Your intuition is correct: the map p : L → <i>S</i><sup>1</sup> is not nullhomotopic.<br> If p : L → <i>S</i><sup>1</sup> was nullhomotopic, then it would factor through the universal cover ℝ of <i>S</i><sup>1</sup>.<br> The inclusion ι : L → <i>S</i><sup>1</sup> × <i>S</i><sup>1</sup> would therefore factor through ℝ × <i>S</i><sup>1</sup>.<br> Let ι' : L → ℝ × <i>S</i><sup>1</sup> denote a lift of ι, and let L<sub>0</sub> := ι'(L).<br> The subset L<sub>0</sub> ⊂ ℝ × <i>S</i><sup>1</sup> is then one of the connected components of the preimage of L in ℝ × <i>S</i><sup>1</sup>.<br> Let us call the other components L<sub><i>n</i></sub> for <i>n</i> ∈ ℤ.<br> Let <i>C</i><sub>0</sub> ⊂ ℝ × <i>S</i><sup>1</sup> be a loop in that separates L<sub>0</sub> from L<sub>1</sub>.<br> We may also assume that <i>C</i><sub>0</sub> is not nullhomotopic in ℝ × <i>S</i><sup>1</sup> [this still needs a small argument...]. The projection <i>C</i> ⊂ <i>S</i><sup>1</sup> × <i>S</i><sup>1</sup> of <i>C</i><sub>0</sub> is then a loop in the complement of L,<br> and represents the element (0,1) of ℤ×ℤ = π<sub>1</sub>(<i>S</i><sup>1</sup> × <i>S</i><sup>1</sup>). Since L lies in the complement of <i>C</i> and L<sub><i>n</i></sub> → L,<br> there exists an <i>n</i> such that L<sub><i>n</i></sub> lies in the complement of <i>C</i>.<br> This contradicts your assumtion that p : L<sub><i>n</i></sub> → <i>S</i><sup>1</sup> is nullhomotopic.