In topology, an open map is a function between two topological spaces which maps open sets to open sets. That is, a function f : X → Y is open if for any open set U in X, the image f(U) is open in Y. A map may be open, closed, both, or neither; in particular, an open map need not be closed and vice versa.
Questions tagged [open-map]
349 questions
70
votes
6 answers
Projection is an open map
Let $X$ and $Y$ be (any) topological spaces. Show that the projection
$\pi_1$ : $X\times Y\to X$
is an open map.
Alisha
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66
votes
4 answers
Open maps which are not continuous
What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to…
math student
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61
votes
2 answers
When is a quotient map open?
Quotient map from $X$ to $Y$ is continuous and surjective with a property : $f^{-1}(U)$ is open in $X$ iff $U$ is open in $Y$.
But when it is open map? What condition need?
Oh hyung seok
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29
votes
6 answers
Example of quotient mapping that is not open
I have the following definition:
Let ($X$,$\mathcal{T}$) and ($X'$, $\mathcal{T'}$) be topological spaces. A surjection $q: X \longrightarrow X'$ is a quotient mapping if $$U'\in \mathcal{T'} \Longleftrightarrow q^{-1}\left( U'\right) \in…
Luc M
- 756
22
votes
2 answers
Intuition for an open mapping
What is an intuitive picture of an open mapping?
The definition of an open mapping (a function which maps open sets to open sets) is simple sounding, but it's really not as easy to picture as the simple language would suggest. When I think of…
echinodermata
- 4,389
21
votes
2 answers
A bijective continuous map is a homeomorphism iff it is open, or equivalently, iff it is closed.
Wikipedia states that "a bijective continuous map is a homeomorphism if and only if it is open, or equivalently, if and only if it is closed.".
How do we prove this fact?
I can prove the obvious direction, but im unsure how to proceed the other ways
terry
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16
votes
2 answers
Show that an open linear map between normed spaces is surjective.
Let $X,Y$ be normed spaces and $T:X\to Y$ is an open linear map. Show that $T$ is surjective.
In order to show $T$ is surjective let's take $y_0\in Y$ and assume the contrary that $Tx\neq y_0\forall x\in X$.
Now taking $x_0\in X\implies Tx_0\neq…
Learnmore
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15
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1 answer
a covering map is open?
$E,B$ are topological spaces and let's say that $p:E\to B$ is a covering map.
Is $p$ open?
i tried to show it as follows:
let $U$ be an open set in $E$, and now for every $x\in U$, $p(x)\in B$. $p$ is a covering map so there is a open neighborhood…
sha
- 1,738
15
votes
1 answer
Find $X,Y\subseteq \Bbb{R}$ such that there are $f\colon X\to Y$ and $g\colon Y\to X$, both bijective and continuous but $X\not\cong Y$.
I came across the following problem:
Let $\Bbb{R}$ be the euclidean space. Find $X,Y\subseteq \Bbb{R}$ such that there are maps $f\colon X\to Y$ and $g\colon Y\to X$, both bijective and continuous but $X\not\cong Y$ ($X$ not homeomorphic to…
Sebastián P. Pincheira
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14
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2 answers
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, continuous, open mapping.
Suppose that $X$ is…
user190570
- 232
13
votes
1 answer
Extending open maps to Stone-Čech compactifications
Let $X$ be a Čech-complete space, and $Y$ a paracompact space. Suppose $f\colon X\to Y$ is a continuous and open surjection.
Since $Y$ is completely regular we have that $\beta(Y)$ is homeomorphic to $Y$ as a dense subset of $\beta Y$ (the…
Asaf Karagila
- 405,794
13
votes
1 answer
Does there exist a continuous, open, and surjective map from $f\colon\mathbb{R}^n\to\mathbb{R}^m$ for $m>n$?
My question is that from above. Here are my approaches so far:
I know that there is no homeomorphism between an open set of $\mathbb R^n$ and an open set of $\mathbb R^m$. So if there is an open set where $ f $ is injective we get a…
Nemesis
- 335
11
votes
3 answers
Is a certain restriction of an open map open?
Let $p:X\rightarrow Y$ be an open map and let $A$ be a subspace of $X$. Then, is it true that $p|_A:A\rightarrow p(A)$ is open?
I think so, but am struggling to show it.
My thoughts: Let $O$ be an open set in $A$. Then, since $A$ has the subspace…
user322548
11
votes
1 answer
Attempted proof of an open mapping theorem for Lie groups
The Classical open mapping theorem for Banach spaces tells that if $T:X \to Y$ is a continuous surjective linear map, then it is open.
I have attempted to essentially "adapt" the proof for Lie groups:
Let $G,H$ be connected Lie groups (embedded in…
Andres Mejia
- 21,467
10
votes
1 answer
If $F:\mathbb R^m\to \mathbb R^m$ is continuous with $|F(x)-F(y)|\geq \lambda|x-y|$, then $F$ is surjective.
I know that this question has been asked for several times (such as this post and the one-dimensional case). However, I still couldn't find an answer without using invariance of domain. The original question is stated as:
If $F:\mathbb R^m\to…
SuperSupao
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