Dgdrxrt

Can this transistor (2N2222) take 6 V on emitter-base? Am I reading the datasheet incorrectly?

Is there a rule of thumb for determining the amount one should accept for a settlement offer?

Is it reasonable to ask other researchers to send me their previous grant applications?

My boss doesn't want me to have a side project

Strange use of "whether ... than ..." in official text

What steps are necessary to read a Modern SSD in Medieval Europe?

Is it okay to majorly distort historical facts while writing a fiction story?

Shortening a title without changing its meaning

Is a distribution that is normal, but highly skewed, considered Gaussian?

Creating a script with console commands

Free fall ellipse or parabola?

How to show a landlord what we have in savings?

Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?

Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?

What difference does it make matching a word with/without a trailing whitespace?

Early programmable calculators with RS-232

Calculating discount not working

Can you teleport closer to a creature you are Frightened of?

Can a PhD from a non-TU9 German university become a professor in a TU9 university?

How seriously should I take size and weight limits of hand luggage?

How badly should I try to prevent a user from XSSing themselves?

How to coordinate airplane tickets?

Why does sin(x) - sin(y) equal this?

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

Show that the $Delta$-complex obtained from $Delta^3$ by performing edge identifications deformation retracts onto a Klein bottle.

8

3

$begingroup$

I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?

algebraic-topology simplicial-complex

share|cite|improve this question

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

$endgroup$

add a comment | 

8

3

$begingroup$

I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?

algebraic-topology simplicial-complex

share|cite|improve this question

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

$endgroup$

add a comment | 

$begingroup$

I am going through some exercises in Hatcher's Algebraic Topology.
You have a $Delta$-complex obtained from $Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]sim[v_1,v_3]$ and $[v_0,v_2]sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?

algebraic-topology simplicial-complex

share|cite|improve this question

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

$endgroup$

share|cite|improve this question

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

share|cite|improve this question

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

edited Nov 6 '18 at 9:51

Batominovski

33.1k33293

asked Jan 18 '12 at 15:07

0986709867

294414

asked Jan 18 '12 at 15:07

0986709867

294414

2 Answers
2

active

oldest

votes

0

$begingroup$

The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.

share|cite|improve this answer

edited Jul 3 '17 at 10:39

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

$endgroup$

add a comment | 

0

$begingroup$

flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.

sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...

edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.

the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)

share|cite|improve this answer

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
  $endgroup$
  – NKS
  Jan 18 '12 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
  $endgroup$
  – 09867
  Jan 18 '12 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NKS yes you do have to cut it up, my bad
  $endgroup$
  – yoyo
  Jan 18 '12 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
  $endgroup$
  – yoyostein
  Jun 1 '16 at 9:28
  
  

  
  
  
  

add a comment | 

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

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

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f100157%2fshow-that-the-delta-complex-obtained-from-delta3-by-performing-edge-ident%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.

share|cite|improve this answer

edited Jul 3 '17 at 10:39

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

$endgroup$

add a comment | 

0

$begingroup$

The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.

share|cite|improve this answer

edited Jul 3 '17 at 10:39

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

$endgroup$

add a comment | 

$begingroup$

The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.

share|cite|improve this answer

edited Jul 3 '17 at 10:39

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

$endgroup$

share|cite|improve this answer

edited Jul 3 '17 at 10:39

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

answered Jun 29 '17 at 14:22

Ka HoKa Ho

62

0

$begingroup$

flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.

sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...

edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.

the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)

share|cite|improve this answer

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
  $endgroup$
  – NKS
  Jan 18 '12 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
  $endgroup$
  – 09867
  Jan 18 '12 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NKS yes you do have to cut it up, my bad
  $endgroup$
  – yoyo
  Jan 18 '12 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
  $endgroup$
  – yoyostein
  Jun 1 '16 at 9:28
  
  

  
  
  
  

add a comment | 

0

$begingroup$

flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.

sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...

edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.

the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)

share|cite|improve this answer

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle.
  $endgroup$
  – NKS
  Jan 18 '12 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square.
  $endgroup$
  – 09867
  Jan 18 '12 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NKS yes you do have to cut it up, my bad
  $endgroup$
  – yoyo
  Jan 18 '12 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks!
  $endgroup$
  – yoyostein
  Jun 1 '16 at 9:28
  
  

  
  
  
  

add a comment | 

$begingroup$

flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle.

sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...

edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $a=[v_0,v_1]=[v_1,v_3]$, $b=[v_0,v_2]=[v_2,v_3]$, and i'm introducing new edges $c$ and $d$. cutting the second triangle into two rectangles (both with edge labels $a,b,c,d$), then regluing along $a$ gives you a rectangle. this is the "standard" klein bottle.

the left two rectangles are what you get by cutting the second triangle along $c,d$. the right two are supposed to indicate regluing along $a$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)

share|cite|improve this answer

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

$endgroup$

share|cite|improve this answer

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

edited Feb 22 at 19:13

Glorfindel

3,41581830

edited Feb 22 at 19:13

Glorfindel

3,41581830

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726

answered Jan 18 '12 at 16:09

yoyoyoyo

6,6211726