$S^n$ and $S^1$ homeomorphic iff $n=1$

34 Views Asked by At

We shall proove that $S^n$ and $S^1$ homeomorphic iff $n=1$. Here is my proof:

Let $S^n$ and $S^1$ homeomorphic via $\varphi: S^n \rightarrow S^1$. Then $S^n-\{0,1\}$ and $S^1-\{\varphi(0), \varphi(1)\}$ are homeomorphic via $\varphi|_{S^n-\{0,1\}}$. Therefore, they are also homotopy equivalent. So we know from the lecture, that $$|\Pi_0(S^n-\{0,1\})| = |\Pi_0(S^1-\{\varphi(0), \varphi(1)\})|$$, where $\Pi_0(...)$ is the set of path components. But we have $$|\Pi_0(S^1-\{\varphi(0), \varphi(1)\})|=2$$ and $$|\Pi_0(S^n-\{0,1\})| = 1$$ for $n \geq 2$. Thus $n=1$.

Is this correct? What makes me doubt is that this is pretty much exactly the same proof as our proof from the lecture that $\mathbb{R}^n$ and $\mathbb{R}$ are homeomorphic iff $n=1$. I only had to make a slight adjustment, so it seems way too easy.