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Let $M$ and $N$ be smooth manifolds and let $S$ be a submanifold of $N$ ($\dim S < \dim N$). Let $\mathfrak S$ be a foliation of $S$. We say that a map between $M$ and $N$ is transverse to $\mathfrak S$ if it is transverse to every leaf of $\mathfrak S$.

Now, suppose $f : M \rightarrow N$ is a smooth map transverse to a foliation $\mathfrak S$ of $S$ at a point $x \in M$ ($f(x) \in S$ and $f$ is transverse to the leaf of $\mathfrak S$ passing through $f(x)$ at the point $x \in M$). Is it possible to find a map $g : M \rightarrow N$, arbitrarily close to $f$ in the strong (Whitney) topology, with $g(x) = f(x)$ and such that $g$ is transverse to $\mathfrak S$ at all points of $M$?

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    $\begingroup$ It is not possible in general. Let $f : M \to N$ be the inclusion $S^1 \subset \mathbb{R}^2$, let $S=\mathbb{R}^2$, let $\mathfrak S$ be the foliation by lines parallel to the horizontal axis, and let $x = (1,0)$. $\endgroup$
    – Lee Mosher
    Dec 13, 2012 at 18:09
  • $\begingroup$ Yes. I forgot to mention $\dim S < \dim N$. $\endgroup$
    – Saurabh T
    Dec 13, 2012 at 18:18

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I originally misunderstood your question. Here is a fairly simple example.

Let $N = \mathbb R^2$, $M=S^2$ and $S$ be the unit circle in $\mathbb R^2$ considered as foliated by its points -- the leaves are $0$-dimensional.

The map $M \to N$ will be a linear projection map. Let's project $S^2$ onto $\mathbb R^2$ in a way so that the image of $S^2$ is a disc in the plane, which half covers the unit circle.

So this map is transverse to the foliation at a point. But no nearby map is transverse to the foliation on all $M$. This is because the image of $M$ is a compact subset of the plane which does not contain the unit circle. So for a sufficiently small perturbation, you similarly do not cover the unit circle. Wherever the boundary of the image intersects the circle, you will not be transverse to the foliation of the circle.

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  • $\begingroup$ @ryan I thought I made it clear, by saying that $f(x) \in S$ and $f$ is transverse to the leaf, that the dimension of $M$ is greater than the codimension of the leaf in $N$. Apparently I was not precise. Anyway, I don't understand your example. Is $M = S^2$, $N = \mathbb R^4$ and $S = S^3$? If yes, how is the map transverse to the leaf of Hopf fibration, this can not happen? $\endgroup$
    – Saurabh T
    Dec 13, 2012 at 21:47
  • $\begingroup$ Okay, I misunderstood your notation. I'll adjust my answer. $\endgroup$ Dec 13, 2012 at 22:55
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    $\begingroup$ @ryan Yes, thank you for your example. Under what conditions do you think it can be done? $\endgroup$
    – Saurabh T
    Dec 14, 2012 at 11:31

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