Here are two problems from Elements of the Representation Theory of Associative Algebras by D. Simson, et. al
(Page $65$).
$1$. Let $Q=(Q_{0},Q_{1},s,t)$ be a quiver. Prove $(KQ)^{op} \cong KQ^{op}$ where $Q^{op}$ denotes the opposite quiver and $KQ$ the path algebra.
I think that the idea is to give the isomorphism by specifying the action on the basis (i.e the paths of certain length) that is given two arrows $\alpha$ from a vertex $1$ to say $2$ and a map $\beta$ from vertex $2$ to $3$ assign the reverse walk $\beta^{-1} \alpha^{-1}$ no? my question is, how do we prove it rigorously? I'm having trouble writing it formally because I think we still have consider cases when the source and target don't coincide.
$2$. Let $Q=(Q_{0},Q_{1})$ be a finite and acyclic quiver. Prove $KQ$ is connected if and only if $KQ/R^{2}$ is connected where $R$ is the arrow ideal of $kQ$.
I think I have this one. Let me mention some results which are proved (page $47$ and $55$)
Result $1$. Let $Q$ be a finite quiver, then $KQ$ is connected if and only if $Q$ is a connected quiver.
Result $2$. Let $Q$ be a finite quiver and let $I$ be an admissible ideal of $KQ$. Then $KQ/I$ is connected if and only if $Q$ is a connected quiver.
So suppose first that $KQ$ is connected, then by by result $1$ it follows that $Q$ is a connected quiver. Now note that $R^{2}$ is an admissible ideal so by result $2$ it follows that $KQ/R^{2}$ is connected.
Now assume $KQ/R^{2}$ is connected, then by result $2$ it follows that $Q$ is a connected quiver so by result $1$ we have that $KQ$ is connected and we are done.
So two questions here:
1) Where do we need that $Q$ is acyclic don't it suffices to ask that $Q$ is finite?
2) It seems that $R^{2}$ is irrelevant, would the result still hold if we replace $R^{2}$ by $R^{m}$ for any $m \geq 2$?