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I'm doing exercise 16 on page 39 in Hatcher:

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

    (a) $X = \mathbb{R}^3$ with $A$ any subspace homeomorphic to $S^1$

    (b) $X = S^1 \times D^2$ with $A$ its boundary torus $S^1 \times S^1$

    (c) $X = S^1 \times D^2$ and $A$ the circle shown in the figure.

entangled circled from Hatcher, p.39

I've done (a) and (b) using proposition 1.17. i.e. I assumed there was a retraction, then the map between the fundamental groups has got to be injective, therefore contradiction.

Now I'm stuck with (c) because according to my understanding the circle on the picture has the same fundamental group as $S^1$ which means it also has the same fundamental group as the solid torus.

What is a different way of proving that there is no retraction (not using prop. 1.17.)? Many thanks for your help!

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    The circle on the picture is entangled (?) and it goes around the hole of the donut so that it cannot be shrunk to a point.2011-03-31
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    You need to study the induced map $i_*$ where $i : A \to S^1 \times D^2$ is inclusion (the target space here is your solid torus).2011-03-31
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    @Matt: that circle *can* be shrunk to a point. There's nothing in the definition of homotopy that stops the thing being deformed passing through itself.2011-03-31
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    @Chris: (your first comment) But the circle goes around the hole. How can you shrink it to a point? I mean, even if you take a normal, non-entangled circle....2011-03-31
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    @Matt: Here is a function from $S^1$ onto $S^1$ which is null homotopic. So it "goes around the hole" yet you can still null-homotope it: $f(e^{i\theta}) = e^{2i\theta}$ if $\theta \in [0,\pi]$ and $f(e^{i\theta})=e^{-2i\theta}$ if $\theta \in [\pi, 2\pi]$.2011-03-31
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    @Ryan: I think I'm confused. I thought that when it goes around a hole you'd have to tear it to shrink it to a point. If I take $S^1$ then $S^1$ is contractible if the $id_{S^1}$ function is null-homotopic. Is the function $h(x,t):= (1-t)x$ a valid homotopy between $id_{S^1}$ and $s_0 \in S^1$ ? Why do I think that a loop must not be torn?2011-03-31
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    @Matt: you appear to be confusing the notion of *homotopy* of a function to a space with an *isotopy* of that space. Notice the function that I gave is *not* the identity function.2011-03-31
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    To explain another way: The loop goes around the torus, but it goes around once one way, and then once the other way, for a total of 0 times around. It's essentially the curve you get when you concatenate the "once around" curve with its inverse to get a nullhomotopic loop.2011-03-31

2 Answers 2

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(i) If $f:X \rightarrow Y$ is a homotopy equivalence then the induced homomorphism $f_* : \pi_1(X, x_0) \rightarrow \pi_1(Y,f(x_0))$ is an isomorphism.

(ii) If $X$ deformation retracts onto $A \subset X$ then $r$, the retraction from $X$ to $A$, is a homotopy equivalence.

claim: There are no retractions $r:X \rightarrow A$

proof: (by contradiction)

Assume there was a retraction. Then by proposition 1.17. (Hatcher p. 36) the homomorphism induced by the inclusion $i_* : \pi_1(A, x_0) \rightarrow \pi_1(X,x_0)$ would be injective.

But $A$ deformation retracts to a point in $X$ so by (i) $i_*(\pi_1(A, x_0))$ is isomorphic to $\{ e \}$, the trivial group. Therefore $i_*$ cannot be injective. Contradiction. There are no retractions $r: X \rightarrow A$.

Can someone tell me if I got it right?

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Proposition 1.17 shows that there is no retraction $r : X \to A$. Unlike the previous examples, as abstract groups you can have an injective map $\pi_1(A) \to \pi_1(A)$, because they are both isomorphic to $\mathbb{Z}$. So you are not done yet.

If $r : X \to A$ is a retraction, then $r_* : \pi_1(A) \to \pi_1(X)$ sends the loop in $A$ on a path in $X$ that is null-homotopic, so $r_*$ is not injective. Inside $X$, you can attach a disk around $A$ (the disk crosses itself but it's not important). Taking the image of this disk by a retraction would give you a disk around $A$ inside $A$, which is not possible.

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    Minor remark: to stick with conventional notation, you probably want to rename your map to $i_* : \pi_1(A) \rightarrow \pi_1(X)$, such that $r_*$ points the other way, from $\pi_1(X) \rightarrow \pi_1(A).$2011-03-31