# Metrics $g_1\leqslant g_2$ implies the Ricci flow $g_1(t)\leqslant g_2(t)$?

Let M be a complete,n dimensional Riemannian manifold without boundary. Suppose $g_1,g_2$ are two metrics on M and $g_1\leqslant g_2$. Suppose that there exists $T>0$ such that for $i=1,2$, the Ricci flow $\frac{\partial g}{\partial t}=-2 Ric$ with initial condition $g(0)=g_i$ exist for $t\in [0,T]$. Do we have $g_1(t)\leqslant g_2(t)$ for $t\in (0,T]$?

When n=2, the metric can be written in the conformal coordinates $$g_i(t)=e^{2u_i(t)}(dr^2+r^2 d\theta^2).$$ $u_i(t)$ satisfies the parabolic equation $$\frac{\partial u_i}{\partial t}=e^{-2u_i}\Delta u_i=-K_i$$ where $K_i$ are the Gauss curvature. We have $u_1(0)\leqslant u_2(0)$, we need to prove $u_1(t)\leqslant u_2(t)$.

Let $v(t)=\min_{x\in \mathbb{R}^2} u_2(t,x)-u_1(t,x)$, suppose the minimum are attained at $x_t$, then $\Delta(u_2-u_1)|_{x_t} \leqslant 0$ and thus $$\frac{\partial v}{\partial t} \leqslant (e^{-2u_2}-e^{-2u_1})\Delta u_1|_{x_t}.$$ It seems that we can't expect that $\frac{\partial v}{\partial t}\leqslant 0$. So how can we prove $u_1(t)\leqslant u_2(t)$ ?

For the compact manifold, @Otis Chodosh already gave a good answer on the comment. If you only assume $$M$$ to be complete, I'm not sure if it is still true since your statement implies the uniqueness of the solution of the Ricci flow (for unbounded domain, the uniqueness of the heat equation is not true because of the famous example of Tychonoff : $$u(x,t)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\frac{d^k}{dt^k}e^{-\frac{1}{t^2}}$$ satisfying $$\frac{\partial}{\partial t}u=\Delta u$$ but $$u(x,0)\equiv0$$).