Hatcher Algebraic Topology 0.24
$begingroup$
This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that
$$X ast Y = S(X wedge Y)$$
by showing
$$X ast Y/(X ast y_0 cup x_0 ast Y) = S(X wedge Y) / S(x_0 wedge y_0),$$
where $ast$ is topological join, $wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.
Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0wedge y_0)$ is suspension of a point which is an interval I. $X wedge Y$ is the space obtained by the quotient $X times Y / X vee Y$ where $X vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have
$$X times Y rightarrow X times Y / X vee Y rightarrow (X times Y / X vee Y) times I rightarrow Z = (X times Y / X vee Y) times I/(text{contract $t=0$ and $t=1$ to a point}) rightarrow Z / S(x_0wedge y_0).$$
$(X ast {y_0} cup {x_0} ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is
$$X times Y times I rightarrow A= X times Y times I/(text{contract $Y$ at $t=0$ and $X$ at $t=1$}) rightarrow A/(X ast {y_0} cup {x_0} ast Y).$$
However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.
Thanks a lot.
algebraic-topology
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
$endgroup$
add a comment |
$begingroup$
This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that
$$X ast Y = S(X wedge Y)$$
by showing
$$X ast Y/(X ast y_0 cup x_0 ast Y) = S(X wedge Y) / S(x_0 wedge y_0),$$
where $ast$ is topological join, $wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.
Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0wedge y_0)$ is suspension of a point which is an interval I. $X wedge Y$ is the space obtained by the quotient $X times Y / X vee Y$ where $X vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have
$$X times Y rightarrow X times Y / X vee Y rightarrow (X times Y / X vee Y) times I rightarrow Z = (X times Y / X vee Y) times I/(text{contract $t=0$ and $t=1$ to a point}) rightarrow Z / S(x_0wedge y_0).$$
$(X ast {y_0} cup {x_0} ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is
$$X times Y times I rightarrow A= X times Y times I/(text{contract $Y$ at $t=0$ and $X$ at $t=1$}) rightarrow A/(X ast {y_0} cup {x_0} ast Y).$$
However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.
Thanks a lot.
algebraic-topology
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
$endgroup$
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13
add a comment |
$begingroup$
This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that
$$X ast Y = S(X wedge Y)$$
by showing
$$X ast Y/(X ast y_0 cup x_0 ast Y) = S(X wedge Y) / S(x_0 wedge y_0),$$
where $ast$ is topological join, $wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.
Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0wedge y_0)$ is suspension of a point which is an interval I. $X wedge Y$ is the space obtained by the quotient $X times Y / X vee Y$ where $X vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have
$$X times Y rightarrow X times Y / X vee Y rightarrow (X times Y / X vee Y) times I rightarrow Z = (X times Y / X vee Y) times I/(text{contract $t=0$ and $t=1$ to a point}) rightarrow Z / S(x_0wedge y_0).$$
$(X ast {y_0} cup {x_0} ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is
$$X times Y times I rightarrow A= X times Y times I/(text{contract $Y$ at $t=0$ and $X$ at $t=1$}) rightarrow A/(X ast {y_0} cup {x_0} ast Y).$$
However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.
Thanks a lot.
algebraic-topology
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
$endgroup$
This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that
$$X ast Y = S(X wedge Y)$$
by showing
$$X ast Y/(X ast y_0 cup x_0 ast Y) = S(X wedge Y) / S(x_0 wedge y_0),$$
where $ast$ is topological join, $wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.
Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0wedge y_0)$ is suspension of a point which is an interval I. $X wedge Y$ is the space obtained by the quotient $X times Y / X vee Y$ where $X vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have
$$X times Y rightarrow X times Y / X vee Y rightarrow (X times Y / X vee Y) times I rightarrow Z = (X times Y / X vee Y) times I/(text{contract $t=0$ and $t=1$ to a point}) rightarrow Z / S(x_0wedge y_0).$$
$(X ast {y_0} cup {x_0} ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is
$$X times Y times I rightarrow A= X times Y times I/(text{contract $Y$ at $t=0$ and $X$ at $t=1$}) rightarrow A/(X ast {y_0} cup {x_0} ast Y).$$
However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.
Thanks a lot.
algebraic-topology
algebraic-topology
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
edited Aug 1 '12 at 22:49
Henry T. Horton
15.2k54665
15.2k54665
asked Aug 1 '12 at 21:04
SinaSina
533
asked Aug 1 '12 at 21:04
SinaSina
533
asked Aug 1 '12 at 21:04
SinaSina
533
533
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13
add a comment |
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)sim (x_2,y,0);$$
$$ (x,y_1,1)sim (x,y_2,1).$$
The reduced version also collapses $x_0ast Y$ and $Xast y_0$.
So the additional relations are:
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y_0,t)sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)sim(x_0,y,0)sim(x_0,y_0,0);$$
$$(x,y,1)sim (x,y_0,1)sim(x_0,y_0,0).$$
The smash product is gotten from $Xtimes Y$ by collapsing $Xtimes y_0$ and $x_0times Y$. The suspension of that can be thought of as $Xtimes Ytimes I$, with the relations:
$$ (x,y_0,t)sim (x_0,y_0,t);$$
$$ (x_0,y,t)sim (x_0,y_0,t);$$
$$ (x,y,1)sim (x_0,y_0,1);$$
$$ (x,y,0)sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)sim (x_0,y_0,0);$$
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y,0)sim (x_0,y_0,0);$$
$$ (x,y,1)sim (x_0,y_0,0).$$
$endgroup$
add a comment |
$begingroup$
Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as 
and the subspace to be collapsed to a point to give the suspension of the smash product is
where the two vertices on the mid line are the base points.
