In Ginzburg's notes [Lectures on Noncommutative Geometry](https://arxiv.org/pdf/math/0506603.pdf), he claim that the path algebra of a quiver is formally smooth (See Section 19.2). 


I have two questions. First, how to show this claim and which criterion of formal smoothness we are going to check? Secondly, does this claim hold for the path algebra of a quiver with relations as well?


Any help or hints would be very appreciated.



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 $A\to kQ$.


