# Every diagrammatic map has only finitely many double points

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??

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.

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 $$\mathbf{S}^1$$.

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 $$\operatorname{codim}(Y)$$.

• I think you don't need the first condition, i.e., $\gamma$ is an immersion; instead $(2)$ along is enough to conclude that $\gamma$ is a self-transverse map. Sep 26 at 16:18
• You are right, I'll remove it from my answer! Sep 26 at 16:23