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
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',
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%2f177775%2fhatcher-algebraic-topology-0-24%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)sim (x_2,y,0);$$
$$ (x,y_1,1)sim (x,y_2,1).$$
The reduced version also collapses $x_0ast Y$ and $Xast y_0$.
So the additional relations are:
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y_0,t)sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)sim(x_0,y,0)sim(x_0,y_0,0);$$
$$(x,y,1)sim (x,y_0,1)sim(x_0,y_0,0).$$
The smash product is gotten from $Xtimes Y$ by collapsing $Xtimes y_0$ and $x_0times Y$. The suspension of that can be thought of as $Xtimes Ytimes I$, with the relations:
$$ (x,y_0,t)sim (x_0,y_0,t);$$
$$ (x_0,y,t)sim (x_0,y_0,t);$$
$$ (x,y,1)sim (x_0,y_0,1);$$
$$ (x,y,0)sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)sim (x_0,y_0,0);$$
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y,0)sim (x_0,y_0,0);$$
$$ (x,y,1)sim (x_0,y_0,0).$$
$endgroup$
add a comment |
$begingroup$
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)sim (x_2,y,0);$$
$$ (x,y_1,1)sim (x,y_2,1).$$
The reduced version also collapses $x_0ast Y$ and $Xast y_0$.
So the additional relations are:
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y_0,t)sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)sim(x_0,y,0)sim(x_0,y_0,0);$$
$$(x,y,1)sim (x,y_0,1)sim(x_0,y_0,0).$$
The smash product is gotten from $Xtimes Y$ by collapsing $Xtimes y_0$ and $x_0times Y$. The suspension of that can be thought of as $Xtimes Ytimes I$, with the relations:
$$ (x,y_0,t)sim (x_0,y_0,t);$$
$$ (x_0,y,t)sim (x_0,y_0,t);$$
$$ (x,y,1)sim (x_0,y_0,1);$$
$$ (x,y,0)sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)sim (x_0,y_0,0);$$
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y,0)sim (x_0,y_0,0);$$
$$ (x,y,1)sim (x_0,y_0,0).$$
$endgroup$
add a comment |
$begingroup$
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)sim (x_2,y,0);$$
$$ (x,y_1,1)sim (x,y_2,1).$$
The reduced version also collapses $x_0ast Y$ and $Xast y_0$.
So the additional relations are:
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y_0,t)sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)sim(x_0,y,0)sim(x_0,y_0,0);$$
$$(x,y,1)sim (x,y_0,1)sim(x_0,y_0,0).$$
The smash product is gotten from $Xtimes Y$ by collapsing $Xtimes y_0$ and $x_0times Y$. The suspension of that can be thought of as $Xtimes Ytimes I$, with the relations:
$$ (x,y_0,t)sim (x_0,y_0,t);$$
$$ (x_0,y,t)sim (x_0,y_0,t);$$
$$ (x,y,1)sim (x_0,y_0,1);$$
$$ (x,y,0)sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)sim (x_0,y_0,0);$$
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y,0)sim (x_0,y_0,0);$$
$$ (x,y,1)sim (x_0,y_0,0).$$
$endgroup$
Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).
The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are:
$$ (x_1,y,0)sim (x_2,y,0);$$
$$ (x,y_1,1)sim (x,y_2,1).$$
The reduced version also collapses $x_0ast Y$ and $Xast y_0$.
So the additional relations are:
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y_0,t)sim (x_0,y_0,0).$$
We can derive further relations too:
$$(x,y,0)sim(x_0,y,0)sim(x_0,y_0,0);$$
$$(x,y,1)sim (x,y_0,1)sim(x_0,y_0,0).$$
The smash product is gotten from $Xtimes Y$ by collapsing $Xtimes y_0$ and $x_0times Y$. The suspension of that can be thought of as $Xtimes Ytimes I$, with the relations:
$$ (x,y_0,t)sim (x_0,y_0,t);$$
$$ (x_0,y,t)sim (x_0,y_0,t);$$
$$ (x,y,1)sim (x_0,y_0,1);$$
$$ (x,y,0)sim (x_0,y_0,0).$$
The reduced suspension adds the relation
$$ (x_0,y_0,t)sim (x_0,y_0,0).$$
Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,
$$ (x,y_0,t)sim (x_0,y_0,0);$$
$$ (x_0,y,t)sim (x_0,y_0,0);$$
$$ (x,y,0)sim (x_0,y_0,0);$$
$$ (x,y,1)sim (x_0,y_0,0).$$
answered Aug 1 '12 at 23:33
user641
add a comment |
add a comment |
$begingroup$
Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as
and the subspace to be collapsed to a point to give the suspension of the smash product is
where the two vertices on the mid line are the base points.
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as
and the subspace to be collapsed to a point to give the suspension of the smash product is
where the two vertices on the mid line are the base points.
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as
and the subspace to be collapsed to a point to give the suspension of the smash product is
where the two vertices on the mid line are the base points.
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as
and the subspace to be collapsed to a point to give the suspension of the smash product is
where the two vertices on the mid line are the base points.
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
edited Dec 23 '18 at 10:16
Glorfindel
3,41981830
3,41981830
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
answered Aug 2 '12 at 16:38
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
add a comment |
add a comment |
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%2f177775%2fhatcher-algebraic-topology-0-24%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
$begingroup$
See proposition 4I.1
$endgroup$
– user641
Aug 1 '12 at 22:57
$begingroup$
Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence.
$endgroup$
– KReiser
Aug 1 '12 at 23:01
$begingroup$
Yes, I agree with KReiser that you are trying to show $X*Ysimeq S(Xwedge Y)$ where $simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $Xtimes Y times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients.
$endgroup$
– Aaron
Aug 1 '12 at 23:13