$X/Awedge Y/B= (Xtimes Y)/(Xtimes Bcup Atimes Y)$

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

In Hatcher's book on vector bundles and K-theory, page 55, in order to extend the external product to the relative form, he uses the following identification:

$X/Awedge Y/B= (Xtimes Y)/(Xtimes Bcup Atimes Y)$,

For $X,Y$ compact Hausdorff spaces with $A,B$ closed subspaces of $X$ and $Y$, respectively.

I guess the identification is done by mapping the pair $([x],[y])$ to $[(x,y)]$.
But I don't know why it is enough, and this identification seems strange to me, since the smash product requires working with pointed spaces!

So, if I understand correctly, $X/Awedge Y/B= fracX/Atimes Y/B([x_0]times Y/B) cup (X/Atimes [y_0]) $ and I don't see the role of $(x_0,y_0)$ in $(Xtimes Y)/(Xtimes Bcup Atimes Y)$.

Thank you very much for any help, this topic is new for me.

general-topology algebraic-topology homology-cohomology vector-bundles topological-k-theory

share|cite|improve this question

asked Jul 21 at 11:32

Kiko

326111

add a comment | 

up vote
1
down vote

favorite

In Hatcher's book on vector bundles and K-theory, page 55, in order to extend the external product to the relative form, he uses the following identification:

$X/Awedge Y/B= (Xtimes Y)/(Xtimes Bcup Atimes Y)$,

For $X,Y$ compact Hausdorff spaces with $A,B$ closed subspaces of $X$ and $Y$, respectively.

I guess the identification is done by mapping the pair $([x],[y])$ to $[(x,y)]$.
But I don't know why it is enough, and this identification seems strange to me, since the smash product requires working with pointed spaces!

So, if I understand correctly, $X/Awedge Y/B= fracX/Atimes Y/B([x_0]times Y/B) cup (X/Atimes [y_0]) $ and I don't see the role of $(x_0,y_0)$ in $(Xtimes Y)/(Xtimes Bcup Atimes Y)$.

Thank you very much for any help, this topic is new for me.

general-topology algebraic-topology homology-cohomology vector-bundles topological-k-theory

share|cite|improve this question

asked Jul 21 at 11:32

Kiko

326111

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

In Hatcher's book on vector bundles and K-theory, page 55, in order to extend the external product to the relative form, he uses the following identification:

$X/Awedge Y/B= (Xtimes Y)/(Xtimes Bcup Atimes Y)$,

For $X,Y$ compact Hausdorff spaces with $A,B$ closed subspaces of $X$ and $Y$, respectively.

I guess the identification is done by mapping the pair $([x],[y])$ to $[(x,y)]$.
But I don't know why it is enough, and this identification seems strange to me, since the smash product requires working with pointed spaces!

So, if I understand correctly, $X/Awedge Y/B= fracX/Atimes Y/B([x_0]times Y/B) cup (X/Atimes [y_0]) $ and I don't see the role of $(x_0,y_0)$ in $(Xtimes Y)/(Xtimes Bcup Atimes Y)$.

Thank you very much for any help, this topic is new for me.

general-topology algebraic-topology homology-cohomology vector-bundles topological-k-theory

share|cite|improve this question

asked Jul 21 at 11:32

Kiko

326111

In Hatcher's book on vector bundles and K-theory, page 55, in order to extend the external product to the relative form, he uses the following identification:

$X/Awedge Y/B= (Xtimes Y)/(Xtimes Bcup Atimes Y)$,

For $X,Y$ compact Hausdorff spaces with $A,B$ closed subspaces of $X$ and $Y$, respectively.

I guess the identification is done by mapping the pair $([x],[y])$ to $[(x,y)]$.
But I don't know why it is enough, and this identification seems strange to me, since the smash product requires working with pointed spaces!

So, if I understand correctly, $X/Awedge Y/B= fracX/Atimes Y/B([x_0]times Y/B) cup (X/Atimes [y_0]) $ and I don't see the role of $(x_0,y_0)$ in $(Xtimes Y)/(Xtimes Bcup Atimes Y)$.

Thank you very much for any help, this topic is new for me.

general-topology algebraic-topology homology-cohomology vector-bundles topological-k-theory

share|cite|improve this question

asked Jul 21 at 11:32

Kiko

326111

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 11:32

Kiko

326111

asked Jul 21 at 11:32

Kiko

326111

asked Jul 21 at 11:32

Kiko

326111

326111

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

In fact, the identification involves pointed spaces, but it has nothing to do with special points $x_0 in X, y_0 in Y$. The space $X/A$ is regarded as a pointed space by taking as a basepoint the common equivalence class $[A]$ of all $a in A$, similarly $Y/B$. The space $X/A wedge Y/B$ has as a basepoint the common equivalence class $e = [eta]$ of all $eta in X/A times [B] cup [A] times Y/B$, the space $X times Y / X times B cup A times Y$ the common equivalence class $x = [xi]$ of all $xi in X times B cup A times Y$.

