# 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?

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It depends on what kind of "manifold" you're talking about. If you mean topological manifold (i.e. a space locally homeomorphic to $\mathbb R$), then the cycloid is in fact a manifold. One way to see this is to note that $f(t) = t-\sin t$ is a continuous, strictly increasing function from $\mathbb R$ to $\mathbb R$, and hence injective; it's easily seen to be surjective as well, so it's a homeomorphism. The map $\psi: \text{Im}(\phi) \to \mathbb R$ given by $\psi(x,y) = f^{-1}(x)$ is a continuous inverse for $\phi$ and thus a homeomorphism. – Jack Lee Nov 7 '12 at 22:17
@JackLee: Yeah, tis true! I'm given a pretty naive definition, essentially that it can be covered by smooth chart. I should have made the distinction that I wanted it locally diffeomorphic to $\mathbb{R}$. – AsinglePANCAKE Nov 8 '12 at 5:56
OK. You could try assuming there is a regular parametrization (i.e., a smooth parametrization whose velocity never vanishes) of the same set, $\psi:\mathbb R\to \mathbb R^2$, written $\psi(t)=(x(t),y(t))$. Then see what you can say about $x'(0)$ and $y'(0)$ to derive a contradiction. – Jack Lee Nov 8 '12 at 15:09
@JackLee I tried, and I'm failing. Anymore hints? Also, how do I check that this inverse is continuous? Finding it explicitly seems...messy. – AsinglePANCAKE Nov 8 '12 at 22:27
What I had in mind about $x'(0)$ and $y'(0)$ was something like this: first, since $y(t)$ takes a minimum at $t=0$, we have $y'(0)=0$. Then use the fact that $\psi'(t)$ has to be parallel to $\phi'(t)$ when $t\ne 0$ to conclude that $y'(t)/x'(t)\to\infty$ as $t\to 0$. This is only possible if $x'(t)\to 0$, which contradicts the assumption that $x'$ is continuous and $\psi'(0)\ne 0$. – Jack Lee Nov 8 '12 at 23:37

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.
@AsinglePANCAKE 'limit invariance' was really a fancy way of saying that taking a limit can be exchanged with composition with a continuous map. Which is more or less the definition of continuity. $\lim_{t\to 0} f(t)=A$ implies $\lim_{t\to 0} f(g(t))=A$ whenever $g$ is continuous with $g(0)=0$. – user53153 Dec 21 '12 at 3:00