Morse polynomial identity

If $M=\mathrm{\varnothing }$, every chain group and Morse number is zero, so the identity holds with $Q=0$, uniquely by coefficient comparison. Suppose $M\mathrm{\ne }\mathrm{\varnothing }$. Its minimum and maximum are critical, so by [L1] there are finitely many distinct critical values $c_1<\cdots <c_{\nu }$ with $\nu \ge 1$. Choose $t_0<\min f$, $t_{\nu }>\max f$, and $c_i<t_i<c_{i+1}$ for $1\le i<\nu$. These are regular values, $M^{t_0}=\mathrm{\varnothing }$, and $M^{t_{\nu }}=M$.

For each $1\le i\le \nu$ the slab $f^{-1}([t_{i-1},t_i])$ is compact, as a closed subset of the compact $M$, and the critical points it contains are exactly the critical points of value $c_i$; they are nondegenerate and share the value $c_i$, and $t_{i-1}<c_i<t_i$ are regular values. Hence [F2] gives, for every $i$ and $j$, ${\dim }_F{H}_j(M^{t_i},M^{t_{i-1}};F)=m_j^{(i)}:=\mathrm{\#}\{p:f(p)=c_i,\mathrm{ind}(p)=j\}\mathrm{.}$

Induction on $i$: the graded vector space $H_{\ast }(M^{t_i};F)$ is finite-dimensional in every degree and vanishes in degrees below $0$ and above $n$. For $i=0$ this is $H_{\ast }(\mathrm{\varnothing };F)=0$. For the step, apply [F1] to the long exact sequence of the pair $(M^{t_i},M^{t_{i-1}})$ from [L2] with $A_k=H_k(M^{t_{i-1}};F)$, $B_k=H_k(M^{t_i};F)$ and $C_k=H_k(M^{t_i},M^{t_{i-1}};F)$: the hypothesis on $A$ is the induction hypothesis, the hypothesis on $C$ is step 2.1 (the sum of the $m_j^{(i)}$ over $j\le n$ is finite and the groups vanish in negative degrees), and [F1] concludes that $B$ is finite-dimensional in each degree.

The same application of [F1] gives, for each $i$, the identity $P_{M^{t_{i-1}}}(t)+P_{M^{t_i},M^{t_{i-1}}}(t)=P_{M^{t_i}}(t)+(1+t)Q_i(t)$ with $Q_i(t)={\sum }_j{q}_j^{(i)}{t}^j\in \mathbb{Z}[t]$, $q_j^{(i)}={\dim }_F\ker (H_j(M^{t_{i-1}};F)\to H_j(M^{t_i};F))\ge 0$, and $P_{M^{t_i},M^{t_{i-1}}}(t)={\sum }_j{m}_j^{(i)}{t}^j$ by step 2.1.

Summing the identities of step 4.1 over $i=1,\dots ,\nu$ telescopes: ${\sum }_i{P}_{M^{t_{i-1}}}-{\sum }_i{P}_{M^{t_i}}=P_{M^{t_0}}-P_{M^{t_{\nu }}}=-P_{M,F}$, since $M^{t_0}=\mathrm{\varnothing }$ and $M^{t_{\nu }}=M$. Hence ${\sum }_{i=1}^{\nu }{P}_{M^{t_i},M^{t_{i-1}}}(t)=P_{M,F}(t)+(1+t)Q(t),Q:={\sum }_{i=1}^{\nu }{Q}_i\in \mathbb{Z}[t],$ and $Q$ has nonnegative coefficients, being a sum of polynomials with nonnegative coefficients.

The left side equals $M_f(t)$: by step 2.1 and [F4], ${\sum }_i{P}_{M^{t_i},M^{t_{i-1}}}(t)={\sum }_{i,j}{m}_j^{(i)}{t}^j$, and the numbers $m_j^{(i)}$ partition the critical points by critical value and index, so ${\sum }_i{m}_j^{(i)}=m_j(f)$ and, by [F3], ${\sum }_j{m}_j(f)t^j=M_f(t)$. Therefore $M_f(t)=P_{M,F}(t)+(1+t)Q(t)$ with $Q\in \mathbb{Z}[t]$ of nonnegative coefficients.

Uniqueness of $Q$: if $Q,Q^{\mathrm{\prime }}\in \mathbb{Z}[t]$ satisfy $(1+t)Q=(1+t)Q^{\mathrm{\prime }}$, then $(1+t)(Q-Q^{\mathrm{\prime }})=0$ and, comparing coefficients, $q_0-q_0^{\mathrm{\prime }}=0$ and $(q_k-q_k^{\mathrm{\prime }})=-(q_{k-1}-q_{k-1}^{\mathrm{\prime }})$ for $k\ge 1$, so all coefficients vanish and $Q=Q^{\mathrm{\prime }}$.

The partial-sum form: writing $M_f-P_{M,F}=(1+t)Q={\sum }_k(q_k+q_{k-1})t^k$ with $q_{-1}:=0$, the coefficient at $t^k$ is $m_k(f)-b_k(M;F)=q_k+q_{k-1}$. Hence for every $k$ ${\sum }_{i=0}^k(-1{)}^{k-i}(m_i(f)-b_i(M;F))={\sum }_{i=0}^k(-1{)}^{k-i}(q_i+q_{i-1})=q_k\ge 0,$ by telescoping; in particular the weak and strong inequalities hold coefficientwise. Evaluating at $t=-1$ using [F5] gives $(M_f-P_{M,F})(-1)=0\cdot Q(-1)=0$, which is the equality of the total alternating sums.