The map $p : X times Y to X/A times Y/B to X/A wedge Y/B$ identifies $X times B cup A times Y$ to $e$. It therefore induces a continuous bijection $p' : X times Y / X times B cup A times Y to X/A wedge Y/B$. The spaces $X times Y / X times B cup A times Y$ as well as $X/A, Y/B$ and $X/A wedge Y/B$ = $(X/A times Y/B) / (X/A times [B] cup [A] times Y/B)$ are compact. Therefore $p'$ is a homeomorphism.

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

add a comment | 

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858427%2fx-a-wedge-y-b-x-times-y-x-times-b-cup-a-times-y%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

In fact, the identification involves pointed spaces, but it has nothing to do with special points $x_0 in X, y_0 in Y$. The space $X/A$ is regarded as a pointed space by taking as a basepoint the common equivalence class $[A]$ of all $a in A$, similarly $Y/B$. The space $X/A wedge Y/B$ has as a basepoint the common equivalence class $e = [eta]$ of all $eta in X/A times [B] cup [A] times Y/B$, the space $X times Y / X times B cup A times Y$ the common equivalence class $x = [xi]$ of all $xi in X times B cup A times Y$.

The map $p : X times Y to X/A times Y/B to X/A wedge Y/B$ identifies $X times B cup A times Y$ to $e$. It therefore induces a continuous bijection $p' : X times Y / X times B cup A times Y to X/A wedge Y/B$. The spaces $X times Y / X times B cup A times Y$ as well as $X/A, Y/B$ and $X/A wedge Y/B$ = $(X/A times Y/B) / (X/A times [B] cup [A] times Y/B)$ are compact. Therefore $p'$ is a homeomorphism.

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

add a comment | 

up vote
2
down vote



accepted

In fact, the identification involves pointed spaces, but it has nothing to do with special points $x_0 in X, y_0 in Y$. The space $X/A$ is regarded as a pointed space by taking as a basepoint the common equivalence class $[A]$ of all $a in A$, similarly $Y/B$. The space $X/A wedge Y/B$ has as a basepoint the common equivalence class $e = [eta]$ of all $eta in X/A times [B] cup [A] times Y/B$, the space $X times Y / X times B cup A times Y$ the common equivalence class $x = [xi]$ of all $xi in X times B cup A times Y$.

The map $p : X times Y to X/A times Y/B to X/A wedge Y/B$ identifies $X times B cup A times Y$ to $e$. It therefore induces a continuous bijection $p' : X times Y / X times B cup A times Y to X/A wedge Y/B$. The spaces $X times Y / X times B cup A times Y$ as well as $X/A, Y/B$ and $X/A wedge Y/B$ = $(X/A times Y/B) / (X/A times [B] cup [A] times Y/B)$ are compact. Therefore $p'$ is a homeomorphism.

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

In fact, the identification involves pointed spaces, but it has nothing to do with special points $x_0 in X, y_0 in Y$. The space $X/A$ is regarded as a pointed space by taking as a basepoint the common equivalence class $[A]$ of all $a in A$, similarly $Y/B$. The space $X/A wedge Y/B$ has as a basepoint the common equivalence class $e = [eta]$ of all $eta in X/A times [B] cup [A] times Y/B$, the space $X times Y / X times B cup A times Y$ the common equivalence class $x = [xi]$ of all $xi in X times B cup A times Y$.

The map $p : X times Y to X/A times Y/B to X/A wedge Y/B$ identifies $X times B cup A times Y$ to $e$. It therefore induces a continuous bijection $p' : X times Y / X times B cup A times Y to X/A wedge Y/B$. The spaces $X times Y / X times B cup A times Y$ as well as $X/A, Y/B$ and $X/A wedge Y/B$ = $(X/A times Y/B) / (X/A times [B] cup [A] times Y/B)$ are compact. Therefore $p'$ is a homeomorphism.

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

In fact, the identification involves pointed spaces, but it has nothing to do with special points $x_0 in X, y_0 in Y$. The space $X/A$ is regarded as a pointed space by taking as a basepoint the common equivalence class $[A]$ of all $a in A$, similarly $Y/B$. The space $X/A wedge Y/B$ has as a basepoint the common equivalence class $e = [eta]$ of all $eta in X/A times [B] cup [A] times Y/B$, the space $X times Y / X times B cup A times Y$ the common equivalence class $x = [xi]$ of all $xi in X times B cup A times Y$.

The map $p : X times Y to X/A times Y/B to X/A wedge Y/B$ identifies $X times B cup A times Y$ to $e$. It therefore induces a continuous bijection $p' : X times Y / X times B cup A times Y to X/A wedge Y/B$. The spaces $X times Y / X times B cup A times Y$ as well as $X/A, Y/B$ and $X/A wedge Y/B$ = $(X/A times Y/B) / (X/A times [B] cup [A] times Y/B)$ are compact. Therefore $p'$ is a homeomorphism.

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 22:38

Paul Frost

3,658420

answered Jul 22 at 22:38

Paul Frost

3,658420

answered Jul 22 at 22:38

Paul Frost

3,658420

3,658420

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858427%2fx-a-wedge-y-b-x-times-y-x-times-b-cup-a-times-y%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email