Let $X$ be a topological space and $pi:mathbb R^2to X$ a covering map.
$begingroup$
Let $B={(x,y)in mathbb R^2mid x^2+y^2leq 1}$ and let $K$ be a compact subset of $X$. Suppose $pi:mathbb R^2setminus Bto Xsetminus K$ is a homeomorphism. Show that $X$ is homeomorphic to $mathbb R^2$.
Conclusions I can make now are as follows:
-The fundamental group is a topological invariant so this means $pi_1(mathbb R^2setminus B)cong pi_1(Xsetminus K)$
-$S_1$ is a deformation retract of $mathbb R^2setminus B$ $Rightarrow$ $pi_1(mathbb R^2setminus B)cong pi_1(S^1)cong mathbb Z$
-Since $mathbb R^2$ is simply connected, each fiber of $pi$ has the same cardinality as the fundamental group of $X$.
It should be enough to show that $X$ is simply connected, but I'm not sure how.
algebraic-topology
$endgroup$
add a comment |
$begingroup$
Let $B={(x,y)in mathbb R^2mid x^2+y^2leq 1}$ and let $K$ be a compact subset of $X$. Suppose $pi:mathbb R^2setminus Bto Xsetminus K$ is a homeomorphism. Show that $X$ is homeomorphic to $mathbb R^2$.
Conclusions I can make now are as follows:
-The fundamental group is a topological invariant so this means $pi_1(mathbb R^2setminus B)cong pi_1(Xsetminus K)$
-$S_1$ is a deformation retract of $mathbb R^2setminus B$ $Rightarrow$ $pi_1(mathbb R^2setminus B)cong pi_1(S^1)cong mathbb Z$
-Since $mathbb R^2$ is simply connected, each fiber of $pi$ has the same cardinality as the fundamental group of $X$.
It should be enough to show that $X$ is simply connected, but I'm not sure how.
algebraic-topology
$endgroup$
add a comment |
$begingroup$
Let $B={(x,y)in mathbb R^2mid x^2+y^2leq 1}$ and let $K$ be a compact subset of $X$. Suppose $pi:mathbb R^2setminus Bto Xsetminus K$ is a homeomorphism. Show that $X$ is homeomorphic to $mathbb R^2$.
Conclusions I can make now are as follows:
-The fundamental group is a topological invariant so this means $pi_1(mathbb R^2setminus B)cong pi_1(Xsetminus K)$
-$S_1$ is a deformation retract of $mathbb R^2setminus B$ $Rightarrow$ $pi_1(mathbb R^2setminus B)cong pi_1(S^1)cong mathbb Z$
-Since $mathbb R^2$ is simply connected, each fiber of $pi$ has the same cardinality as the fundamental group of $X$.
It should be enough to show that $X$ is simply connected, but I'm not sure how.
algebraic-topology
$endgroup$
Let $B={(x,y)in mathbb R^2mid x^2+y^2leq 1}$ and let $K$ be a compact subset of $X$. Suppose $pi:mathbb R^2setminus Bto Xsetminus K$ is a homeomorphism. Show that $X$ is homeomorphic to $mathbb R^2$.
Conclusions I can make now are as follows:
-The fundamental group is a topological invariant so this means $pi_1(mathbb R^2setminus B)cong pi_1(Xsetminus K)$
-$S_1$ is a deformation retract of $mathbb R^2setminus B$ $Rightarrow$ $pi_1(mathbb R^2setminus B)cong pi_1(S^1)cong mathbb Z$
-Since $mathbb R^2$ is simply connected, each fiber of $pi$ has the same cardinality as the fundamental group of $X$.
It should be enough to show that $X$ is simply connected, but I'm not sure how.
algebraic-topology
algebraic-topology
asked Jan 15 at 23:54
user475040
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $Xsetminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
$endgroup$
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
|
show 3 more comments
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075137%2flet-x-be-a-topological-space-and-pi-mathbb-r2-to-x-a-covering-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $Xsetminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
$endgroup$
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
|
show 3 more comments
$begingroup$
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $Xsetminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
$endgroup$
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
|
show 3 more comments
$begingroup$
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $Xsetminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
$endgroup$
Hint: Suppose that the inverse image of every point has at least two elements. At least one of these must lie in the unit ball. Choose a countable sequence that has no limit points in $Xsetminus K$. Looking at the preimages in $B$ and the "evenly covered" property gives a contradiction.
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
edited Jan 16 at 0:27
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
answered Jan 16 at 0:10
Cheerful ParsnipCheerful Parsnip
21.2k23699
21.2k23699
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
|
show 3 more comments
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
How do we know such a countable sequence with no limit points exist in $Xsetminus K$?
$endgroup$
– user475040
Jan 16 at 17:03
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Well it is homeomorphic to $mathbb R^2setminus B$, where it is easy to find such a sequence that goes out to infinity.
$endgroup$
– Cheerful Parsnip
Jan 16 at 17:18
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
Ok I see that now. I am still having trouble reaching a contradiction. Let ${y_i}_{iinmathbb N}$ be a countable sequence that has no limit points in $Xsetminus K$. By assumption, $|pi^{-1}({y_i})|>1$. Only one preimage of $y_i$ can lie in $mathbb R^2setminus B$ due to the homeomorphism. So all other preimages must lie in $B$.
$endgroup$
– user475040
Jan 16 at 18:08
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
$begingroup$
And since $B$ is compact, those preimages must have a limit point in $B$.
$endgroup$
– Cheerful Parsnip
Jan 16 at 19:44
1
1
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
$begingroup$
Okay I think I see it. So since $p$ is a limit point$V_k$ $textit{must} $ contain another preimage of $y_i$.
$endgroup$
– user475040
Jan 16 at 21:53
|
show 3 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075137%2flet-x-be-a-topological-space-and-pi-mathbb-r2-to-x-a-covering-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
