Why is this a versal extension? On page 10 of Sernesi's book "Deformations of Algebraic Schemes", he claims that given an $A$-algebra $R$ such that $$R = \frac{P}{I} = \frac{A[x_1,\ldots,x_n]}{I}$$ the following short exact sequence is a versal extension $$ 0 \to \frac{I}{I^2} \to \frac{P}{I^2} \to R \to 0 $$ meaning, given any square-zero extension $$0\to J \to R' \to R \to 0$$ there is a morphism from the first short exact sequence to the second. Why is this true?

I presume $R'$ is also an $A$-algebra. Since $P$ is a polynomial ring over $A$, the projection map $P\to R$ factors through $R'\to R$ via an algebra map $\phi:P\to R'$ (just pick suitable images for $x_1,\ldots,x_n$). All one has to do is to show that $\phi(I^2)={0}$ in $R'$, for then one gets a map $P/I^2\to R'$, etc. Let $f:R'\to R$ denote the map from the second exact sequence. If $a$, $b\in I$ then $f(\phi(a))=0$ and $f(\phi(b))=0$. Then $\phi(a)$, $\phi(b)\in J$ so $\phi(ab)=0$. This does it.