[proofplan]
We use the smooth equivalence theorem for Lott--Sturm--Villani curvature-dimension bounds on weighted Riemannian manifolds: for a weighted manifold $(M,g,e^{-V}\operatorname{vol}_g)$, the condition $CD(K,N)$ is equivalent to the Bakry--Emery tensor inequality $\operatorname{Ric}_{V,N}\ge Kg$. In the unweighted case $V=0$ and with $N=n=\dim M$, the Bakry--Emery tensor reduces exactly to the ordinary Ricci tensor. Thus the forward and reverse implications are obtained by specializing the smooth weighted theorem and identifying the tensor appearing in the specialization.
[/proofplan]
[step:Recall the smooth weighted equivalence theorem for $CD(K,N)$]
We use the following standard smooth equivalence theorem of Lott--Sturm--Villani and Sturm for weighted Riemannian manifolds, stated here because the result is not yet available as a separate theorem in the wiki: for a connected complete smooth $n$-dimensional Riemannian manifold $(M,g)$, a function $V\in C^\infty(M)$, the weighted measure $m=e^{-V}\operatorname{vol}_g$, and a parameter $N\in[n,\infty]$, the metric-measure space $(M,d_g,m)$ satisfies the Lott--Sturm--Villani condition $CD(K,N)$ with the corresponding distortion coefficients if and only if
\begin{align*}
\operatorname{Ric}_{V,N}(v,v)\ge K g_p(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$.
Here, for $N\in(n,\infty)$, the $N$-Bakry--Emery Ricci tensor is
\begin{align*}
\operatorname{Ric}_{V,N}:=\operatorname{Ric}_g+\nabla^2 V-\frac{1}{N-n}\nabla V\otimes \nabla V,
\end{align*}
while for $N=\infty$ it is
\begin{align*}
\operatorname{Ric}_{V,\infty}:=\operatorname{Ric}_g+\nabla^2 V.
\end{align*}
For $N=n$, the convention is that $CD(K,n)$ can hold only when $\nabla V=0$, and in that case
\begin{align*}
\operatorname{Ric}_{V,n}:=\operatorname{Ric}_g+\nabla^2 V.
\end{align*}
Equivalently, in the unweighted case $V=0$, one has
\begin{align*}
\operatorname{Ric}_{0,n}=\operatorname{Ric}_g.
\end{align*}
[guided]
The only external input in this proof is the smooth weighted characterization of the Lott--Sturm--Villani curvature-dimension condition. Its content is that the global optimal-transport convexity inequality defining $CD(K,N)$ is exactly equivalent, on a smooth complete weighted Riemannian manifold, to a pointwise tensor lower bound.
Concretely, let $V\in C^\infty(M)$ be a smooth weight and define the measure $m$ by
\begin{align*}
m(A):=\int_A e^{-V(x)}\,d\operatorname{vol}_g(x)
\end{align*}
for every Borel set $A\subset M$. The theorem says that $(M,d_g,m)$ satisfies $CD(K,N)$ precisely when the Bakry--Emery tensor satisfies
\begin{align*}
\operatorname{Ric}_{V,N}(v,v)\ge K g_p(v,v)
\end{align*}
for every point $p\in M$ and tangent vector $v\in T_pM$.
This theorem is the rigorous form of the Jacobian-comparison argument. In one direction, the Ricci lower bound controls determinants of Jacobi fields along optimal-transport geodesics, and those determinant estimates give the distortion coefficients $\tau_{K,N}^{(t)}$ in the Lott--Sturm--Villani inequality. In the reverse direction, localizing the $CD(K,N)$ inequality to very small transports and expanding the determinant distortion to second order recovers the infinitesimal curvature tensor inequality. For the present theorem, we only need the specialization $V=0$ and $N=n$.
[/guided]
[/step]
[step:Specialize the weighted theorem to the Riemannian volume measure]
Define the smooth weight function
\begin{align*}
V:M&\to\mathbb R
\end{align*}
\begin{align*}
p&\mapsto 0.
\end{align*}
Then $V\in C^\infty(M)$ and
\begin{align*}
e^{-V}\operatorname{vol}_g=\operatorname{vol}_g.
\end{align*}
Moreover, for every $p\in M$, the gradient $\nabla V(p)\in T_pM$ is $0$, and the Hessian $\nabla^2V(p):T_pM\times T_pM\to\mathbb R$ is the zero bilinear form. Hence the $n$-Bakry--Emery Ricci tensor in the unweighted case satisfies
\begin{align*}
\operatorname{Ric}_{0,n}(v,v)=\operatorname{Ric}_g(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$.
[/step]
[step:Derive $CD(K,n)$ from the Ricci lower bound]
Assume that
\begin{align*}
\operatorname{Ric}_g(v,v)\ge K g_p(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$. By the identity $\operatorname{Ric}_{0,n}=\operatorname{Ric}_g$, this is exactly
\begin{align*}
\operatorname{Ric}_{0,n}(v,v)\ge K g_p(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$.
The hypotheses of the smooth weighted equivalence theorem are satisfied: $(M,g)$ is connected, complete, smooth, and $n$-dimensional; the weight $V:M\to\mathbb R$ is smooth; and the measure $e^{-V}\operatorname{vol}_g$ is precisely $\operatorname{vol}_g$. Therefore the theorem gives that
\begin{align*}
(M,d_g,\operatorname{vol}_g)
\end{align*}
satisfies the Lott--Sturm--Villani curvature-dimension condition $CD(K,n)$ with distortion coefficients $\tau_{K,n}^{(t)}$.
[/step]
[step:Recover the Ricci lower bound from $CD(K,n)$]
Conversely, assume that the metric-measure space
\begin{align*}
(M,d_g,\operatorname{vol}_g)
\end{align*}
satisfies the Lott--Sturm--Villani curvature-dimension condition $CD(K,n)$ with distortion coefficients $\tau_{K,n}^{(t)}$.
Using the same smooth weight $V:M\to\mathbb R$, $p\mapsto 0$, we have
\begin{align*}
\operatorname{vol}_g=e^{-V}\operatorname{vol}_g.
\end{align*}
Applying the reverse implication of the smooth weighted equivalence theorem with $N=n$ gives
\begin{align*}
\operatorname{Ric}_{0,n}(v,v)\ge K g_p(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$. Since $\operatorname{Ric}_{0,n}=\operatorname{Ric}_g$, this becomes
\begin{align*}
\operatorname{Ric}_g(v,v)\ge K g_p(v,v)
\end{align*}
for every $p\in M$ and every $v\in T_pM$.
[/step]
[step:Conclude the equivalence]
The previous two steps prove both implications:
\begin{align*}
\operatorname{Ric}_g\ge Kg
\end{align*}
as quadratic forms on $TM$ implies $CD(K,n)$ for $(M,d_g,\operatorname{vol}_g)$, and $CD(K,n)$ for $(M,d_g,\operatorname{vol}_g)$ implies
\begin{align*}
\operatorname{Ric}_g\ge Kg.
\end{align*}
This is exactly the claimed equivalence.
[/step]
