Formally smooth map

Remarks

The lift $v$ need not be unique. If the lift always exists and is unique, $A$ is formally étale over $R$; if it is always unique, $A$ is formally unramified over $R$.

Example: let $A = R[x_i : i \in I]$ be a polynomial ring in any set of variables. Given $u \colon A \to C/J$, choose for each $i$ an element $c_i \in C$ with $\pi(c_i) = u(x_i)$, and let $v$ be the $R$-algebra map with $v(x_i) = c_i$. Then $\pi \circ v$ and $u$ agree on the variables, so they are equal.

Non-example: let $k$ be a field and $A = k[x]/(x^2)$. Take $C = k[x]/(x^3)$ and $J = (x^2)/(x^3)$, so $J^2 = 0$ and $C/J = A$, and let $u$ be the identity of $A$. A lift $v$ would send $x$ to $x + a x^2$ for some $a \in k$, and then $v(x)^2 = x^2 \ne 0$ in $C$, while $v(x^2) = v(0) = 0$. So no lift exists, and $A$ is not formally smooth over $k$.