# 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

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 $V_k$.