# quasi-circle has trivial homology group

As I am reviewing my algebraic topology material, I got confused by something seems really basic.

For a quasi-circle $$Q$$ formed by $$y=\sin(1/x)$$ with $$x \in [0,1]$$, $$[-1,1]$$ on y-axis, and an arc $$c$$ connecting the $$(0,0)$$ and $$(1,0)$$ (as in Hatcher's book 1.3.7), I want to prove that it has trivial homology groups.

I have two ways to do it, which seems to give me different results.

Let $$A$$ contain the arc $$c$$, the verticle $$[-1,1]$$, and a bit of the topologist's sine curve on both ends. Let $$B$$ contain the topologist's sine curve (not including the verticle part). Clearly, both $$A$$ and $$B$$ are contractible, thus have trivial homogology groups.

In this case, we get $$\tilde{H}_n(A) \oplus \tilde{H}_n(B) = 0$$ for all $$n$$, which forces $$\tilde{H}_n(Q) \cong \tilde{H}_n(A \cap B)$$. In this case, $$\tilde{H}_n(A \cap B) = 0$$ for all $$n$$ except for $$\tilde{H}_0(A \cap B) \cong \mathbb{Z}$$, as it has two connected components.

However, I can also choose $$A$$ such that it only contains some topologist's sine curve on the 1 side but not on the 0 side. In which case I would think their interior also covers the quaso circle, and we would have all homology groups are trivial.

It seems I made a simple mistake somewhere, can anyone point it out? Is there an easier way to find the homology group than I did?

$$\textbf{Edit}$$ After seeing the comment, I realized that when $$A$$ contains both the vertical line segment and the topologist sine curve, it can not be contracted. So I thought let $$A$$ contain the verticle segment and the arc, and let $$B$$ contain the topologist sine curve and the arc. That would give result same as the second approach above.

• Why is $A$ contractible? Nov 13, 2021 at 5:09
• @EricWofsey I thought I could have the homeomorphism between $\sin(1/x)$ and the line segment and contract everything to the origin. However, I am aware of the question here about quaso-circle is not contractible. I suppose that might force $A$ to contain only the arc and the verticle line segment. Nov 13, 2021 at 5:45
• In Mayer-Vietoris, the interiors of $A$ and $B$ must cover the space. Not so for the sets you proposed in the edit. In particular, the points on the vertical segment aren't in the interior of $A$. Nov 13, 2021 at 6:44
• @EthanDlugie But if I have both topologist sine curve and the verticle segment in $A$, I get contractibility issue. What should I do then? Nov 13, 2021 at 6:57
• This space is also known as the Warsaw circle. Nov 13, 2021 at 9:29

Claim: for any singular simplex $$\sigma:\Delta^n \to Q$$, there exists some $$\delta>0$$ such that $$\sigma(\Delta^n) \cap ((0,\delta) \times \mathbb{R})$$ is empty. Given this fact, any given simplex lies in a contractible subspace of $$Q$$. Since homotopic maps are homologous, this gives that every singular chain (and so every singular cycle) is nullhomologous. In other words $$\tilde H_n(Q)=0$$ for all $$n$$.
Proof of claim: if the claim were not true, then there would be a sequence of points $$p_k \in \Delta^n$$ such that the $$x$$-coordinates of $$\sigma(p_k)$$ approach $$0$$ monotonically. Since the $$n$$-simplex is compact, this limit is achieved. I.e. we have $$p_k \to p_*$$ in $$\Delta^n$$ with $$\sigma(p_*) \in \{0\} \times [-1,1]$$. But $$\Delta^n$$ is path connected, so there is a path $$p_1 \rightsquigarrow p_2 \rightsquigarrow \dotsb \rightsquigarrow p_*$$ in $$\Delta^n$$ whose image in $$Q$$ connects a point on the sine curve with the vertical segment. (We can go far enough out in the path to make sure it's actually "jumping the gap" rather than going round the bottom arc of the quasi circle.) This is impossible.
• You really need to say more here to explain how you're getting your path $p_1 \rightsquigarrow p_2 \rightsquigarrow \dotsb \rightsquigarrow p_*$. In particular, you really need to use the fact that $\Delta^n$ is locally path-connected, not that is is path-connected. After all, $Q$ itself is path-connected and compact, but your argument could not work with $Q$ in place of $\Delta^n$ since then $\sigma$ could just be the identity map. Nov 14, 2021 at 1:45
• The easiest way is probably to observe that by continuity, there is a neighborhood $U$ of $p_*$ such that $\sigma(U)$ does not include the bottom arc. By local path-connectedness, we may assume this $U$ is path-connected. Then there is some $p_k\in U$, and then you can consider a path from $p_k$ to $p_*$ in $U$ to get a contradiction. Nov 14, 2021 at 1:51