How to show Warsaw circle is non-contractible?

Question

The Warsaw circle is defined as a subset of $\mathbb{R}^2$： $$\left\{\left(x,\sin\frac{1}{x}\right): x\in\left(0,\frac{1}{2\pi}\right]\right\}\cup\left\{(0,y):-1\leq y\leq1\right\}\cup C\;,$$ where $C$ is the image of a curve connecting the other two pieces.

A map from Warsaw circle to a single point space seems to be a well-known example showing weak homotopy equivalence is indeed weaker than homotopy equivalence. I am trying to see why the Warsaw circle is non-contractible. It seems intuitively reasonable since two 'ends' of it are connected in some sense, but I failed to give a proof. Any hint would be appreciated. Thank you very much.

Answer 1

Let $W$ denote the Warsaw circle. By collapsing the interval piece down to a point, the quotient space is homeomorophic to $S^1$. This gives a map $f:W\rightarrow W/{\sim} \cong S^1$.

I claim this map is not null homotopic. Believing this for a second, note that for any contractible space $X$ any continuous $f:X\rightarrow Y$ is null homotopic, so this will show that $W$ is not contractible.

So, assume $f$ is null homotopic. Then we can lift it to get a map $\hat{f}:W\rightarrow \mathbb{R}$. Now, $f$ is $1-1$, except on the interval piece. This implies the lift is $1-1$, except, perhaps, on the interval piece. But since the interval piece is connected and it must map into the fiber $\mathbb{Z}$ of the map $\mathbb{R}\rightarrow S^1$, this implies that $\hat{f}$ is 1-1, except that it collapses the interval to a point.

Said another way, $\hat{f}$ descends to an injective map on $W/{\sim}$. Since $W/{\sim}$ is homeomorphic to $S^1$, $\hat{f}$ gives an injective map from $S^1$ to $\mathbb{R}$. But using the intermediate value theorem twice, it's easy to see that there is no injective continuous map from $S^1$ to $\mathbb{R}$.

Answer 2

Its Cech cohomology is equal to $\mathbb Z$ in degree $1$. You can either check this directly, or you could apply Alexander Duality (which relates the reduced Cech cohomology with the reduced homology of the complement.) In this case, since the complement has two path components, the Cech cohomology of the Warsaw circle must be $\mathbb Z$ in degree 1.

Answer 3

Let $W$ denote the Warsaw circle and $C$ is the space as in the question. By collapsing $C$ down to a point we find the map $f:W\rightarrow W/C \cong S^1$.

$\underline{Claim}$: f is not null homotopic.

If we assume that the $\underline{Claim}$ holds we find the thesis. Indeed, any continuous $f:X\rightarrow Y$ is null homotopic if its domain is contractible; so this will show that $W$ is not contractible.

So, we have done if we prove the $\underline{Claim}$.

$\textit{Proof}(\underline{Claim})$ Assume by absurd that $f$ is null homotopic. Let $e:\mathbb{R}\to S^1:r\mapsto (cos2\pi r,sin2\pi r)$ be the usual helic covering space of $S^1$ and let $H:W\times I\to S^1$ (with I the interval $[0,1]$) be an homotopy between $f$ and a constant map $c:W\to S^1:w\mapsto (0,1)$. Since $e$ is surjective there exists a lifting $c_0 :W\to \mathbb{R}$ of $c$ along $e$. Now since every covering space is a Hurewicz fibration (see Proposition 1.30 Hatcher Algebraic Topology https://pi.math.cornell.edu/~hatcher/AT/AT.pdf) there is a lifting of $H_0:W\times I\to \mathbb{R}$ of $H$ along $e$ such that $e\circ H_0=H$ and $c_0=i\circ H_0$ where $i:W\cong W\times\{0\}\to W\times I$ is the obviously inclusion. In particular $f_0(-):=H_0(-,1):W\to \mathbb{R} $ is a lifting of $f$ along $e$.

Recall that $C$ is the space as in the question.Let $C$ be the space as in the question. The map $f_{|C}~$ is injective and the commutative equation $ f_{|C} =e \circ f_{0|C}~$ implies that $f_{0|C}~$ is injective too. Moreover $f_{0|W/C}$$ ~:W/C\to \mathbb{R}~$ has image in a connected component of $e^{-1}\{(0,1)\}~$, then it is a constant map. Hence by the universal property of quotient $f_0~$ defines a unique, continuous and injective map $\hat{f}_0:S^1\cong W/C\to \mathbb{R}$. Here we get the absurd since there isn't such map. Indeed if by absurd such map $\hat{f}_0~$ exists then $Im(\hat{f}_0)~$ is a compact, path connected subspace of $\mathbb{R}$ because $S^1~$ is a path connected, compact space (with the topology of subspace of $\mathbb{R}^2$). But compact, connected subspace of $\mathbb{R}$ is simply a closed, path connected interval of $\mathbb{R}~$ $[a,b]$. Finally, we find this second absurd since this implies that $S^1~$ is homeomorphism to a closed, connected interval of $\mathbb{R}$ (i.e. $S^1\cong Im{\hat{f}_0}\cong [a,b]$): this is impossibile since $\pi_1(S^1)\cong\mathbb{Z}\neq 0\cong\pi_{1}([a,b])$.