Post History
#7: Post undeleted
by
Snoopy
·
2023-02-05T14:04:52Z (almost 2 years ago)
#6: Post deleted
by
Snoopy
·
2023-02-05T12:47:42Z (almost 2 years ago)
#5: Post edited
by
Snoopy
·
2023-02-05T00:12:56Z (almost 2 years ago)
- <sup>I figured out an answer after posting the question for a while. I would like to record it here.</sup>
- ---
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
- ---
Proof. Define $g:X/{\sim}$ as in the standard proof. Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
- <sup>I figured out an answer after posting the question for a while. I would like to record it here.</sup>
- ---
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
- ---
- Proof. Define $g:X/{\sim}$ as in the standard proof. Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $\pi^{-1}(g^{-1}(U))=f^{-1}(U)$ is open by the continuity of $f$ and thus $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ by the definition of quotient topology. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
#4: Post edited
by
Snoopy
·
2023-02-04T23:57:54Z (almost 2 years ago)
- <sup>I figured out an answer after posting the question for a while. I would like to record it here.</sup>
- ---
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
- ---
Proof. Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
- <sup>I figured out an answer after posting the question for a while. I would like to record it here.</sup>
- ---
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
- ---
- Proof. Define $g:X/{\sim}$ as in the standard proof. Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
#3: Post edited
by
Snoopy
·
2023-02-04T20:55:12Z (almost 2 years ago)
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$.- ---
Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$.
- <sup>I figured out an answer after posting the question for a while. I would like to record it here.</sup>
- ---
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$, which, of course, could have been used in the simple proof via continuity of functions between topological spaces.
- ---
- Proof. Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$. Since $[x]$ is arbitrary, the proof is complete.
#2: Post edited
by
Snoopy
·
2023-02-04T17:20:51Z (almost 2 years ago)
The problem with the argument in OP is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$.
- ---
- Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$.
- The problem with the mentioned argument is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
- One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$.
- ---
- Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$.
#1: Initial revision
by
Snoopy
·
2023-02-04T17:19:43Z (almost 2 years ago)
The problem with the argument in OP is that one attempted to prove the continuity of $g$ at one point $[x]$ using merely the continuity of $f$ at $x$ instead of using the global continuity of $f$ on $X$.
One way that I figured out to save the mentioned argument is that one should exploit the fact that $\pi^{-1}(g^{-1}(V))=f^{-1}(V)$.
---
Without loss of generality, suppose $U$ is an _open_ neighborhood of $g([x])$. Then $g^{-1}(U)$ is an open neighborhood of $[x]$ in $X/\sim$ because of the identity above and continuity of $f$. Since $g(g^{-1}(U))\subset U$, we have shown that $g$ is continuous at $[x]$.