[幾何] homotopy

看板Math作者 (Steve)時間6年前 (2019/04/15 21:55), 6年前編輯推噓2(2推 0噓 3→)
留言5則, 3人參與, 6年前最新討論串1/1
Let S^n be the n-sphere and let f:S^n→S^n be a mapping such that f(x)≠-f(-x) for each x\in S^n. Then a mapping F:S^n ×[0,1]→S^n is well defined by tf(x)+(1-t)f(-x) F(x,t)=---------------------- ∥tf(x)+(1-t)f(-x)∥ . We can see from the above that f(x) is homotopic to f(-x). Now, I want to ask you a question. Is f(-x) homotopic to F(x,1/2)? The answer is yes, but why is that? Is it possible to apply the definition to determine whether f(-x) and F(x,1/2) are homotopic? I mean, can we construct a homotopy between f(-x) and F(x,1/2)? -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.233.124 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1555336530.A.37B.html
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04/15 22:22, 6年前 , 1F
t改成2t不行嗎
04/15 22:22, 1F
Did you mean sf(x)+(1/2-s)f(-x) F(x,2s)=------------------------ ?? ∥sf(x)+(1/2-s)f(-x)∥ Sorry, I don't see any difference. What I'm going to do is to find a homotopy between f(-x) and the mapping [f(x)+f(-x)]/2 ------------------ ∥[f(x)+f(-x)]/2∥ .
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04/15 23:47, 6年前 , 2F
If f(x)=-f(-x), then F(x,1/2) is not well-define
04/15 23:47, 2F
Sorry, I missed a minus sign. After correcting the sign error, we know the denominator is never zero for any x and any t. The mapping is well-defined.
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04/16 03:05, 6年前 , 3F
應該是他要避開的情況寫錯了吧，f(-x)=f(x)不會矛盾
04/16 03:05, 3F
Thx. ※ 編輯: rtyxn (140.112.233.124), 04/16/2019 07:40:19
推
04/16 11:46, 6年前 , 4F
The homotopy is simply F(x,t/2).
04/16 11:46, 4F
!!!!!!!!!!!!!!!!!!!!!!!!! Thank you! ※ 編輯: rtyxn (140.112.233.124), 04/16/2019 14:16:35
推
04/16 17:13, 6年前 , 5F
In fact, F(x,s) ~ F(x,r), for any s,r in [0,1]
04/16 17:13, 5F
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