Mixing
“Tearing apart†in terms of open sets and preimages
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When one considers a continuous map $f: D^2 to S^1$, there comes an argument that we should "tear apart" the disk $D^2$ at some point, thus making $f$ discontinuous. How to show this using usual definition of continuity of function, that the preimages of open sets in $S^1$ should be open sets in $D^2$, where this condition breaks?
general-topology
edited 53 mins ago
AOrtiz
8,83621237
asked 2 hours ago
Entsy
253
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up vote
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When one considers a continuous map $f: D^2 to S^1$, there comes an argument that we should "tear apart" the disk $D^2$ at some point, thus making $f$ discontinuous. How to show this using usual definition of continuity of function, that the preimages of open sets in $S^1$ should be open sets in $D^2$, where this condition breaks?
general-topology
edited 53 mins ago
AOrtiz
8,83621237
asked 2 hours ago
Entsy
253
It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
1
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When one considers a continuous map $f: D^2 to S^1$, there comes an argument that we should "tear apart" the disk $D^2$ at some point, thus making $f$ discontinuous. How to show this using usual definition of continuity of function, that the preimages of open sets in $S^1$ should be open sets in $D^2$, where this condition breaks?
general-topology
edited 53 mins ago
AOrtiz
8,83621237
asked 2 hours ago
Entsy
253
When one considers a continuous map $f: D^2 to S^1$, there comes an argument that we should "tear apart" the disk $D^2$ at some point, thus making $f$ discontinuous. How to show this using usual definition of continuity of function, that the preimages of open sets in $S^1$ should be open sets in $D^2$, where this condition breaks?
general-topology
edited 53 mins ago
AOrtiz
8,83621237
asked 2 hours ago
Entsy
253
edited 53 mins ago
AOrtiz
8,83621237
edited 53 mins ago
AOrtiz
8,83621237
edited 53 mins ago
AOrtiz
8,83621237
8,83621237
asked 2 hours ago
Entsy
253
asked 2 hours ago
Entsy
253
asked 2 hours ago
Entsy
253
253
It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
1
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago
add a comment |Â
It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
1
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago
It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
1
1
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago
add a comment |Â
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It's not clear what you are asking or even what you mean. Add some context or details please.
– DanielWainfleet
1 hour ago
1
You should clarify your question. Since constant functions are continuous, perhaps you want to restrict the class of functions you are considering, such as functions $f$ with $f|_S^1 = mathrmidentity$?
– AOrtiz
53 mins ago
Do you mean, if we "tear apart" something, why is that discontinuous? Let's say "tear apart" at a point $x$ means that for every open set around $x$, there is some point in that open set that goes "far away" from $f(x)$. More precisely, let's say that there is some open set $V$ around $f(x)$ such that, for every open set $U$ around $x$, some point in $U$ does not land in $V$. This means precisely that the preimage of $V$ under $f$ does not contain any open neighborhood of $x$; in other words, $f^-1(V)$ is not open.
– CJD
46 mins ago
I think you're asking why the definition of continuity implies that there is no continuous map from $D^2$ to $S^1$. Is this correct? If so, note that this isn't actually true: consider e.g. any constant map. We can even get surjective continuous maps (think about taking a circle of paper, wrapping it into a sorta-cylinder, and projecting it). What is true is that there is no homeomorphism from $D^2$ to $S^1$.
– Noah Schweber
11 mins ago