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Here is a problem and my attempt at the solution. If my conclusion or the proof is incorrect I would appreciate a pointer in the right direction. Thanks in advance.

Let $S^1$ be the unit circle in the xy plane in $\mathbb{R^3}$ and let $E^1_+$ and $E^1_-$ be two of its semicircles. Find the homology groups (with integer coefficients) of a) $\mathbb{R^3}\setminus E^1_+$ and b) $\mathbb{R^3}\setminus \ S^1$.

Attempt of solution:

a) $H_n(\mathbb{R^3},S^1)\cong H_n(\mathbb{R^3}\setminus E^1_+, E^1_-)$, by excision. Then $H_n(\mathbb{R^3}\setminus E^1_+, E^1_-) \cong H_n(\mathbb{R^3}\setminus E^1_+, {pt})$, since $E^1_-$ is contractible. Then we know $H_n(\mathbb{R^3}\setminus E^1_+, {pt}) \cong \tilde{H_n}(\mathbb{R^3}\setminus E^1_+)\cong H_n(\mathbb{R^3}\setminus E^1_+)$ for $n>0$. So we get $H_n(\mathbb{R^3}\setminus E^1_+) \cong H_n(\mathbb{R^3},S^1)$. From the long exact sequence of relative homology we know that $H_n(\mathbb{R^3},S^1) \cong H_{n-1}(S^1)$, so $H_n(\mathbb{R^3}\setminus E^1_+)\cong \mathbb{Z}$ for $n=1,2$. For $n=0$ we also get $\mathbb{Z}$ since $\mathbb{R^3}\setminus E^1_+$ is arcwise connected.

b) the same reasoning but starting with $H_n(\mathbb{R}^3,E^2_+)$, where $E^2_+$ is the upper hemisphere of $S^2$, gives $H_n(\mathbb{R}^3 \setminus S^1) \cong H_{n-1}(E^2_+)$, but $E^2_+$ is contractible so $H_n(\mathbb{R}^3 \setminus S^1)\cong \mathbb{Z}$ for $n=1$ and for $n=0$ it is also $\mathbb{Z}$ since $\mathbb{R}^3 \setminus S^1$ is arcwise connected. For $n>1$, $H_n(\mathbb{R}^3 \setminus S^1)=0$.

EDIT @msteve pointed out that b) is wrong. Excision can't be used like in a) since $S^1$ isn't in the interior of $E^2_+$. So a different approach is needed here.

user198182
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  • I think $H_2(\mathbb{R}^3 - S^1) = \mathbb{Z}$? You have at least one free generator by taking a sphere enclosing the removed unit circle, so the second homology is nonzero. – msteve Mar 02 '15 at 02:45
  • @msteve you're right. Excision can't be used in b) like in a) since $S^1$ isn't in the interior of $E^2_+$. So a different approach is needed here. Thanks for catching that. – user198182 Mar 02 '15 at 03:56
  • I don't see how you used excision in A) – Mariano Suárez-Álvarez Mar 02 '15 at 04:11
  • (the circle has empty interior as a subset of R^3) – Mariano Suárez-Álvarez Mar 02 '15 at 04:14
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    $\Bbb R^3 - S^1$ deformation retracts onto $S^2 \vee S^1$. Use the homotopy equivalence of homology to conclude the proof. You don't need to use this much machinery at all. – Balarka Sen Mar 02 '15 at 07:15
  • @BalarkaSen how do you prove that it does retract to that? I don't see that. – user198182 Mar 02 '15 at 07:55
  • @MarianoSuárez-Alvarez I see. I was wrong. Could you suggest another approach? – user198182 Mar 02 '15 at 07:58
  • @user198182 Have a look at page 46 of Hatcher. – Balarka Sen Mar 02 '15 at 08:05
  • @BalarkaSen, that is, provided you can actually prove that retraction does exist! :-) (And a picture does not count..) – Mariano Suárez-Álvarez Mar 02 '15 at 08:11
  • @MarianoSuárez-Alvarez Well, if you want to explicitly write up the retract, be my guest :P. Some deformation retracts are extremely hard to algebraically derive. For example, can you show that $\Bbb R^3 - {0, 1}$ deformation retracts onto the wedge of two circles? It's impossible to write down a formula, but it's easy to "see" it :) – Balarka Sen Mar 02 '15 at 08:15
  • Well, it was you who suggested to use the existence of that retract. Notice that the first homology is also $\mathbb Z$ if the $S^1$ is knotted in $\mathbb R^3$. Can you also "see" a retraction in that case? – Mariano Suárez-Álvarez Mar 02 '15 at 08:21
  • I assumed $S^1$ is not knotted inside $\Bbb R^3$? At least, the question makes it clear. It's the unit circle on the xy-plane. For general knots, there is no such retract (or at least, it's not apparent that there is). – Balarka Sen Mar 02 '15 at 08:23
  • @BalarkaSen I see the deformation retraction now. – user198182 Mar 02 '15 at 10:35

1 Answers1

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Ok, here we go:

a) Set $U_1 := \mathbb{R}^3 \setminus S^1$. Let $1/5 > \varepsilon > 0$ and consider $S^1 \times D_{\varepsilon}$ a solid torus. (The product of $S^1$ with the disc of radius $\varepsilon$) If $S^1$ were knotted, we could still find a small enough $\varepsilon$ and our proof would be exactly the same. In particular, the homology of $\Bbb R^3 \setminus K$ is the same for every knot.

Set $U_2 := int(S^1 \times D_\varepsilon)$. We have the following facts:

  • $\mathbb{R}^3$ is covered by the open sets $U_1$ and $U_2$.

  • $U_2 \simeq S^1$, where $\simeq$ means homotopy equivalence.

  • $U_1 \cap U_2 \simeq T^2$, where $T^2$ is the torus.

So applying the Mayer-Vietoris sequence, we get

$$\dots \to H_{n+1}(\Bbb R^3) \to H_n(U_1 \cap U_2) \to H_n(U_1) \oplus H_n(U_2) \to {H}_{n}(\Bbb R^3 ) \to H_{n-1}(U_1 \cap U_2) \to \dots$$ and using our facts, we get $$\dots \to H_{n+1}(\Bbb R^3) \to H_n(T^2) \to H_n(U_1) \oplus H_n(S^1) \to {H}_{n}(\Bbb R^3 ) \to H_{n-1}(T^2) \to \dots$$

Now $\Bbb R^3$ is contractible, so for $n>0$ we have the isomorphisms $H_n (T^2) \simeq H_n(U_1) \oplus H_n(S^1)$. Therefore $H_n(U_1) =0$ for $n>2$, $H_2(U_1) \simeq H_1(U_1) \simeq \Bbb Z$ and since $U_1$ is path-connected, $H_0(U_1) \simeq \Bbb Z$. This completes the proof of a).

b) Clearly $\Bbb R^3 \setminus E^1_+ \simeq \Bbb R^3 \setminus {x}$, where $x \in E^1_+$ is just a point. Now $\Bbb R^3$ retracts onto $S^2$ and we know its homology, so we are done.

Pedro A. Castillejo
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