Proof that $(0,1)\times S^{1}$ is not homeomorphic to $\mathbb{R}^{2}$.

Question

I want to prove that:

$$\text{The open cylinder }(0,1)\times S^{1}\text{ is not homeomorphic to } \mathbb{R}^{2}.$$

I proved that $(0,1)\times S^{1}$ is homeomorphic to $\mathbb{R}^{2}\setminus\{0\}$ and I know that $\mathbb{R}^{2}\setminus\{0\}$ is not simply connected. Thus, I can get a proof. The problem is: I know how to prove that $\mathbb{R}^{2}\setminus\{0\}$ is not simply connected using Homotopy Theory, but I cannot use it now. Can someone help me to find another proof?

Answer 1

In $\mathbb{R}^2$, the following property holds: if $K\subset\mathbb{R}^2$ is compact, then there is a compact $K^\star\subset\mathbb{R}^2$ such that $K\subset K^\star$ and furthermore $\mathbb{R}^2\setminus K^\star$ is connected (simply take a closed disk containing $K$). Therefore, if $M$ is a topological space homeomorphic to $\mathbb{R}^2$, then $M$ has this property too: if $\psi\colon\mathbb{R}^2\longrightarrow M$ is a homeomorphism and $K\subset M$ is compact, then $\psi^{-1}(K)$ is compact too and therefore, if $K^\star\subset\mathbb{R}^2$ is a compact containing $\psi^{-1}(K)$ and such that $\mathbb{R}^2\setminus K^\star$ is connected, then $\psi(K^\star)$ is compact, $K\subset\psi(K^\star)$ and $M\setminus K^\star=\psi(\mathbb{R}^2\setminus K^\star)$, which is connected.

However, this property doesn't hold for $(0,1)\times S^1$: $\left\{\frac12\right\}\times S^1$ is compact, but no compact $K^\star\subset(0,1)\times S^1$ containing $\left\{\frac12\right\}\times S^1$ is such that $\bigl((0,1)\times S^1\bigr)\setminus K^\star$ is connected.

Answer 2

If $ (0, 1) \times S^1 $ were homeomorphic to $\mathbb{R}^2$, then $\pi_1((0, 1) \times S^1) \cong \pi_1((0, 1)) \times \pi_1(S^1) \cong 1 \times \pi_1(S^1) \cong \mathbb{Z}$ would be isomorphic to $\pi_1(\mathbb{R}^2) \cong 1$, but $\mathbb{Z} \not \cong 1$.

Answer 3

This is probably overkill, but assume by contradiction that $\mathbb R^2 \backslash \{ (0,0) \}$ is simply connected.

Consider the vector field $$F=(P,Q)=\Bigl(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\Bigr)$$

Than, since $$\frac{\partial P}{\partial y}=\frac{-1(x^2+y^2)+2y^2}{(x^2+y^2)^2}\\ \frac{\partial Q}{\partial x}=\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}\\ \frac{\partial P}{\partial y}=\frac{\partial{Q}}{\partial x}$$ and the domain is $\mathbb R^2 \backslash \{ (0,0) \}$ is simply connected, it follows that $F$ is conservative in this domain, and hence $$\int_{x^2+y^2=1} -\frac{y}{x^2+y^2}dx +\frac{x}{x^2+y^2}dy =0$$

But, with the standard parametrisation $x=\cos(t), y=\sin(t)$ you have $$\int_{x^2+y^2=1} -\frac{y}{x^2+y^2}dx +\frac{x}{x^2+y^2}dy =\int_0^{2 \pi}\frac{\sin^2(t)+\cos^2(t)}{\sin^2(t)+\cos^2(t)}dt =2 \pi$$