The cellular approximation theorem states that given a continuous map between two CW complexes $f : X \to Y$, then $f$ is homotopic to a cellular map - that is some map $f'$ with $f'(X_n) \subset Y_n$ for all $n$ where $X_n$ is the $n$-skeleton of $X$.

The simplicial approximation theorem states that given a continuous map between two simplicial complexes $f : K \to L$ then $f$ is homotopic to map $f'$ such that there are subdivisions of the simplicial structures on $K$ and $L$ and such that $f'$ is simplicial with respect to the new simplicial decompositions.

I frequently think about manifolds and I typically consider my manifolds as being given with handle decompositions. Let $M$ and $N$ be two (smooth compact closed connected if you like) manifolds together with handle decompositions $M = h_1^0 \cup \cdots \cup h_l^n$ and $N = j_1^0 \cup \cdots \cup j_k^m$. Denote by $M_i$ the union of all of the handles of index less than or equal to $i$ of $M$ and similarly for $N$. Consider a (smooth) map $f : M \to N$.

Is $f$ homotopic to a map $f'$ so that $f'(M_i) \subset N_i$ for all $i$? Or better, do there exist handle decompositions of $M$ and $N$ (possibly different from those given) and a map $f'$ homotopic to $f$ so that $f'(M_i) \subset N_i$ for these new handle decompositions?

I have a feeling that the answer is no and I have been trying to come up with a conterexample for $M$ and $N$ both closed compact connected 3-manifolds. I believe in this case one would want to consider a map of degree greater than 1. Any thoughts on how to do this? Thanks!

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    $\begingroup$ If you do not assume any regularity conditions, and if the two handlebodies are ordered by increasing indices (as it is a CW-complex), then this easily follows from cellular approximation, by considering the CW-complex given by handle cores and smooth approximation too. $\endgroup$ – Daniele Zuddas Jun 9 '17 at 1:30

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