Most theorems stating that transversality is an open condition look like:

We have manifolds $M$, $N$, and a submanifold $A\subseteq N$. Then assuming $A$ is closed in $N$, in the appropriate topology on the space of smooth maps $C^\infty(M,N)$ (weak/compact-open topology if $M$ is compact, or strong/Whitney topology if $M$ is noncompact), the maps transverse to $A$ form an open dense subset.

However, I don't understand why this condition $A$ closed in $N$ is needed. I was trying to see if there are counterexamples for it.

For reference, here is Hirsch's statement (Differential Topology, Chapter 3): enter image description here

And also from this bachelor thesis https://www.math.harvard.edu/media/ThesisXFinal.pdf:

enter image description here

Obs: This post (Reference for transversality -- relative version) seems related, but I'm not sure it answers this question. The answer in it seems to explain why $A$ closed in $N$ matters, but only if we're not asking transversality in all of $M$.


2 Answers 2


Let $M=\mathbb{R}$, $N=\mathbb{R}^2$, and $A \subset N$ be the 1-submanifold consisting of the $x$-axis and a sequence of open vertical lines segments approaching the $y$-axis. Consider the element of $C^\infty(M,N)$ which just embeds $M$ as the $y$-axis. This map is transverse to $A$.

enter image description here

However, any open set in $C^\infty(M,N)$ around this map will include embeddings of $M$ that intersect one of the vertical line segments tangentially. So the transverse maps do not form an open subset of $C^\infty(M,N)$. Note thought that they are still dense, as any of the embeddings of $M$ that shares a vertical tangency with $A$ can be nudged a bit to become transverse again.


Easier example: Take $A$ to be the positive real axis in $N=\Bbb R^2$. The function $f\colon \Bbb R\to \Bbb R^2$ given by $f(t)=(t,t^2)$ is transverse to $A$, but no neighborhood of $f$ in any topology will stay transverse to $A$. (You can easily modify this with maps $S^1\to \Bbb R^2$, if you want a compact domain.)

If I remember correctly, closedness isn't required for denseness.

  • $\begingroup$ I think you mean transverse to $A$? Also the parabola is not transverse to the positive real axis, right? $\endgroup$ Commented Nov 6, 2021 at 17:43
  • $\begingroup$ Positive means positive, not non-negative. So, yes, it's transverse by default. Yes, I meant $A$. I'll edit that. $\endgroup$ Commented Nov 6, 2021 at 17:54

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