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Suppose (Ln) is a sequence of loops in a torus S1 × S1 converging in the Hausdorff metric to some set L in the torus. Suppose also that for each loop Ln the projection map p:S1 × S1 -> S1 defined by p(x,y) =x when restricted to Ln is not null-homotopic, then can we conclude that the restriction of the projection map p to L is also not null homotopic? Or are there counter examples?

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2 Answers 2

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Your intuition is correct: the map p : L → S1 is not nullhomotopic.

If p : L → S1 was nullhomotopic, then it would factor through the universal cover ℝ of S1.
The inclusion ι : L → S1 × S1 would therefore factor through ℝ × S1.
Let ι' : L → ℝ × S1 denote a lift of ι, and let L(0) := ι'(L).
The subset L(0) ⊂ ℝ × S1 is then one of the connected components of the preimage of L in ℝ × S1.
Let us call the other components L(n) for n ∈ ℤ.

Let C0 ⊂ ℝ × S1 be a loop in that separates L(0) from L(1).
We may also assume that C0 is not nullhomotopic in ℝ × S1 [this still needs a small argument...].

The projection CS1 × S1 of C0 is then a loop in the complement of L,
and represents the element (0,1) of ℤ×ℤ = π1(S1 × S1).

Since L lies in the complement of C and Ln → L in the Hausdorff metric,
there exists an n such that Ln lies in the complement of C.
This contradicts your assumtion that p : LnS1 is nullhomotopic.

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  • $\begingroup$ Thanks for the answer. It appears that the notation L<sub>n</sub> is being used in two different contexts. could you please clarify. $\endgroup$
    – arun s
    Commented May 22, 2010 at 17:02
  • $\begingroup$ I fixed the mismatch of notations. $\endgroup$ Commented May 22, 2010 at 20:17
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EDITED. The following example should not be conisdered as a contre-example, it is just a well known example of a pathology that can happen.

EXAMPLE. $L$ is composed of a union of the vertical circle $(0, S^1)$ and a disjoint $R^1$ that is emdedded in $T^2$ in such a way, that it accumulates to $(0, S^1)$ from both sides. This $R^1$ projects one-to one to the horisontal circle without a point and can be represented a as graph (function of $x$), than if is given by the following formula:

$y= (sin(1/x))$

The point is that such $L$ can be Hausodrff approximated by a sequence of circles $s_n$ that are not null-homotopic ($s_n$ wiggles more an more near the vertical circle $(0, S^1)$ as $n\to \infty$).

The problem with this example, is that the topology on the union of $S^1$ and $R^1$ that we should take (I guess) is the toplogy induced from $T^2$. And for this induced topology, I guess the projection to the horisontal cicle is non null-homotopic...

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  • $\begingroup$ This is not a counterexample. $\endgroup$ Commented May 22, 2010 at 16:16
  • $\begingroup$ I dont understand why the inclusion composed with the proyection is nullhomotopic. $\endgroup$
    – rpotrie
    Commented May 22, 2010 at 16:24

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