same partition as $sim$.
general-topology equivalence-relations quotient-spaces
asked Jan 6 at 16:05
TJCrowTJCrow
665
$endgroup$
Is the following theorem true? It seems straightforward but I haven't seen it published anywhere, not even as a corollary, so I'm concerned I've missed something. Discussions that introduce quotient spaces all seem to dance around this very simple and useful fact. Why don't they just come right out and say it?
Let $X$ and $Y$ be a topological spaces. Let $sim$ be an equivalence relation on $X$. Then $Y$ is homeomorphic to the quotient space $X/{sim}$ iff there exists a quotient map $f:X to Y$ that induces the same partition as $sim$.
general-topology equivalence-relations quotient-spaces
general-topology equivalence-relations quotient-spaces
asked Jan 6 at 16:05
TJCrowTJCrow
665
asked Jan 6 at 16:05
TJCrowTJCrow
665
asked Jan 6 at 16:05
TJCrowTJCrow
665
asked Jan 6 at 16:05
TJCrowTJCrow
665
asked Jan 6 at 16:05
TJCrowTJCrow
665
665
$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
$begingroup$
Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
$begingroup$
@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
$endgroup$
– TJCrow
Jan 6 at 17:34
$begingroup$
Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37
|
show 1 more comment
$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
$begingroup$
Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
$begingroup$
@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
$endgroup$
– TJCrow
Jan 6 at 17:34
$begingroup$
Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37
$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
$begingroup$
Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
$begingroup$
@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
$endgroup$
– TJCrow
Jan 6 at 17:34
$begingroup$
@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
$endgroup$
– TJCrow
Jan 6 at 17:34
$begingroup$
Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37
$begingroup$
Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37
|
show 1 more comment
1 Answer
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$begingroup$
It is well known (and stated in most textbooks) that a continuous surjection $f : X to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y to Z$:
$g$ is continuous if and only if $g circ f : X to Z$ is continuous.
An obvious corollary is this.
Given two quotient maps $f : X to Y, f' : X to Y'$ and a bijection $phi : Y to Y'$ such that $phi circ f = f'$. Then $phi$ is a homeomorphism.
Each quotient map $f : X to Y$ induces an equivalence relation on $X$ by defining $x sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X to X/sim$.
In fact, this does not frequently occur as an explicit statement in the literature.
$endgroup$
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
add a comment |
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$begingroup$
It is well known (and stated in most textbooks) that a continuous surjection $f : X to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y to Z$:
$g$ is continuous if and only if $g circ f : X to Z$ is continuous.
An obvious corollary is this.
Given two quotient maps $f : X to Y, f' : X to Y'$ and a bijection $phi : Y to Y'$ such that $phi circ f = f'$. Then $phi$ is a homeomorphism.
Each quotient map $f : X to Y$ induces an equivalence relation on $X$ by defining $x sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X to X/sim$.
In fact, this does not frequently occur as an explicit statement in the literature.
$endgroup$
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
add a comment |
$begingroup$
It is well known (and stated in most textbooks) that a continuous surjection $f : X to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y to Z$:
$g$ is continuous if and only if $g circ f : X to Z$ is continuous.
An obvious corollary is this.
Given two quotient maps $f : X to Y, f' : X to Y'$ and a bijection $phi : Y to Y'$ such that $phi circ f = f'$. Then $phi$ is a homeomorphism.
Each quotient map $f : X to Y$ induces an equivalence relation on $X$ by defining $x sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X to X/sim$.
In fact, this does not frequently occur as an explicit statement in the literature.
$endgroup$
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
add a comment |
$begingroup$
It is well known (and stated in most textbooks) that a continuous surjection $f : X to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y to Z$:
$g$ is continuous if and only if $g circ f : X to Z$ is continuous.
An obvious corollary is this.
Given two quotient maps $f : X to Y, f' : X to Y'$ and a bijection $phi : Y to Y'$ such that $phi circ f = f'$. Then $phi$ is a homeomorphism.
Each quotient map $f : X to Y$ induces an equivalence relation on $X$ by defining $x sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X to X/sim$.
In fact, this does not frequently occur as an explicit statement in the literature.
$endgroup$
It is well known (and stated in most textbooks) that a continuous surjection $f : X to Y$ is a quotient map if and only if the following is satisfied for all functions $g : Y to Z$:
$g$ is continuous if and only if $g circ f : X to Z$ is continuous.
An obvious corollary is this.
Given two quotient maps $f : X to Y, f' : X to Y'$ and a bijection $phi : Y to Y'$ such that $phi circ f = f'$. Then $phi$ is a homeomorphism.
Each quotient map $f : X to Y$ induces an equivalence relation on $X$ by defining $x sim x'$ iff $f(x) = f(x')$. Now apply the corollary to $f$ and the quotient map $p : X to X/sim$.
In fact, this does not frequently occur as an explicit statement in the literature.
answered Jan 6 at 16:46
community wiki
Paul Frost
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
add a comment |
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
$begingroup$
Thanks @PaulFrost, for confirming that it does not get explicitly stated. And that bugs me because it's actually this exact theorem that gets used whenever someone wants to show how a space is homoemorphic to a quotient space. For example, when an author wants to show that $[0,1]/{sim}$ where $sim$ glues the endpoints together is homeomorphic to $S^1$, they define $f:[0,1] to S^1$, $f(t)=(cos t, sin t)$ and show that (1) $f$ induces the same partition on $[0,1]$ as $sim$ does and (2) that $f$ is a quotient map.
$endgroup$
– TJCrow
Jan 6 at 18:19
add a comment |
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$begingroup$
Absolutely true. Actually, you can find the theorem in Basic Topology by Armstrong.
$endgroup$
– user549397
Jan 6 at 16:12
$begingroup$
Thank you @Philip for your encouraging comment. However, I looked in Armstrong and did not find the theorem. The closest I found was Theorem 4.2 on page 67. But this isn't quite the same as the theorem in the original post.
$endgroup$
– TJCrow
Jan 6 at 17:08
$begingroup$
That's the theorem I am talking about. If you are not familiar with it at the beginning, just refer to Paul's post in the answer zone. Crystal clear for a beginner I think.
$endgroup$
– user549397
Jan 6 at 17:18
$begingroup$
@Philip, I respectfully disagree that they are the same theorems. Armstrong's Theorem 4.2(a) only applies to the equivalence relation induced by $f$. The theorem in the op applies to any equivalence relation.
$endgroup$
– TJCrow
Jan 6 at 17:34
$begingroup$
Yeah, I agree, but the necessary condition in the op is covered by the previous discussion in Armstrong's so I just take it for granted.
$endgroup$
– user549397
Jan 7 at 0:37