Timeline for Maps of balls with fixed value along boundary
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2015 at 14:32 | vote | accept | ali elgindi | ||
| Feb 24, 2015 at 13:35 | comment | added | Allen Hatcher | On the other hand, if you wish to allow only homotopies $f_t$ and $g_t$ such that $f_t=g_t$ on $S^2$ for all $t$, then such pairs $(f,g)$ are equivalent to maps $h:S^3\to X$. Homotopy classes of maps $h:S^3\to X$ without basepoint conditions are in one-to-one correspondence with orbits of the action of $\pi_1(X)$ on $\pi_3(X)$, and this set of orbits can be nontrivial. | |
| Feb 24, 2015 at 13:23 | comment | added | Allen Hatcher | If you do not require $f=g$ on $S^2$ during homotopies, then the problem becomes trivial since every map $B^3\to X$ is homotopic to a constant map. Since we may assume $X$ is path-connected, this means $f$ and $g$ are homotopic. | |
| Feb 24, 2015 at 12:13 | comment | added | ali elgindi | Thank you very much for your help Professor Hatcher. Your books were an excellent guide during my topology classes and great reference books for the (budding) research mathematician. I understand now that homotopies of maps $B^3 \rightarrow X$ which are all fixed along the boundary $S^2$ by a specific map is equivalent to $\pi_3 (M)$. In my problem, I have two specific maps $f,g:B^3 \rightarrow X$ that agree along the boundary. All I need is that they are homotopic, not necessarily through maps all agreeing along the boundary. Would this relaxed condition make a difference? @AllenHatcher | |
| Feb 23, 2015 at 14:00 | history | answered | Allen Hatcher | CC BY-SA 3.0 |