Post History
#11: Post undeleted
by
Snoopy
·
2023-02-05T14:04:47Z (about 1 year ago)
#10: Post deleted
by
Snoopy
·
2023-02-05T12:47:47Z (about 1 year ago)
#9: Post edited
by
Snoopy
·
2023-02-04T23:51:43Z (about 1 year ago)
- A typical textbook theorem about quotient space is as follows:
>**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} ightarrow Y$ such that $f=g \circ \pi$.It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here or do I need to change the argument completely?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem** (Gamelin-Greene _Introduction to Topology_ p.106): Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} ightarrow Y$ such that $f=g \circ \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$. One such proof is done in the mentioned textbook as follows:
- > Proof. Since $f$ is constant on each equivalence class, one can define $g:X/{\sim}\to Y$ with $g([x])=f(x)$. One can easily check that $f=g \circ \pi$. To show that $g$ is continuous, consider any open set $V$ in $Y$. Since $f^{-1}(V)$ is open in $X$ by continuity of $f$ and $\pi^{-1}(g^{-1}(V))=(g\circ \pi)^{-1}(V)=f^{-1}(V)$, by the definition of the quotient topology on $X/{\sim}$, $g^{-1}(V)$ is open and the proof is complete.
- ---
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which I think is supposed to be easy too, I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/{\sim}$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here or do I need to change the argument completely?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#8: Post edited
by
Derek Elkins
·
2023-02-04T23:33:20Z (about 1 year ago)
Improve kerning of quotient set notation.
Universal property of quotient spaces
- A typical textbook theorem about quotient space is as follows:
>**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim ightarrow Y$ such that $f=g{\circ} \pi$.- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here or do I need to change the argument completely?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X/{\sim} ightarrow Y$ such that $f=g \circ \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here or do I need to change the argument completely?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#7: Post edited
by
Snoopy
·
2023-02-04T17:22:27Z (about 1 year ago)
Universal property of quotient space
- Universal property of quotient spaces
#6: Post edited
by
Snoopy
·
2023-02-04T17:22:11Z (about 1 year ago)
University property of quotient space
- Universal property of quotient space
#5: Post edited
by
Snoopy
·
2023-02-04T15:12:30Z (about 1 year ago)
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
**Question**: How can I go on from here?- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here or do I need to change the argument completely?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#4: Post edited
by
Snoopy
·
2023-02-04T15:06:09Z (about 1 year ago)
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$.- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$ with the well-defined $g([x]):=f(x)$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#3: Post edited
by
Snoopy
·
2023-02-04T15:05:30Z (about 1 year ago)
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces).- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces) because $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$ for any open set $V$ in $Y$.
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#2: Post edited
by
Snoopy
·
2023-02-04T15:03:59Z (about 1 year ago)
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces).
But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too I got stuck.- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
How can I go on from here?- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
- A typical textbook theorem about quotient space is as follows:
- >**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
- It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces).
- But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too. But I got stuck.
- Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
- Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
- It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
- **Question**: How can I go on from here?
- ---
- I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.
#1: Initial revision
by
Snoopy
·
2023-02-04T15:02:58Z (about 1 year ago)
University property of quotient space
A typical textbook theorem about quotient space is as follows:
>**Theorem**: Let $f$ be a continuous function from a topological space $X$ to a topological space $Y$. Let $\sim$ be an equivalence relation on $X$ such that $f$ is constant on each equivalence class. Then there exists a continuous function $g: X / \sim \rightarrow Y$ such that $f=g{\circ} \pi$.
It is very easy to prove it by using the definition of [continuous functions between topological spaces](https://en.wikipedia.org/wiki/Continuous_function#Continuous_functions_between_topological_spaces).
But when I try to use the definition of [continuity at a point](https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point), which is supposed to be easy too I got stuck.
Let $x\in X$ and denote $[x]=\pi(x)$. I want to show that $g$ is continuous at $[x]$. Now let $U\subset Y$ be a neighborhood of $g([x])$. The goal is to find a neighborhood $O$ of $[x]$ in $X/~$ such that $g(O)\subset U$.
Since $f(x)=g([x])$, and $f$ is continuous (at $x$), there exists a neighborhood $N\subset X$ such that $x\in N$ and $f(N)\subset U$.
It suffices to show that $\pi(N)$ is a neighborhood of $[x]$ since $g(\pi(N))=f(N)\subset U$. But now I don't see how to show that $[x]$ is an interior point of $\pi(N)$.
How can I go on from here?
---
I suppose I would either need to change $\pi(N)$ to something else or apply some property of $\pi$ that I have not used yet.