Showing something isn't a manifold So I'm following some notes that are introducing manifolds with pretty minimal prerequisites.  What I want to do is show where the image of $\phi: \mathbb{R}\rightarrow \mathbb{R^2}$ $t\mapsto (t-\sin(t),1-\cos(t))$ isn't a manifold.  Since this is the standard cycloid, it's pretty clear that things go bad at the cusps, but how do I rigorously show that $\phi(\mathbb{R})$ isn't a manifold there?  The derivative of $\phi$ isn't 1:1, but that's not enough is it?  How do I know there's not some better parametrization of $\phi(\mathbb{R})$ around the cusps?
 A: tl;dr version: skip the first two paragraphs.
Let's focus on the question first. We are told to prove that the set $X=\phi(\mathbb R)$ "is not a smooth manifold". What does this mean? The set already has a topological manifold structure. Are we to prove that it does not admit a differentiable structure? But that would be false. In fact one can put a differentiable structure on $X$ simply by declaring that $\phi$ is a diffeomorphism (i.e., smooth charts on $X$ would be defined as compositions of $\phi$ with smooth charts on $\mathbb R$). The set $X$ can be a smooth manifold. [Added] I'll elaborate: suppose $X$ be a topological space and $\phi:\mathbb R\to X$ is a homeomorphism. We can introduce differentiable structure on $X$ be declaring $\phi$ to be a smooth chart. The result is a smooth manifold with an atlas of one chart. (If your definition of manifold involves a maximal atlas, Zorn's lemma provides a maximal atlas containing $\phi$.)    
But $X$ cannot be an submanifold of $\mathbb R^2$. Being a submanifold  requires that the inclusion map $\imath :X\to\mathbb R^2$ be smooth with derivative of maximal rank (in this case, 1). Let's assume this and get a contradiction. Let $u : X\to \mathbb R$ be a local coordinate at a cusp such that $u(0)=0$. Then $\imath\circ u^{-1}$ is a smooth map from $\mathbb R$ to $\mathbb R^2$ with nonzero derivative at $0$. The differentible structure on $X$ is now out of the way. (It did not actually have to get in the way, if one uses a more direct definition of "submanifold of $\mathbb R^n$".)
It remains to prove that there is no smooth map $x=X(s),y=Y(s)$ that takes values in $X$, sends $0$ to cusp $(0,0)$, and has nonzero (rank 1) derivative at $0$. Since $Y$ has a minimum at $0$, it follows that $Y'(0)=0$. The derivative requirement implies $X'(0)\ne 0$. Invoking the limit definition of the derivative, we see that $\lim_{s\to 0}\frac{Y(s)}{X(s)}=0$. Our original parametrization $t\mapsto (x,y)$ also sends $0$ to $(0,0)$. Hence, when $t$ is small, the point $(x(t),y(t))$ is close to $(0,0)$. This implies $(x(t),y(t))=(X(s),Y(s))$ where $s=s(t)\to 0$ as $t\to 0$. Therefore, $\lim_{t\to 0}\frac{y(t)}{x(t)}=0$. But calculation shows that $\lim_{t\to 0}\frac{1-\cos t}{t-\sin t}$ does not exist. 
