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 Posted: Mon Mar 05, 2007 11:27 pm
grey wanderer bluewolfcub Ok well I got question2 done eventually, but I've nooo idea about questions 1 or 3. I mean I know kinda how to test them, I just can't figure out what explicit retract would make the thing work. Any hints? sweatdrop Which assignment? Sorry I've been away for awhile... The only post I see with links to assignments appears to be from December. Last post was indeed december, but there's no due date on these assignments - only the end of the year. There's another one I have to hand in after this as well sweatdrop I know I'm leaving it late, but the guy said it might be worth 50% of the mark and I want to have it right. Noone else has handed them in either, apparently...
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Posted: Tue Mar 06, 2007 2:45 pm
EDIT: I foolishly forgot that slashes are dealt with differently on this board.. I've replaced them with -
When I first encountered these definitions I found it easier to think of the function f as a continuous family of maps. I'll use f_t if I want to talk about the map indexed by i.
Under this convention, we're looking for a continuous family of maps f_t such that:
1) Each f_t has the punctured space R^n - {0} as both domain and codomain
2) Each f_t is the identity on the unit sphere S^{n-1} living inside R^n - {0}
3) f_0 is the identity on R^n - {0}
4) f_1(R^n) =S^{n-1}
Hmmmm..... hmmmm......
Question Number One:
Where are the "holes" in R^n-{0}. I don't mean that in any technical sense... I just mean where does your intutition say the space is "flexible"
Answer:
0 and infinity
Extended Question Number Two:
How can we visualize a strong deformation retract of S^1 in R^2 -{0}?
I imagine having a dial with all real numbers from 0 to 1
When the dial is pointing at 0, I see a red circle centered on the black dot at the origin, and the rest of R^2 painted white.
As I turn the dial nothing seems to be happening until I reach 1 when all the outer area disappears and I'm left with a red circle with black interior and exterior. If I mark some points with gray paint and watch how they move as I turn the dial, I see them being pulled towards the red circle-- almost as if the circle had a gravitational pull. The closer I get to 1 the more quickly (and extreme) the movement of these gray dots.
The dots on the interior of the circle move radially outward from the center and the dots on the exterior of the circle move radially inward.
Just before I get to the 1 on the knob the the gray dots become a blinding blur of motion, and then as I turn the knob the last little bit the chaos collapses into nothingness and I am left with only the red circle.
Question Number 3:
What sort of symmetries did my example contain?
Answer:
Distance from the origin! All the point movied *RADIALLY* (hint! hint! hint!) Can you think of any UNIQUE property shared by ALL the points in S^{n-1}.
Question Number 4:
What sort of transformations would be spoiled when applied to a point whose distance from the origin is zero? Is this why we have to leave out the point {0}?
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Posted: Tue Mar 06, 2007 3:10 pm
As far as #3... well we could talk about it, but the Wikipedia article on Mapping cylinder actually has the homotopy in a rather unclear form (so it's more of a hint than an answer). Writing down their map carefully, filling in the missing details, and making it a rigorous proof would still require some work so giving you the reference isn't a freebie. Good Luck!
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Posted: Wed Mar 07, 2007 12:07 am
I'm confused...
I thought I'd just have to figure out some equ of x and t and prove it's an sdr by subbing in t=0 and make that = to, what, x? and t=1 should = something like x as well, and then sub in something else that I forget...
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Posted: Wed Mar 07, 2007 8:22 am
bluewolfcub I'm confused... I thought I'd just have to figure out some equ of x and t and prove it's an sdr by subbing in t=0 and make that = to, what, x? and t=1 should = something like x as well, and then sub in something else that I forget... That's mostly right. Your strong deformation retract is map from f:R^n - 0 cross I -> R^n - 0 (which is essentially #1 from my post) so f(x,0) = x (which was #3 in my post)) f(y,t) = y whenever y in S^{n-1} (which was #2 in my post) f(x,1) in S^{n-1} for all x in R^n - 0 (which was #4 in my post) I was just giving you hints as to which map would do the job.
