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Pointwise positive Ricci curvature does not imply compactness or a finite diameter. The complete positively curved paraboloid $z=x^2+y^2$ has induced metricand Gaussian curvatureFor example, this follows from the graph formula $K=\det D^{2}z/(1+|Dz{|}^2{)}^2$. The metric dominates the Euclidean plane metric, so it is complete by upward stability of Riemannian completeness. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the Ricci tensor is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.

There is no positive-Ricci metric on $S^2\times S^1$. If one existed, compactness of the unit tangent bundle would give ${\mathrm{Ric}}_{\mathrm{tr}}\ge cg$ for some $c>0$. The lifted metric on the universal cover is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover compact. A covering fiber is closed and discrete and therefore finite, implying a finite fundamental group. Butwhich is infinite. This is the finite fundamental group from a uniform positive Ricci bound obstruction; it rules out every possible metric, not just the standard product metric.