Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – Rudy the Reindeer Mar 31 '11 at 18:07
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). – Ryan Budney Mar 31 '11 at 18:19
@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. – Chris Eagle Mar 31 '11 at 18:28
@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 $\theta \in [\pi, 2\pi]$. – Ryan Budney Mar 31 '11 at 18:53
@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. – Ryan Budney Mar 31 '11 at 22:00

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.

share|cite|improve this answer
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 $\pi_1(X) \rightarrow \pi_1(A).$ – Gerben Mar 31 '11 at 21:00
up vote 2 down vote accepted

(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?

share|cite|improve this answer
Can someone tell me if I got it right? Ta! – Rudy the Reindeer May 4 '11 at 9:35
Nope. You did not get it right. Try again! – Rudy the Reindeer Dec 29 '12 at 19:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.