Solution (source code)

= Solution

Again assume the intended trace <Ricci curvature> bound $\operatorname{Ric}\geq(n-1)kg$, and put $D=\pi/\sqrt k$. Compactness from the <Bonnet-Myers theorem> ensures that the diameter is attained: choose $p,q$ with $d(p,q)=D$. For $0<r<D$, the open balls $B(p,r)$ and $B(q,D-r)$ are disjoint, because a point in both would violate the <triangle inequality>.

The clearly stated <Bishop volume comparison with a positive-curvature model> from part (b), applied between radius $r$ and radius $D$, gives
$$
V_p(r)\geq\frac{v_k(r)}{v_k(D)}\operatorname{Vol}M,\qquad V_q(D-r)\geq\frac{v_k(D-r)}{v_k(D)}\operatorname{Vol}M.
$$
Here both radius-$D$ ball volumes equal the total volume, as shown in part (b). The symmetry $s_k(D-t)=s_k(t)$ gives
$$
v_k(r)+v_k(D-r)=v_k(D).
$$
The two lower bounds therefore sum to $\operatorname{Vol}M$, while disjointness bounds that same sum above by $\operatorname{Vol}M$. Both inequalities must be equalities. In particular
$$
\frac{V_p(r)}{v_k(r)}=\frac{\operatorname{Vol}M}{v_k(D)}\qquad(0<r<D).
$$
Letting $r\downarrow0$, the local Euclidean volume asymptotic gives one on the left. Thus $\operatorname{Vol}M=v_k(D)$, and the detailed equality proof in part (b) applies. We obtain
$$
\boxed{\operatorname{diam}M=\pi/\sqrt k\quad\Longrightarrow\quad M\cong S_k^n.}
$$
This proves <maximal diameter rigidity from disjoint comparison balls> directly, including the required equality argument.

\b[The literal trace interpretation again has a counterexample.] On $M=S^2(r)\times S^2(r)$ with $r^2=1/(2k)$, the <Ricci curvature> equals $2kg\geq kg$. For the <product Riemannian metric>, distances satisfy $d^2=d_1^2+d_2^2$: every product curve has length at least $\sqrt{d_1^2+d_2^2}$ by the integral <triangle inequality>, and a pair of constant-speed minimizing factor <geodesics> attains it. Consequently
$$
\operatorname{diam}M=\sqrt{(\pi r)^2+(\pi r)^2}=\frac{\pi}{\sqrt k}.
$$
The mixed <sectional curvature> is zero, so this manifold is not $S_k^4$. The stronger bound, or <normalized Ricci curvature> convention, is essential.