All Questions
Tagged with second-countable quotient-spaces
5 questions
2
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0
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213
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Quotient topology, product topology and subspace topology [closed]
Let $S$ be the set $\mathbb R^2\setminus\{(0, y) \mid y \neq 0\}$.
Let $\tau_1$ denote the subspace topology on $S$ induced from the usual topology of $\mathbb{R}^2$.
Now, consider the surjective map $...
- 47
1
vote
1
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67
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Prove that $S^2$ has a countable basis $\{U_n\}$ implies that $P^2$ has a countable basis $\{p(U_n)\}$.
The Problem: Let $p: S^2\to P^2$ be the quotient map from the $2$-sphere $S^2$ to its projective plane $P^2$. Then $S^2$ has a countable basis $\{U_n\}$ implies that $P^2$ has a countable basis $\{p(...
- 1,487
4
votes
1
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595
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The continuous image of a First Countable Space need not be First Countable (Willard 16.B.1)
In Stephen Willard's General Topology appears the following exercise:
A quotient of a second countable space need not be second countable (for each $n\in \mathbb{N}$, let $I_n$ be a copy of $[0,1]$ ...
- 2,793
14
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2
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When is the quotient space of a second countable space second countable?
I am a bit confused about this concept because I have read that the quotient space is second countable if the quotient map is open. However, I thought the definition of a quotient map was a surjective,...
- 232
2
votes
1
answer
634
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If a topological space $S$ is second-countable, must necessarily every quotient space of $S$ be second-countable?
Let $S$ be a second countable topological space. Let $S^*$ be a quotient space of $S$ with quotient map $\pi$. If $\pi$ is open, it's easy to show that it transfers a basis of $S$ into a basis of $S^*$...
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