Hatcher Ex $0.25$, the homotopy equivalence of suspension of a CW complex












1












$begingroup$


Exercise $0.25$:



If $X$ is a CW complex with components $X_alpha$, show that the suspension $SX$ is homotopy equivalent to $Ylor_alpha SX_alpha$ for some graph $Y$.



In the case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and $2$-spheres.



Edit: It's not hard actually, from the basic examples to find what graph $Y$ is.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
    $endgroup$
    – John Hughes
    Oct 30 '18 at 17:57












  • $begingroup$
    @JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
    $endgroup$
    – Andrews
    Oct 31 '18 at 8:44






  • 2




    $begingroup$
    If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
    $endgroup$
    – Arnaud D.
    Oct 31 '18 at 11:15










  • $begingroup$
    @ArnaudD. OK.Thanks.
    $endgroup$
    – Andrews
    Oct 31 '18 at 11:21






  • 1




    $begingroup$
    I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
    $endgroup$
    – John Hughes
    Oct 31 '18 at 15:40
















1












$begingroup$


Exercise $0.25$:



If $X$ is a CW complex with components $X_alpha$, show that the suspension $SX$ is homotopy equivalent to $Ylor_alpha SX_alpha$ for some graph $Y$.



In the case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and $2$-spheres.



Edit: It's not hard actually, from the basic examples to find what graph $Y$ is.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
    $endgroup$
    – John Hughes
    Oct 30 '18 at 17:57












  • $begingroup$
    @JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
    $endgroup$
    – Andrews
    Oct 31 '18 at 8:44






  • 2




    $begingroup$
    If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
    $endgroup$
    – Arnaud D.
    Oct 31 '18 at 11:15










  • $begingroup$
    @ArnaudD. OK.Thanks.
    $endgroup$
    – Andrews
    Oct 31 '18 at 11:21






  • 1




    $begingroup$
    I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
    $endgroup$
    – John Hughes
    Oct 31 '18 at 15:40














1












1








1





$begingroup$


Exercise $0.25$:



If $X$ is a CW complex with components $X_alpha$, show that the suspension $SX$ is homotopy equivalent to $Ylor_alpha SX_alpha$ for some graph $Y$.



In the case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and $2$-spheres.



Edit: It's not hard actually, from the basic examples to find what graph $Y$ is.










share|cite|improve this question











$endgroup$




Exercise $0.25$:



If $X$ is a CW complex with components $X_alpha$, show that the suspension $SX$ is homotopy equivalent to $Ylor_alpha SX_alpha$ for some graph $Y$.



In the case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and $2$-spheres.



Edit: It's not hard actually, from the basic examples to find what graph $Y$ is.







general-topology algebraic-topology cw-complexes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:18







Andrews

















asked Oct 30 '18 at 17:44









AndrewsAndrews

3551317




3551317












  • $begingroup$
    Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
    $endgroup$
    – John Hughes
    Oct 30 '18 at 17:57












  • $begingroup$
    @JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
    $endgroup$
    – Andrews
    Oct 31 '18 at 8:44






  • 2




    $begingroup$
    If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
    $endgroup$
    – Arnaud D.
    Oct 31 '18 at 11:15










  • $begingroup$
    @ArnaudD. OK.Thanks.
    $endgroup$
    – Andrews
    Oct 31 '18 at 11:21






  • 1




    $begingroup$
    I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
    $endgroup$
    – John Hughes
    Oct 31 '18 at 15:40


















  • $begingroup$
    Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
    $endgroup$
    – John Hughes
    Oct 30 '18 at 17:57












  • $begingroup$
    @JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
    $endgroup$
    – Andrews
    Oct 31 '18 at 8:44






  • 2




    $begingroup$
    If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
    $endgroup$
    – Arnaud D.
    Oct 31 '18 at 11:15










  • $begingroup$
    @ArnaudD. OK.Thanks.
    $endgroup$
    – Andrews
    Oct 31 '18 at 11:21






  • 1




    $begingroup$
    I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
    $endgroup$
    – John Hughes
    Oct 31 '18 at 15:40
















$begingroup$
Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
$endgroup$
– John Hughes
Oct 30 '18 at 17:57






$begingroup$
Consider the case where $X$ is two points -- say $+1$ and $-1$ on the number line. The suspension adds two more points, one above the line and one below, and connects each to the points of $X$, yielding a diamond, which is topologically a circle. So it's certainly possible. You might want to think through the case where $X$ consists of just 3 points on the number line. Unrelated: What does that subscript "alpha" on the join $U vee_alpha SX_alpha$ mean?
$endgroup$
– John Hughes
Oct 30 '18 at 17:57














$begingroup$
@JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
$endgroup$
– Andrews
Oct 31 '18 at 8:44




$begingroup$
@JohnHughes Thanks for your time and help. I misunderstood the definition of $SX$.
$endgroup$
– Andrews
Oct 31 '18 at 8:44




2




2




$begingroup$
If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
$endgroup$
– Arnaud D.
Oct 31 '18 at 11:15




$begingroup$
If you have found your answer, you can write an answer to your own question (and even accept it after a delay). It could be useful for others. In any case, avoid modifying the title like you did.
$endgroup$
– Arnaud D.
Oct 31 '18 at 11:15












$begingroup$
@ArnaudD. OK.Thanks.
$endgroup$
– Andrews
Oct 31 '18 at 11:21




$begingroup$
@ArnaudD. OK.Thanks.
$endgroup$
– Andrews
Oct 31 '18 at 11:21




1




1




$begingroup$
I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
$endgroup$
– John Hughes
Oct 31 '18 at 15:40




$begingroup$
I agree,,,writing up your answer (without completely solving the problem, which would lessen the value of Hatcher's book to future students) would be a great way to give back to MSE (and to solidify your own understanding of the point on which you were confused).
$endgroup$
– John Hughes
Oct 31 '18 at 15:40










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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2977996%2fhatcher-ex-0-25-the-homotopy-equivalence-of-suspension-of-a-cw-complex%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2977996%2fhatcher-ex-0-25-the-homotopy-equivalence-of-suspension-of-a-cw-complex%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Berounka

Fiat S.p.A.

Type 'String' is not a subtype of type 'int' of 'index'