Timeline for "Ambient homotopy" between preimages under a fiber bundle?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2015 at 12:48 | comment | added | Dimitri Chikhladze | @RyanBudney Yeah, I did not mean anything specific by the "ambient homotopy". Your argument however shows that the answer is negative when one choses it to be the ambient isotopy. | |
| Nov 18, 2015 at 12:38 | vote | accept | Dimitri Chikhladze | ||
| Nov 18, 2015 at 6:04 | comment | added | Gustavo Granja | @Ryan Budney: I don't think the OP is using "ambient homotopy" in any technical sense. | |
| Nov 17, 2015 at 21:51 | comment | added | Ryan Budney | @PeterLeFanuLumsdaine: but it will certainly fail for embeddings also. For example, a 1-parameter family of embeddings usually does not come from an ambient homotopy (ambient isotopy in the topological category). For example, co-dimension two knots can all be "pulled tight" into unknots, via 1-parameter families of embeddings. | |
| Nov 17, 2015 at 20:38 | answer | added | Gustavo Granja | timeline score: 4 | |
| Nov 17, 2015 at 17:50 | comment | added | Dimitri Chikhladze | @GustavoGranja Can you add more details? | |
| Nov 17, 2015 at 16:28 | comment | added | Gustavo Granja | Now that the question has been changed, the answer is yes. If $f_t$ is an isotopy between $f$ and $g$ (say $f_0=f, f_1=g$) then $X_{f_t}$ will embed for each $t$. By homotopy invariance of bundles, for each $t$, $X_{f_t}$ is equivalent to $X_f$ as a bundle over $X$. The choice of identification is a choice of isomorphism between the pullback via the isotopy and the bundle $X_f \times[0,1]$. | |
| Nov 17, 2015 at 15:26 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
added 1 character in body; edited tags
|
| Nov 17, 2015 at 14:33 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
added 106 characters in body
|
| Nov 17, 2015 at 14:28 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
added 106 characters in body
|
| Nov 17, 2015 at 10:06 | comment | added | Mark Grant | There is a notion of when maps $f: Y\to X$ and $f': Y'\to X$ are homotopically equivalent. This means there is a homotopy equivalence $h: Y\to Y'$ making the evident triangle commute. I think that is the best you can hope for here. | |
| Nov 17, 2015 at 1:31 | comment | added | Peter LeFanu Lumsdaine | @RyanBudney: if $f, g$ are supposed to be embeddings, then that rules out the constant map. | |
| Nov 17, 2015 at 1:02 | comment | added | Ryan Budney | I don't there there's any reasonable (geometrically interesting) definition where your question would have an affirmative answer. Let $p$ be the double cover of the circle. Let $f$ and $g$ be maps from $S^0$ to $S^1$, one being a constant map and the other giving antipodal points. Any isomorphism of the pull-back's domains require choices that can not be done by some single isotopy in the target circle. You can compound this example by letting $f$ and $g$ have higher-dimensional domains. | |
| Nov 17, 2015 at 0:54 | history | edited | Dimitri Chikhladze | CC BY-SA 3.0 |
added 560 characters in body
|
| Nov 17, 2015 at 0:42 | comment | added | Dimitri Chikhladze | I mean maps $X_f \rightarrow E$ and $Y_f \rightarrow E$ which by Peter's comment still have different domains. In fact the question subsumes defining the appropriate notion of the ambient homotopy. | |
| Nov 17, 2015 at 0:26 | comment | added | Peter LeFanu Lumsdaine | A priori, the pullbacks of f and g along p have different codomains $f^*E$, $g^*E$. To make sense of the question, one has to identify these somehow; one certainly can do so, since they’re pullbacks of homotopic maps into a bundle, but the answer may be sensitive to how one does so, and how nice your notion of ambient homotopy is. | |
| Nov 17, 2015 at 0:25 | comment | added | Ryan Budney | Something in your question does not parse for me. The two pull-backs are maps $f^*p : E_f \to X$ and $g^*p : E_g \to X$ these are maps from two different spaces to $X$, so your definition of ambient homotopy does not make sense. | |
| Nov 16, 2015 at 22:30 | history | asked | Dimitri Chikhladze | CC BY-SA 3.0 |