# Topological transversality

Warmup question:
Let us say that two continuous functions $f,g:[0,1]\to \mathbb R$ are topologically transverse if their difference $f-g$ has only finitely many zeros, and each zero separates an interval where $f>g$ from an interval where $f<g$ (and let's say that we also impose $f(0)\not = g(0)$ and $f(1)\not = g(1)$).

Let $f_1,\ldots,f_n\in C^0([0,1])$ be continuous functions. Is it true that set $$\big\{\,g\in C^0([0,1])\,\,\big|\,\,\,\forall i,\,\,\, g\,\, \text{ is topologically transverse to } f_i\big\}$$ is dense in $C^0([0,1])$?

What I really need:
Let $\gamma_1,\ldots,\gamma_n$ be a finite collection of curves (Jordan Arcs) in $\mathbb R^2$. No transversality assumed amongst the $\gamma_i$. Is the set of curves that are topologically transverse to all the $\gamma_i$ dense in the $C^0$ topology?
Even more generally:
Let $M_1,\ldots,M_k$ be a finite collection of topological submanifolds of $\mathbb R^n$ (of various dimensions, say). Is the set of (let's say compact) topological submanifolds of $\mathbb R^n$ that are topologically transverse to all the $M_i$ dense among all submanifolds of $\mathbb R^n$, with respect to the $C^0$-topology?

(Here, I'm not exactly sure which "$C^0$-topology" on the set of all submanifolds of $\mathbb R^n$ is best adapted to my problem. The "$C^0$-distance" between two submanifolds $M,N\subset\mathbb R^n$ could be taken to mean:
(1) the Hausdorff distance (probably not what I want).
(2) $\inf_f\sup_{x\in M}|f(x)-x|$, where $f$ runs over all homeomorphism from $M$ to $N$.
(3) $\inf_f\sup_{x\in \mathbb R^n}|f(x)-x|$, where $f$ runs over all homeomorphism of $\mathbb R^n$ that map $M$ to $N$.)

• Just to be sure: in your warm-up question you take the sup norm on $C^0([0,1])$ ? And then for $n=1$ the answer is simple: given $g$, approximate $g-f$ with a polynomial function $p$ (say) transverse to the horizontal axis, then $f+p$ does the job. Am I right ? – Mathieu Baillif Jul 16 '14 at 8:35
• @Mathieu: Yes. What you say is indeed correct. – André Henriques Jul 16 '14 at 9:06
• For Jordan curves you can follow the proof in the Appendix of Epstein, D. B. A., Curves on 2-manifolds and isotopies. Acta Math. 115 (1966) 83–107. In higher dimensions the result, I think is still true provided submanifolds are tame (some people will take this as a definition of a submanifold). However, I have to check the literature. (The most difficult case of dimensions 4 and 5 is probably treated in Quinn's papers.) – Misha Jul 16 '14 at 18:08
• The paper to read is Quinn's projecteuclid.org/download/pdf_1/euclid.bams/1183554528. You have to assume existence of a normal microbundle to $M_i$'s, otherwise the transversality theorem fails. – Misha Jul 16 '14 at 19:44
• Do I understand correctly that "existence of a normal microbundle" is the same as "tame" is the same as "looks locally like the standard inclusion of $\mathbb R^k\subset \mathbb R^n$"? Also, the paper by Quinn that you cite only does the case of a single manifold $M$. Does the case of a finite family $M_1,\ldots,M_k$ follow from the case of a single submanifold? – André Henriques Jul 16 '14 at 20:11

## 1 Answer

The answer to your first two questions is yes. For the third, I don't think I understand the correct generalization of topological transversality to higher dimensions. I'll prove the second. While the first doesn't strictly follow (one needs to show that transversality can be achieved in the space of graphical curves), the proof is similar.

Let $\gamma\colon[0,1]\to\mathbb R^2$ be a continuous map. We begin by making $\gamma$ topologically transverse to $\gamma_1$ in a way that is well-adapted to the other $\gamma_i$. Let $K_{ij}=\partial(\gamma_i\cap\gamma_j)$, where the boundary is taken in $\gamma_i$, and let $K_i=\bigcup K_{ij}$. We prove first that $\gamma$ can be perturbed in a $C^0$-small way to be topologically transverse to $\gamma_1$ and disjoint from $K_1$.

After an automorphism of $\mathbb R^2$, we may assume that $\gamma_1=[0,1]\times\mathbb R$. To just get transversality with $\gamma_1$, we can modify $\gamma$ to be piecewise-linear near $\gamma_1$ and unchanged elsewhere. Since $K_{1i}$ is nowhere dense in $\gamma_1$ for all $i$, so is $K_1$, and from there it's easy to see that a piecewise-linear map can be perturbed to avoid $K_1$.

By induction, assume that $\gamma$ has been perturbed to be topologically transverse to $\gamma_i$ for $i<k$ while avoiding $K_i$, and consider $\gamma_k$. Since $K_{ik}=K_{ki}$, we know that $\gamma\cap\gamma_k$ lies in $U_{ki}:=\gamma_k\setminus K_{ki}$. Note that $U_{ki}$ has the property that all it can only meet $\gamma_i$ by coinciding with $\gamma_i$ on some connected component. Thus, all intersections of $\gamma$ with $\gamma_k$ which are also intersections of $\gamma$ with $\gamma_i$ for some $i<k$ are transverse to $\gamma_k$. Let $$N_k=(\gamma\cap\gamma_k)\setminus\bigcup_i(\gamma_k\cap\gamma_i)=(\gamma\cap\gamma_k)\setminus\bigcup_i\mathrm{interior}(\gamma_k\cap\gamma_i).$$ Now $N_k$ is a closed subset of $\mathbb R^2$ and avoids $\bigcup_{i<k}\gamma_i$, so it has a neighborhood $V_k$ which separates it from the earlier curves. Now we just repeat what we did with $\gamma_1$ and $K_1$ for $\gamma_k$ and $K_k$, but constrain ourselves to only perturbing $\gamma$ in $V_k$.

Edit: I assume this is the type of induction Misha was talking about. It seems like it should work with no changes for equidimensional submanifolds in higher dimensions, but the general case might need a little more.

• Great! This indeed answers the question "I really need". I'm still curious to see whether something similar applies to higher dimensions. – André Henriques Jul 17 '14 at 7:49