In our class today we said today that an 1-times continuously differentiable immersion $\phi:[0,1] \rightarrow \mathbb{R}^n$ is a submanifold of dimension 1 if it is closed such that $\phi(0)=\phi(1)$ (but this should be the only point where the curve is not injective) and $\phi'(0)=\phi'(1)$. I still do not get it why we want that the curve to be closed. what is wrong with a curve that is not closed?
Is this only because it may not be differentiable at the end points or is there more in it?
On
My understanding of a k-submanifold $C$ of $\mathbb R^n $ is that there must be an open set $U$ in $\mathbb R^n $ , and some chart map $\gamma$ , so that $\gamma (C\cap U)=(x_1,x_2,..,x_k, 0,0,..,0)$, i.e., C is embedded in $\mathbb R^n$ in the same way as the "standard embedding" of $\mathbb R^k$ in $\mathbb R^n$. In the case of a curve $C$, you need an open set $V$ , and a chart map $\gamma$ with $\gamma (V \cap C)=(x,0,0,..,0)$. But: every intersection of an open set $V \subset \mathbb R^n$ with the non-closed curve $C$ will be of the form $(a,b]$, i.e., a half-open interval (when seen in $\mathbb R$. An a half-open interval subspace cannot be homeomorphic to a copy of $\mathbb R$ in $\mathbb R^n$, since $\mathbb R$ itself is homeomorphic to $(a,b); a\neq b$, but is not homeomorphic to $(a,b]$
If the curve is not closed, i.e. if $\phi(0)\neq\phi(1)$, then there is no neighborhood $U_0$ ($U_1$) of $\phi(0)$ (or $\phi(1)$) in $\mathbb{R}^n$ and a neighborhood chart $\psi:U_0\to\mathbb{R}^n$ that carries the image of $\phi$ to the $x_1$-axis. Thus $\phi[0,1]$ cannot be a submanifold of $\mathbb{R}^n$.