Connection between framed cobordisms and zero sets Let $W\subseteq M\times[0,1]$ be a framed submanifold (a framed cobordism in $M$) and $2w<m-1$ where $w,m=\dim W,M$. Assume that $M$ is compact and that $W\cap M\times \{0\}$ is the zero set of a smooth $f: M\simeq M\times\{0\}\to \mathbb{R}^n$ transversal to $0$ ($n=m-w)$. Does it follow that $W\cap M\times\{1\}$ is also the zero set $g^{-1}(0)$ for some smooth $g: M\simeq M\times\{1\}\to\mathbb{R}^n$ transversal to $0$? 
If it matters, assume that $M$ has a boundary and $W$ is disjoint from $\partial M\times [0,1]$.

This is closely related to this question; what I understood from the answers there, it should be possible (in the indicated dimension range) to "project" the framed cobordism $W\subseteq M\times[0,1]$ to a framed manifold $\pi(W)\subseteq M$ s.t. $\partial \pi(W)=M_1\sqcup M_2$ and $M_1$ is the zero set of some map. Then one can show that $M_2$ is the zero set of some $\mathbb{R}^n$-valued map as well (this part I believe I can formalize). I don't see how to do this "projection of the framed manifold"; I need to find a perturbation of the inclusion $W\hookrightarrow M\times[0,1]$ to remove double points (easy) and vertical vectors (less trivial for me) and somehow define a framing on the perturbation and project it to an $(n-1)$-framing.. I was not able to formalize this: is it the right direction of thoughts and does the statement hold at all? 
 A: Let's start by summarising the conclusions reached in the previous discussions: Since, in particular, $W \cap (M \times \{1\})$ is framed null-cobordant by assumption, the same is also true for $V := W \cap (M \times \{0\})$. Let $\nu V \subset M$ denote a tubular neighbourhood, and write $M' := M \setminus \nu V$. By the answers to your previous question, $V \subset M$ is the regular zero-set of a $\mathbb{R}^n$-valued function if and only if $V' \subset \partial M'$, obtained as a constant section in the null-cobordant framing, is framed null-cobordant inside $M'$. Here the framing of $V' \subset \partial M'$ is chosen to be the one induced by the framing of $V$ in the obvious way.
In conclusion, we need to construct the above framed null-cobordism in the case when $2w < m-1$. We let $X \subset M'\times I \subset M \times I$ denote a framed null-cobordism of $V'$, which exists by a general position-argument. Here we used that
$$2 \dim X =2w < m+1= \dim (M \times I)$$ together with the fact that $V$ is framed null-cobordant in $M \times I$.
By hand, we may arrange so that the canonical projection $\pi \colon M' \times I \to M'$ restricts to an embedding in some collar neighbourhood of $\partial X \subset X$. Again, a general-position argument shows that, after a compactly supported perturbation of $X \setminus \partial X$, the restriction $\pi|_X \colon X \to M'$ is an embedding. This follows from the proof of the weak Whitney embedding theorem, since
$$ \dim (M \times I) = m+1 > 2w+2 > 2\dim W+1.$$
In fact, the above argument immediately shows something even stronger: The one-parameter family of maps $$(\operatorname{id}_M, (1-t) \operatorname{id}_I+t) \colon M' \times I \to M' \times I, \:\: t \in [0,1],$$ induces a one-parameter family of embeddings in $M' \times I$, starting at $X$ and ending at $\pi(X) \times \{1\}$. There is a compactly supported isotopy of $(M' \setminus \partial M') \times I$ inducing this family of embeddings. In particular, the existence of the isotopy implies that $\pi(X)$ has a trivial normal-bundle as well, at least when considered as a submanifold of $M' \times I$.
What remains is thus to check that we can arrange so that the section $\partial_s$, where $s$ denotes the coordinate on the $I$-factor of $M' \times I$, coincides with (say) the first section $f_1$ of the frame on $\pi(X) \subset M' \times I$. Again, this can be arranged by hand to hold near $\partial \pi (X)$. To arrange this in all of $\pi(X)$ we argue as follows. Since $w < n - 2$ (recall that $n=m+1-w$ is the dimension of the normal bundle of $\pi(X) \subset M' \times I$), a general position argument shows that $f_1(x)$ can be assumed to be nowhere inside $-\mathbb{R}_+\partial _s$. There is hence a unique element in $SO(n)$ above any point $ x \in X$ mapping $f_1(x)$ to $\partial_s$ while fixing $\langle f_1(x),\partial_s \rangle^\perp \subset \mathbb{R}^n$. This construction induces a continuous map $X \to SO(n)$ which does the job (by acting pointwise on the frame).
