In Hatcher's book, exercise: 1.1.16-c:

  1. Show that there are no retractions $r :X \rightarrow A$ in the following cases:

(c) $X = S_1 × D_2$ and A the circle shown in the figure.

Page 39

By the inclusion, the induced map from the fundamental group of A to the fundamental group of $X$ , and the generator of the former is sent to just the homotopy class of the knotted circle inside the solid torus. But is that true that"the knot can be homotoped away by pulling the ends through each other,"? Why? Anyway who can give me a careful proof of this one? Thank you very much!


I deleted my first answer, which was missing the point of this question (sorry about that).

There is no requirement in the definition of homotopy that the path cannot "pass through itself" when you contract it to a point. One way of seeing how this path is trivial in the fundamental group is as follows: Pick two points $x_0$ and $x_1$ on $A$ where the path loops around itself (one on either side of the loop). Parameterize the path that goes around $A$ once in two pieces: $\ell_1$ that starts at $x_0$ and goes to $x_1$, and $\ell_2$ that starts at $x_1$ and goes back to $x_0$ along the "other half" of $A$. It is then easy to see that you can perterb $\ell_1$ slightly so that it becomes $\ell_2$ parameterized backwards. Then you can see the rest of the homotopy by bringing $x_1$ back to $x_0$ along $\ell_2$.


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