Exercise about Lagrangian submanifolds I am trying to solve the following exercise:
Let $(M,\omega)$ be a symplectic manifold and $L$ a compact Lagrangian submanifold such that $H^{1}(L)=0$. Let $\{L_{t}\}_{t\in(-1,1)}$ be a smooth family of Lagrangian submanifolds of $(M,\omega)$ with $L_{0}=L$. Does there exist $\epsilon>0$ such that for all $t\in(-\epsilon,\epsilon)$: $L\cap L_{t}\neq\emptyset$?  
So far, all counterexamples I could find have nontrivial $H^{1}$, like for instance a family of circles in $T^{*}S^{1}\cong S^{1}\times\mathbb{R}$. This makes me believe that the statement is true. Any suggestions how to prove it?
 A: By the symplectic tubular neighborhood theorem, there is a neighborhood $U$ of $L_0$, symplectomorphic to $T^*L_0$ by a symplectomorphism that takes $L_0$ to the zero section of $T^*L_0$. For sufficiently small $t$, $L_t \subset U$ (this is nothing more than the fact that $L_t$ varies continuously: the map $f_t: L \times I \to M$ has $f_t^{-1}(U)$ open, and hence by the so-called tube lemma, and the compactness of $L$, there is an open set of $t$ including zero such that $L_t \subset U$.
A similar argument will prove that, for small $t$, $L_t$ is a section of the cotangent bundle: for consider the projection $\pi \circ f_t : L \times I \to L_0$. For $t=0$, this is a diffeomorphism (the identity). By continuity of the derivative, $f_t$ is also a submersion for small $t$ (and hence a diffeomorphism, since it's homotopic to the identity). That this projection is a diffeomorphism means we can identify $L_t$ as the image of a section of the cotangent bundle; call this section $\alpha_t$. 
Now Lagrangian sections of the cotangent bundle can be identified with closed forms. Now we invoke $H^1(L) = 0$; $\alpha_t = dg_t$ for some $g_t$. (One could take that to be continuous in $t$, but that doesn't actually matter unless you're interested in how the intersection points move around.) Now $g_t$ has a maximum somewhere ($x$, say) on $L_0$, and hence at that point $0 = dg_t(x) = \alpha_t(x)$. Hence $L_t \cap L_0$ is nonempty: it contains $x$.