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Posted: Wed Mar 07, 2007 8:56 am
My brain is fried today, sorry sweatdrop
er, would something along the lines of (1-t)x + t.(x/ |x|) work?
tryna think about some kinda unit vector thing
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Posted: Wed Mar 07, 2007 9:29 am
bluewolfcub My brain is fried today, sorry sweatdrop er, would something along the lines of (1-t)x + t.(x/ |x|) work? tryna think about some kinda unit vector thing Seems plausible. You just have to show that it satisfies all the criterion.
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Posted: Wed Mar 07, 2007 9:49 am
grey wanderer bluewolfcub My brain is fried today, sorry sweatdrop er, would something along the lines of (1-t)x + t.(x/ |x|) work? tryna think about some kinda unit vector thing Seems plausible. You just have to show that it satisfies all the criterion. I can't get what the third one to check is sub in something for x, instead of t? ok so t=0, you have x, thats fine, t=1, you have... x/|x|... er... whats that then? 1, innit? that's the same in both?
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Posted: Wed Mar 07, 2007 10:49 am
bluewolfcub grey wanderer bluewolfcub My brain is fried today, sorry sweatdrop er, would something along the lines of (1-t)x + t.(x/ |x|) work? tryna think about some kinda unit vector thing Seems plausible. You just have to show that it satisfies all the criterion. I can't get what the third one to check is sub in something for x, instead of t? ok so t=0, you have x, thats fine, t=1, you have... x/|x|... er... whats that then? 1, innit? that's the same in both? x/|x| isn't 1. It's a vector whose length is 1. Consider the pt (3,4) It has length 5 (standard Euclidean of course), then f_1(3,4) = (3/5,4/5) That's a point on the unit circle... as intended. On the other hand you need to show (switching back to the *other* notation) that f(x,t) = x whenever x is in S^{n-1} in other words f(x,t) = x whenever |x| = 1 no matter what t is. And... as you've shown |f(x,1)| = 1 so f(x,1) in S^{n-1}
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Posted: Wed Apr 04, 2007 11:58 am
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Posted: Thu Apr 05, 2007 8:43 am
I've been bogged down with other assignments and tests so I *still* havent finished this sweatdrop oh but I got a first in my last assignment! 75%! biggrin
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Posted: Fri Apr 06, 2007 9:49 am
bluewolfcub I've been bogged down with other assignments and tests so I *still* havent finished this sweatdrop oh but I got a first in my last assignment! 75%! biggrin Congratulations! And good luck on the second assignment.
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Posted: Mon Apr 16, 2007 8:16 am
grey wanderer On the other hand you need to show (switching back to the *other* notation) that f(x,t) = x whenever x is in S^{n-1} in other words f(x,t) = x whenever |x| = 1 no matter what t is. And... as you've shown |f(x,1)| = 1 so f(x,1) in S^{n-1} oh. (1-t)x + t(x/|x|) doesnt do that... oh! ( (1-t)+t )x ___________ |x| does work.... yay or does it... f(x,0) still leaves me with x/|x| i hope thats ok sweatdrop
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Posted: Wed May 09, 2007 7:15 am
right! can anyone help with the cone question
X is contractible <=> X is a retract of CX
I dont veen know where to start sweatdrop
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Posted: Thu May 10, 2007 4:45 pm
bluewolfcub right! can anyone help with the cone question X is contractible <=> X is a retract of CX I dont veen know where to start sweatdrop It's another definition game. I'd start by carefully writing out what it means for X to be a retract of CX. (in particlar, it means there is a function from f: CX->X with certain properties.) Then I would carefully write out the definition of CX. I claim that this function, f, tells you how to build a homotopy between X and a single point. Once you understand the proof in this direction, it should be fairly straight forward to go back in the other direction. (I know my hints can be hard to follow sometimes, so feel free to ask for more details if this isn't enough to get you started)
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