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EDIT: Let $M,N$ be compact connected smooth Riemannian manifolds. Let us assume that $N$ is closed, while $M$ might have a geodesically convex boundary.

Given $f\colon M\to N$ be an $\varepsilon$-almost Riemannian submersion, i.e. a submersion such that for any point $x\in M$ and any tangent vector $v\in T_xM$ which is orthogonal to the tangent space $T_xF$ to the fiber $F=f^{-1}(f(x))$ one has $$(1-\varepsilon)|v|\leq |df_x(v)|\leq (1+\varepsilon)|v|.$$

Assume in addition that the restriction $f|_{\partial M}\colon \partial M\to N$ is an $\varepsilon$-almost Riemannian submersion.

Let us denote by $D$ the maximal diameter of the fibers of $f$.

Question. Is it true that if $0\leq \varepsilon < 1/100$ and $D< inj(N)/100$ then the fibers of $f$ are connected?

Remark. I think I can prove this in two cases: (1) for $\varepsilon =0$, (2) for closed $M$ and small positive $\varepsilon$.

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    $\begingroup$ Something wrong with the definition of submersion... $\endgroup$ Commented Feb 18 at 18:50
  • $\begingroup$ You are right. Corrected. $\endgroup$
    – asv
    Commented Feb 18 at 21:13

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In the definition of submersion you should assume that $d_xf\colon \textrm{T}_x\to \textrm{T}_{f(x)}$ is onto. In this case there is an almost horizontal lift $\phi_x\colon\textrm{T}_{f(x)}\to\textrm{T}_x$ such that $d_xf\circ\phi_x=\mathrm{id}$.

Moreover, by applying the partition of unity, you can assume that $\phi$ depends continuously on $x$. This makes it possible to lift homotopy from $N$ to $M$.

Since $\textrm{diam}\, F_x$ is small, we can connect any two of its points (say $x$ and $y$) by a short curve. Project this curve to $N$; it is a loop based at $f(x)$. Then shrink the loop to $f(x)$. By lifting this homotopy to $M$, we get a curve from $x$ to $y$ that lies in the fiber.

(This argument is nearly the same as here)

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  • $\begingroup$ Where did you use $\phi$? $\endgroup$
    – asv
    Commented Feb 19 at 19:57
  • $\begingroup$ @asv the homotopy in $N$ can be described by a time-dependent vector field on $[0,1]$. (More precicesly to the bundle on $[0,1]$ induced by the map $[0,1]\to N$). One can use $\phi$ to lift this vector field to $M$. (More precicesly to the bundle on $[0,1]$ induced by the map $[0,1]\to M$). $\endgroup$ Commented Feb 19 at 20:47
  • $\begingroup$ Do you need $\phi$ just to prove essentially that $f$ is a locally trivial fibration? Then one can use the covering homotopy lemma. $\endgroup$
    – asv
    Commented Feb 19 at 21:32
  • $\begingroup$ @asv it is closely related, but I did not do that. $\endgroup$ Commented Feb 19 at 22:44

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