Do continuous injections preserve open sets?Axiomatizing topology through continuous mapsDiscontinuous maps taking compacts to compactsOne point compactification of uncountable set with discrete topology cannot be imbedded into $mathbbR^2$?(Proof-verification) Constant maps are the only maps that are always continuousProve some properties of continuous functions whose composition is an identity mapMust every subset of $mathbb R$ contain $2$ homeomorphic distinct open sets?$f:[0,2pi]rightarrowmathbbS$, $f(t)=(cos t, sin t)$. Find a base for the finest Topology that makes $f$ continuousSurjective, open, and continuous map on a 2nd countable topological spaceMeaning of “the weakest topology such that <blank> is continuous”Does a continuous function preserve metrizability?
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Do continuous injections preserve open sets?
Axiomatizing topology through continuous mapsDiscontinuous maps taking compacts to compactsOne point compactification of uncountable set with discrete topology cannot be imbedded into $mathbbR^2$?(Proof-verification) Constant maps are the only maps that are always continuousProve some properties of continuous functions whose composition is an identity mapMust every subset of $mathbb R$ contain $2$ homeomorphic distinct open sets?$f:[0,2pi]rightarrowmathbbS$, $f(t)=(cos t, sin t)$. Find a base for the finest Topology that makes $f$ continuousSurjective, open, and continuous map on a 2nd countable topological spaceMeaning of “the weakest topology such that <blank> is continuous”Does a continuous function preserve metrizability?
$begingroup$
Do continuous injections preserve open sets?
I'm pretty sure that's true in euclidean space.
If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?
If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?
Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbbR^m$ to $mathbbR^n$" in my title deleted it. Should I start a new one?
general-topology real-numbers open-map
asked 2 days ago
user3146user3146
746
$endgroup$
add a comment |
$begingroup$
Do continuous injections preserve open sets?
I'm pretty sure that's true in euclidean space.
If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?
If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?
Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbbR^m$ to $mathbbR^n$" in my title deleted it. Should I start a new one?
general-topology real-numbers open-map
asked 2 days ago
user3146user3146
746
$endgroup$
3
$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago
add a comment |
$begingroup$
Do continuous injections preserve open sets?
I'm pretty sure that's true in euclidean space.
If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?
If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?
Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbbR^m$ to $mathbbR^n$" in my title deleted it. Should I start a new one?
general-topology real-numbers open-map
asked 2 days ago
user3146user3146
746
$endgroup$
Do continuous injections preserve open sets?
I'm pretty sure that's true in euclidean space.
If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?
If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?
Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $mathbbR^m$ to $mathbbR^n$" in my title deleted it. Should I start a new one?
general-topology real-numbers open-map
general-topology real-numbers open-map
asked 2 days ago
user3146user3146
746
asked 2 days ago
user3146user3146
746
edited 2 days ago
user3146
asked 2 days ago
user3146user3146
746
asked 2 days ago
user3146user3146
746
asked 2 days ago
user3146user3146
746
746
3
$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago
add a comment |
3
$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago
3
3
$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No they do not. It's not true in euclidean space.
The function $$f(x)=begincasesx& xleq 0\ x+1 & x>0endcases$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.
answered 2 days ago
5xum5xum
92k394162
$endgroup$
add a comment |
$begingroup$
Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^-1$ is continuous, then $f$ is an open map.
answered 2 days ago
JosuéJosué
3,51242672
$endgroup$
add a comment |
$begingroup$
Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.
I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).
answered 2 days ago
L.DeRL.DeR
487
$endgroup$
add a comment |
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3 Answers
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3 Answers
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active
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$begingroup$
No they do not. It's not true in euclidean space.
The function $$f(x)=begincasesx& xleq 0\ x+1 & x>0endcases$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.
answered 2 days ago
5xum5xum
92k394162
$endgroup$
add a comment |
$begingroup$
No they do not. It's not true in euclidean space.
The function $$f(x)=begincasesx& xleq 0\ x+1 & x>0endcases$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.
answered 2 days ago
5xum5xum
92k394162
$endgroup$
add a comment |
$begingroup$
No they do not. It's not true in euclidean space.
The function $$f(x)=begincasesx& xleq 0\ x+1 & x>0endcases$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.
answered 2 days ago
5xum5xum
92k394162
$endgroup$
No they do not. It's not true in euclidean space.
The function $$f(x)=begincasesx& xleq 0\ x+1 & x>0endcases$$ is an injection, however, $f((-1,1)) = (-1, 0]cup (1, 2)$ which is not open.
answered 2 days ago
5xum5xum
92k394162
answered 2 days ago
5xum5xum
92k394162
answered 2 days ago
5xum5xum
92k394162
answered 2 days ago
5xum5xum
92k394162
92k394162
add a comment |
add a comment |
$begingroup$
Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^-1$ is continuous, then $f$ is an open map.
answered 2 days ago
JosuéJosué
3,51242672
$endgroup$
add a comment |
$begingroup$
Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^-1$ is continuous, then $f$ is an open map.
answered 2 days ago
JosuéJosué
3,51242672
$endgroup$
add a comment |
$begingroup$
Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^-1$ is continuous, then $f$ is an open map.
answered 2 days ago
JosuéJosué
3,51242672
$endgroup$
Whether $f:Xto Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^-1$ is continuous, then $f$ is an open map.
answered 2 days ago
JosuéJosué
3,51242672
answered 2 days ago
JosuéJosué
3,51242672
answered 2 days ago
JosuéJosué
3,51242672
answered 2 days ago
JosuéJosué
3,51242672
3,51242672
add a comment |
add a comment |
$begingroup$
Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.
I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).
answered 2 days ago
L.DeRL.DeR
487
$endgroup$
add a comment |
$begingroup$
Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.
I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).
answered 2 days ago
L.DeRL.DeR
487
$endgroup$
add a comment |
$begingroup$
Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.
I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).
answered 2 days ago
L.DeRL.DeR
487
$endgroup$
Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.
I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).
answered 2 days ago
L.DeRL.DeR
487
edited yesterday
answered 2 days ago
L.DeRL.DeR
487
answered 2 days ago
L.DeRL.DeR
487
answered 2 days ago
L.DeRL.DeR
487
487
add a comment |
add a comment |
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$begingroup$
A bijective map is open iff its inverse is continuous.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything?
$endgroup$
– user3146
2 days ago