# Paths transversal to countable collection of submanifolds

I am interested in the following claim, which is a bit counter-intuitive to me. I am wondering whether it is correct.

Claim. Let $$Y$$ be an open (path-)connected subset of $$\mathbb{R}^d$$, and let $$\{Z_i\}_{i\in I}$$ be a countable family of closed submanifolds of $$Y$$. Then, for any path $$\gamma:[0,1]\to Y$$ there exists a smooth path $$\gamma':[0,1]\to Y$$ such that (1) $$\gamma$$ and $$\gamma'$$ are path-homotopic and (2) $$\gamma'$$ is transversal to $$Z_i$$ for every $$i\in I$$.

For the case of a single submanifold $$Z_i$$, the question has an answer here: Tranversal paths between two points

There, a user refers to the following result in Guillemin and Pollack's Differential Topology (p.73), which immediately implies the claim for a singleton $$Z_i$$.

Corollary. If, for $$f:X\to Y$$, the boundary map $$\partial f:\partial X\to Y$$ is transversal to $$Z$$, then there exists a map $$g:X\to Y$$ homotopic to $$f$$ such that $$\partial g=\partial f$$ and $$g \pitchfork Z$$.

My (very rough) understanding is as follows: The corollary follows from the Transversality Homotopy Extension Theorem (p.72), which is proven from the Transversality Theorem. The constructed homotopy in the Extension Theorem is independent of the particular submanifold $$Z$$, thus all proofs should still go through with a countable family because a countable intersection of full measure subsets (of the unit ball '$$S$$') has full measure.

Is this reasoning correct?

• To be more precise, I should probably add a condition like: $\gamma(0),\gamma(1)\notin\bigcup_{i\in I} Z_i$. Commented Apr 18, 2021 at 17:46