= Solution
The usual compact closed-manifold hypothesis is needed for the first assertion, or a proper bounded-below Morse exhaustion in an appropriate noncompact version. As literally printed without any such condition, $f(x)=x$ on $\mathbb R$ is a <Morse function> with no <critical points>, while $b_0=1$. We prove the intended closed-manifold statement, which applies to the <surface> product later in the question.
By the <Morse handle-attachment theorem>, crossing an index-$i$ <critical point> attaches an $i$-handle, and no topological change occurs between critical levels. Retraction to the <core disks of handles> gives a <CW complex> model with $c_i$ cells in dimension $i$. Its <cellular chain complex> has $C_i\cong\mathbb Z^{c_i}$. Tensor with $\mathbb Q$, which computes the ranks $b_i$ of integral <homology>. Since <homology> is a quotient of a <vector subspace> of $C_i\otimes\mathbb Q$, \b[$b_i\le c_i$]. This argument has not used the <Morse inequalities>.
The strong <Morse inequalities> and top-dimensional equality are
$$
\boxed{\sum_{i=0}^j(-1)^{j-i}c_i\ge\sum_{i=0}^j(-1)^{j-i}b_i\quad(0\le j\le m),\qquad\sum_{i=0}^m(-1)^ic_i=\sum_{i=0}^m(-1)^ib_i.}
$$
Their weak form is $c_i\ge b_i$. To see the algebra behind the strong form, put $r_i=\operatorname{rank}_{\mathbb Q}\partial_i$ in the <Morse chain complex>, with $r_0=r_{m+1}=0$. The kernel-image dimensions give $c_i=b_i+r_i+r_{i+1}$, so the alternating difference at $j$ is exactly $r_{j+1}\ge0$. Conversely, given the strong inequalities, denote those alternating differences by $S_j$, with $S_{-1}=0$. Then $c_j-b_j=S_j+S_{j-1}\ge0$, proving that they imply every weak inequality.
They are \b[not equivalent as numerical restrictions]. For the <Betti numbers> $b=(1,0,0,1)$ of $S^3$, the formal counts $c=(2,0,0,2)$ satisfy all weak inequalities and the Euler equality, but at $j=1$ give $c_1-c_0=-2<-1=b_1-b_0$. They therefore cannot be the counts of a <Morse function> on $S^3$; the strong inequalities detect that additional obstruction.
If all the strong inequalities are equalities, all $r_i$ vanish and $c_i=b_i$. An integer boundary <matrix> with rational rank zero is the zero <matrix>. Thus the integral complex itself has zero differential, not merely its rationalization. Consequently
$$
\boxed{H_i(M;\mathbb Z)\cong\mathbb Z^{c_i}=\mathbb Z^{b_i}\quad(0\le i\le m),}
$$
and higher <homology> vanishes. In particular there is no <homology> torsion. This is an integral consequence of a <Perfect Morse function>, rather than just an assertion about ranks.
For $\Sigma_g$, the integral <homology> is free with <Betti numbers> $(1,2g,1)$. The <Künneth theorem> therefore has no torsion terms, and the product's <Poincaré polynomial> is
$$
(1+2gt+t^2)^2=1+4gt+(4g^2+2)t^2+4gt^3+t^4.
$$
Summing the weak inequalities yields the lower bound
$$
\boxed{\#\operatorname{Crit}(F)\ge4g^2+8g+4=(2g+2)^2\quad\text{for Morse }F.}
$$
Choose a <surface> <Morse function> $h$ with one minimum, $2g$ saddles and one maximum. It can be built from one <closed disk>, the standard $2g$ <one-handles>, and a closing <closed disk>, using the local forms $u^2+v^2$, $-u^2+v^2$, and $-u^2-v^2$ at the <handle> centres. Regular collars interpolate between the <handles> without new <critical points>. Now set $F(x,y)=h(x)+h(y)$. Its <critical points> are the ordered pairs of those of $h$, and its <Hessian matrix> is a block direct sum; the indices add. This <product of Morse functions> has exactly \b[$(2g+2)^2$ <critical points>] and realizes every displayed Betti count. Small bumps constant near the critical pairs can separate their values without altering the <Hessians> or creating new <critical points>.
The same minimum does \b[not] apply to arbitrary <smooth functions>. Here is an explicit finite-critical-point counterexample, avoiding any assumption that degeneracy leaves the Morse lower bound intact. On $T^2=(\mathbb R/\pi\mathbb Z)^2$, take
$$
h_0(u,v)=\sin u\sin v\sin(u-v).
$$
It is well defined because shifting either coordinate by $\pi$ reverses two of its factors. Its derivatives are $\sin v\sin(2u-v)$ and $\sin u\sin(u-2v)$. If either $\sin u$ or $\sin v$ vanishes, both derivatives vanish only at $(0,0)$. Otherwise $2u-v$ and $u-2v$ must be multiples of $\pi$, giving exactly $(\pi/3,2\pi/3)$ and $(2\pi/3,\pi/3)$ in addition to the origin. At the first point the <Hessian> is $\sqrt3\bigl(\begin{smallmatrix}1&-1/2\\-1/2&1\end{smallmatrix}\bigr)$, positive definite; at the second it is its negative. They are a minimum and maximum. The origin is an isolated degenerate <monkey saddle>, with leading term $uv(u-v)$. Thus this <three-critical-point function on a torus> has exactly three <critical points>.
For $g>1$, perform <connected-sum gluing of functions> between this <torus> and a perfect Morse <surface> of genus $g-1$: delete a small neighbourhood of the <torus> maximum and of the other <surface>'s minimum, and join their regular boundary circles by a monotone cylinder. Collar coordinates ensure nonzero differential throughout the neck, so precisely those two <critical points> are lost and none are added. The result on $\Sigma_g$ has $3+[2(g-1)+2]-2=2g+1$ <critical points>, retaining the degenerate point. For $g=1$, simply use $h_0$. Calling this <smooth function> $\widetilde h$ and the earlier perfect <Morse function> $h$, the smooth product $\widetilde h(x)+h(y)$ has
$$
\boxed{(2g+1)(2g+2)<(2g+2)^2}
$$
<critical points>. This explicitly proves the negative answer to the unrestricted smooth-minimum question; it does not assert that this smaller illustrative count is sharp.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-25-critical-points.png]
{title=The explicit three-critical-point torus function and the local merger of two Morse saddles into one degenerate saddle}
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