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I need to know the following: let $f:\rightarrow {\mathbb R}$ be a real-analytic function defined in a neighbourhood of a point in an analytic manifold. If the fiber $f^{-1}(0)$ is simply connected, is that so for any fiber $f^{-1}(\epsilon)$ for small $\epsilon$?

In the complex case I gather this is a trivial consequence of Grauert's semicontinuity theorem but what about the real case?

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Try $f(x,y)=x^2+y^2$, $f \colon \mathbb{R}^2 \to \mathbb{R}$.

If you want connected constant dimensional fibers, try $f(x,y,z)=z$ on the cylinder with one point deleted $X=\{(x,y,z)|x^2+y^2=1\}-\{(1,0,0)\}$.

For a more natural example, take the map $f(x,y)=y^3-p-qx-rx^2-x^3$ for suitable constants $p, q, r$, so that you get a family of cubic curves, and arrange that the curve $f^{-1}(0)$ is a cusp, while $f^{-1}(\varepsilon)$ has two components for $\varepsilon > 0$. Then you really see jumping of fundamental group. But I still don't have a nice example with all fibers compact.

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    $\begingroup$ 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. $\endgroup$
    – pfortuny
    Commented Jun 29, 2017 at 11:04
  • $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Jun 29, 2017 at 12:49
  • $\begingroup$ Yes. Sorry again, (0,...,0) is an isolated singularity of (f=0), so that fiber is not a differentiable manifold. $\endgroup$
    – pfortuny
    Commented Jun 29, 2017 at 15:40
  • $\begingroup$ You didn't say you wanted the fibers to be manifolds. $\endgroup$
    – Ben McKay
    Commented Jun 29, 2017 at 16:13

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