Every diagrammatic map has only finitely many double points

Question

Definition: A smooth map $\gamma\colon \Bbb S^1\to \Bbb R^2$ is called diagrammatic if the following conditions are satisfied:

$(1)$ the map is an immersion, i.e., derivative at each point is non-zero,

$(2)$ if $z\neq w\in \Bbb S^1$ satisfy $\gamma(z)=\gamma(w)$, then $\gamma'(z)$ and $\gamma'(w)$ are linearly independent,

$(3)$ given any $p\in \gamma(\Bbb S^1)$, the preimage $\gamma^{-1}(p)$ consists of either one or two points.

Furthermore, any $p\in \gamma(\Bbb S^1)$ for which there exist $z\neq w\in \Bbb S^1$ with $\gamma(z)=p=\gamma(w)$ is called a double point of $\gamma$.

I am solving the following Problem:

Problem: Every diagrammatic map has only finitely many double points.

My Idea: Suppose not, then there is a sequence $\{p_n\}$ of distinct double points of $\gamma$. Passing to a subsequence, if needed, and using compactness of $\gamma(\Bbb S^1)$, we may assume that $p_n\to \ell\in \gamma(\Bbb S^1)$. Write $\{z_n,w_n\}=\gamma^{-1}(p_n)$. Passing to subsequence and compactness of $\Bbb S^1$, let $z_n\to z$. Thus $\gamma(z)=\ell$. Now, $\gamma'(z_n)\to \gamma'(z)$ and $\gamma'(w_n)\to \gamma'(z)$ as $\gamma$ is $C^\infty$-smooth. Since $\gamma'(z_n)$ and $\gamma'(w_n)$ are linearly independent, $\gamma'(z)=0$, a contradiction to the assumption that $\gamma$ is an immersion.

Is my idea correct??

Answer 1

I do not get how you conclude that $\gamma'(z)=0$ from $\gamma'(w_n)$ and $\gamma'(z_n)$ being linearly independent. Otherwise, your proof seems fine.

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Here is the standard way to proceed.

Let $\Delta$ be the diagonal of $\mathbf{S}^1\times\mathbf{S}^1$, then $\Delta$ is a codimension one submanifold, but from assumption (2), $\gamma$ is a self-transverse map. Therefore, $(\gamma,\gamma)$ is transverse to $\Delta$ and $(\gamma,\gamma)^{-1}(\Delta)$ is a codimension one submanifold of $\mathbf{S}^1$. In conclusion, $(\gamma,\gamma)^{-1}(\Delta)$, which is nothing less than the set of double points of $\gamma$, is discrete in $\mathbf{S}^1$, thus finite by compactness of $\mathbf{S}^1$.

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Reminders.

Definition. Two smooth maps $f\colon X\to Z$ and $g\colon Y\to Z$ are \emph{transverse} whenever for all $x\in X$ and all $y\in Y$ such that $f(x)=z=g(y)$, then $T_xf(T_xX)+T_yg(T_xY)=T_xZ$.

If $Y$ is a submanifold of $Z$, then $f$ is transverse to $Y$ when it is transverse to the inclusion of $Y$ in $Z$.

The following proposition is a standard result.

Proposition. Let $\Delta$ be the diagonal of $Z$, then $f$ and $g$ are transverse if, and only if, $(f,g)$ is transverse to $\Delta$.

Proof. You can find a proof on MSE here. $\Box$

In particular, transversality is relevant to differential topology because of the following theorem (which is a consequence of the inverse function theorem).

Theorem If $f$ is transverse to $Y$, then $f^{-1}(Y)$ is submanifold of $X$ of codimension $\operatorname{codim}(Y)$.