The meaning of $pi_1(S^1,(1,0))$
$begingroup$
I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:
“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?
Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.
algebraic-topology homotopy-theory
asked Jan 14 at 15:33
User12239User12239
341216
$endgroup$
add a comment |
$begingroup$
I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:
“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?
Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.
algebraic-topology homotopy-theory
asked Jan 14 at 15:33
User12239User12239
341216
$endgroup$
3
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
1
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52
add a comment |
$begingroup$
I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:
“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?
Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.
algebraic-topology homotopy-theory
asked Jan 14 at 15:33
User12239User12239
341216
$endgroup$
I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:
“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?
Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Jan 14 at 15:33
User12239User12239
341216
asked Jan 14 at 15:33
User12239User12239
341216
edited Jan 14 at 15:39
User12239
asked Jan 14 at 15:33
User12239User12239
341216
asked Jan 14 at 15:33
User12239User12239
341216
asked Jan 14 at 15:33
User12239User12239
341216
341216
3
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
1
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52
add a comment |
3
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
1
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52
3
3
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
1
1
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3073349%2fthe-meaning-of-pi-1s1-1-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3073349%2fthe-meaning-of-pi-1s1-1-0%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
3
$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41
$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41
1
$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49
$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52