The matrix ring $k^{n\times n}$ can be realized in many ways as a quotient of a path algebra: For example choose the quiver $1\leftrightarrows 2 \leftrightarrows \cdots \leftrightarrows n-1\leftrightarrows n$ and impose the relations $i\to i\pm 1\to i = e_i$ where $e_i$ is the idempotent associated to the vertex $i$. So in this sense loops are trivial. The isomorphism is given by $e_i\mapsto E_{ii}$ and $(i\to i\pm 1) \mapsto E_{i,i\pm 1}$.

In fact one can show that every path algebra over a strongly connected finite quiver has a matrix ring as a quotient: Factoring out all "loops" (i.e. imposing the relation $i_1\to i_2 \to \ldots \to i_k \to i_1=e_{i_1}$ for every closed loop in the quiver) gives a matrix algebra where the vertex idempotents become the diagonal idempotents in the matrix ring.

It is not even necessary to quotient out all loops as can be seen be the example with the linear quiver. One has to quotient out just enough loops that all other loops lie in the ideal generated by that.

This feels like a homotopy condition: Consider the quiver as one dimensional CW-complex and glue in a 2-cell to some loops you want to get rid of. If the resulting complex is contractible, the corresponding path algebra should be a matrix ring.

Now my questions:

• This naive form of the statement seems not to be true because it does not take into account the orientations of the loops. Is there a way to make this homotopy feeling precise? (If the quiver is symmetric in the sense that edges come in pairs of opposite orientation then I think I have proved the statement.)
• Is there a more general version of this line of thinking? Maybe some result that says that if two quivers are homotopy equivalent and the relations are in compatible in some sense then the corresponding path algebra quotients are morita equivalent?