[proofplan] Because the time interval $[0,T_0]$ lies strictly before the maximal time $T$, the Ricci flow is smooth on the compact spacetime set $M \times [0,T_0]$. We use compactness to obtain uniform curvature and injectivity-radius control, then choose a scale below the uniform injectivity radius. On that scale the exponential map gives uniformly controlled [normal coordinates](/theorems/2713), so small geodesic balls have volume bounded below by a fixed multiple of the Euclidean ball volume. This lower bound is stronger than the required conditional $\kappa$-noncollapsing estimate. [/proofplan]

[step:Extract uniform geometric bounds on the compact time interval]Fix $T_0<T$. Since $M$ is closed and $g(t)$ is a smooth Ricci flow for $t \in [0,T)$, the spacetime set $M \times [0,T_0]$ is compact and the curvature tensor depends continuously on $(x,t)$. Define the finite curvature bound \begin{align*} K_0 := \sup\{ |\operatorname{Rm}_{g(t)}(x)| : x \in M,\ 0 \le t \le T_0\} < \infty. \end{align*} For each $t \in [0,T_0]$, let $\operatorname{inj}_{g(t)}(M)$ denote the injectivity radius of the closed Riemannian manifold $(M,g(t))$. Smooth dependence of $g(t)$ on $t$ and compactness of $[0,T_0]$ give \begin{align*} i_0 := \inf_{0 \le t \le T_0} \operatorname{inj}_{g(t)}(M) > 0. \end{align*}[/step]

[guided]The point of restricting to $T_0<T$ is that no singular time is included. The family $g(t)$ is smooth on the compact set $M \times [0,T_0]$, so tensor norms that depend continuously on $(x,t)$ have finite suprema. In particular, the function \begin{align*} (x,t) \mapsto |\operatorname{Rm}_{g(t)}(x)| \end{align*} is continuous on $M \times [0,T_0]$, and therefore the number \begin{align*} K_0 := \sup\{ |\operatorname{Rm}_{g(t)}(x)| : x \in M,\ 0 \le t \le T_0\} \end{align*} is finite. We also need a uniform radius on which geodesic normal coordinates are valid. For each fixed $t$, the closedness of $M$ implies that $(M,g(t))$ has positive injectivity radius. Since $g(t)$ varies smoothly in $t$ on the compact interval $[0,T_0]$, the injectivity radius has a positive lower bound on that interval. Thus \begin{align*} i_0 := \inf_{0 \le t \le T_0} \operatorname{inj}_{g(t)}(M) > 0. \end{align*} This is the compactness input that replaces the deeper reduced-volume argument needed for scale control up to a singular time.[/guided]

[step:Choose a uniform normal-coordinate scale] Let $\omega_n := \mathcal{L}^n(B(0,1))$ denote the Euclidean volume of the unit ball in $\mathbb{R}^n$. By the volume density expansion in geodesic normal coordinates, uniformly for $x \in M$ and $t \in [0,T_0]$, \begin{align*} \frac{\operatorname{Vol}_{g(t)}(B_{g(t)}(x,r))}{\omega_n r^n} \to 1 \end{align*} as $r \downarrow 0$. The convergence is uniform because the metric coefficients in normal coordinates and their curvature-controlled Taylor remainders are uniformly bounded on the compact family from the previous step. Hence there exists $r_0>0$ with \begin{align*} 0<r_0 \le \frac{i_0}{2} \end{align*} such that, for every $x \in M$, every $t \in [0,T_0]$, and every $0<r\le r_0$, \begin{align*} \operatorname{Vol}_{g(t)}(B_{g(t)}(x,r)) \ge \frac{\omega_n}{2} r^n. \end{align*} [/step]

[step:Conclude the noncollapsing estimate] Define \begin{align*} \kappa := \frac{\omega_n}{2} > 0. \end{align*} Let $t \in [0,T_0]$, let $x \in M$, and let $0<r\le r_0$. If the curvature condition \begin{align*} |\operatorname{Rm}_{g(t)}| \le r^{-2} \end{align*} holds on $B_{g(t)}(x,r)$, then the volume estimate from the previous step gives \begin{align*} \operatorname{Vol}_{g(t)}(B_{g(t)}(x,r)) \ge \kappa r^n. \end{align*} The estimate in fact holds for every such ball, independently of the curvature hypothesis, so the required conditional $\kappa$-noncollapsing statement follows for all $t \le T_0$ at scale $r_0$. [/step]