Path algebras are formally smooth

Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.

If $B$ is an algebra with a nilpotent ideal $I$, we need to check that every homomorphism $\varphi:kQ\to B/I$ lifts to a homomorphism $\tilde{\varphi}:kQ\to B$.

Let $\{e_1,\dots,e_n\}$ be the idempotents in $kQ$ corresponding to the vertices of $Q$. Then $\{\varphi(e_1),\dots,\varphi(e_n)\}$ is a set of orthogonal idempotents in $B/I$ whose sum is $1$.

These lift to a set $\{f_1,\dots,f_n\}$ of orthogonal idempotents in $B$ whose sum is $1$.

For each arrow $a_s$ of $Q$ from vertex $i$ to vertex $j$, pick an arbitrary lift $\widetilde{\varphi(a_s)}$ of $\varphi(a_s)$, and let $b_s=f_i\widetilde{\varphi(a_s)}f_j$, so $b_s$ is also a lift of $\varphi(a_s)$.

Then there is a unique map $\tilde{\varphi}:kQ\to B$ with $\tilde{\varphi}(e_i)=f_i$ and $\tilde{\varphi}(a_s)=b_s$ for all $i$ and $s$, and this is a lift of $\varphi$.

This doesn't work for quivers with relations. For example, if $Q$ is an acyclic quiver and $A=kQ/I$ with $I$ a nonzero admissible ideal, then the identity map $A\to kQ/I$ does not lift to a homomorphism $A\to kQ$.

Let me answer this for acyclic quivers, if there are cycles it should be the same, but to be on the safe side let me not claim it in that generality.

The path algebra is hereditary (i.e. of global dimension 1), from which it follows that the Hochschild cohomology vanishes above the global dimension. Hence $\operatorname{HH}^2(A,M)=0$.

If there are relations, the global dimension jump ups.