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Jun 29, 2017 at 16:13 comment added Ben McKay You didn't say you wanted the fibers to be manifolds.
Jun 29, 2017 at 15:40 comment added pfortuny Yes. Sorry again, (0,...,0) is an isolated singularity of (f=0), so that fiber is not a differentiable manifold.
Jun 29, 2017 at 13:57 history edited Ben McKay CC BY-SA 3.0
added another example
Jun 29, 2017 at 12:49 comment added Ben McKay If the fibers are all compact, and $f$ is a submersion, then all fibers near any given fiber are diffeomorphic, by usual argument: construct (using partition of unity) a vector field on $X$ projecting to the usual translation field on $\mathbb{R}$, and flow fibers.
Jun 29, 2017 at 12:46 history edited Ben McKay CC BY-SA 3.0
added constant dimensional fiber case
Jun 29, 2017 at 11:04 comment added pfortuny Oh, sorry. This comes from thinking without examples...... Thanks. The fibers have constant dimension, in my case. But I have to look at the problem in more detail.
Jun 29, 2017 at 11:03 history answered Ben McKay CC BY-SA 3.0