Let $A$ be ring, then $A[x]$ is formally smooth over $A$. I'm having trouble with the following example of the polynomial ring $A[x]$ for some ring $A$ being formally smooth:

Let $h \colon C \to C /J$ with $J \subset C$ an ideal such that $J^2 = 0$. Given the commutative diagram of rings,
\begin{array}{ccc}
A & \xrightarrow f & C \\
 \downarrow  g & & \downarrow h \\
 A[x] & \xrightarrow {\overline{f}}  & C/J
\end{array}
choose $c_i \in C$ such that $h(c_i) = \overline{f}(x_i)$. Define $F \colon A [x] \to C$ as the $A$-algebra map $x_i \mapsto c_i$. Then $F \circ g = f$ and $h \circ F = \overline{f}$.

My question/trouble is that I do not see how we're using $J$.

$A[x]$ satisfies a much stronger universal property, such that $J^2=0$ is not needed. In fact, all we need is $C\rightarrow C/J$ is surjective so that the image of $x$ in $C/J$ has a preimage in $C$.
