[step:Apply the equality case in reduced volume monotonicity]Let $n := \dim M$. Let $(p,0)$ denote the regular basepoint for the reduced distance, meaning that Perelman's reduced distance based at $(p,0)$ is defined and has the standard smoothness and cut-locus structure along minimizing $\mathcal L$-geodesics. Let $l: M \times I \to \mathbb{R}$ be that reduced distance. For each $\tau \in I$, define the reduced-volume density $v_\tau: M \to [0,\infty)$ by
\begin{align*}
v_\tau(x) = (4\pi\tau)^{-n/2} e^{-l(x,\tau)}.
\end{align*}
This density is taken with respect to the Riemannian volume measure $d\mu_{g(\tau)}$. Define the reduced volume $\widetilde V: I \to [0,\infty)$ by
\begin{align*}
\widetilde V(\tau) = \int_M v_\tau(x)\,d\mu_{g(\tau)}(x).
\end{align*}
We invoke Perelman's reduced-volume monotonicity and equality-case rigidity theorem as a packaged equality-case result; the present proof reconstructs the soliton form from that equality-case identity rather than reproving Perelman's monotonicity formula. In the form used here, for a complete backward Ricci flow with bounded curvature on compact time subintervals, reduced distance based at a regular point, and finite reduced volume, $\widetilde V$ is non-increasing. Moreover, if its [distributional derivative](/page/Distributional%20Derivative) vanishes on a compact subinterval, then the theorem gives all five conclusions needed below: the reduced-distance function is smooth on the pre-cut regular set; the $\mathcal L$-cut locus has zero contribution to the reduced-volume measure; the pre-cut regular set is carried smoothly by the unique minimizing $\mathcal L$-geodesic parametrization; this parametrization is invariant under the vector-field transport used below, so transported curves starting in the pre-cut region cannot hit the $\mathcal L$-cut locus before the target time; and the non-positive monotonicity density vanishes pointwise there. In particular, on the regular set of spacetime points reached by unique minimizing $\mathcal L$-geodesics before their cut time, the pointwise identity
\begin{align*}
\operatorname{Ric}_{g(\tau)} + \nabla^{g(\tau)}\nabla^{g(\tau)} l(\cdot,\tau) - \frac{1}{2\tau}g(\tau) = 0
\end{align*}
holds. Here a regular basepoint means that the reduced exponential map is smooth away from its cut locus, the $\mathcal L$-cut locus is the first failure set of unique minimizing $\mathcal L$-geodesics, and the pre-cut regular set is the image of the reduced exponential map before that failure time. The hypotheses of this theorem are exactly the completeness, compact-time curvature bound, regular-basepoint, and finiteness assumptions in the statement. Since $\widetilde V$ is constant on $I$, its distributional derivative on every compact subinterval $J \subset I$ is zero. The equality case therefore gives, on the regular set of points reached by unique minimizing $\mathcal L$-geodesics before cut time,
\begin{align*}
\operatorname{Ric}_{g(\tau)} + \nabla^{g(\tau)}\nabla^{g(\tau)} l(\cdot,\tau) - \frac{1}{2\tau}g(\tau) = 0.
\end{align*}
Here $\nabla^{g(\tau)}\nabla^{g(\tau)} l(\cdot,\tau)$ denotes the Hessian of the function $x \mapsto l(x,\tau)$ with respect to $g(\tau)$.[/step]