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Let $M$ be a closed manifold. If $f$ is a Morse function on $M$, then by Morse inequalities we know that $f$ must have at least $\sum_i\beta_i(M;\mathbb{Z}_2)$ critical points. When is it possible to find such a function with exactly this number of critical points? If one can find such a function, then the boundary maps in the Morse complex must be zero. Does there exist any sort of topological property that can prevent from boundary maps being zero?

For instance, on a closed surface, this is possible since if we take a Morse function with only one max point and one min point, then the boundary maps in the Morse complex must be zero and we have $\beta_1(M;\mathbb{Z}_2)$ critical points of index $1$, one with zero index and one with index $2$.

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Permit me to talk about the Morse complex over $\mathbb {Z}$. For $\mathbb{Z}_2$ a similar argument works.

A function whose Morse complex has trivial boundary maps is called a perfect Morse function. Finding perfect Morse functions is hard; To my best knowledge it is as open problem to characterize manifolds that admit perfect Morse functions.

Simple examples of manifold which will never have perfect Morse functions can be made as follows: Take a group $G$, whose abelianization is trivial. Construct a manifold $M$, with fundamental group $G$ (This is always possible if $G$ is finitely presented group for a manifold $M$ of dimension $4$). Then $H_1(M;\mathbb{Z})=G_{\mathrm{ab}}=0$. A perfect Morse function would not have critical points of index $1$. But since $\pi_1(M)=G$, this cannot be the case.

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