# Opposite and connected quivers problem

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 $m \geq 2$?

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What does is mean for a $K$-algebra to be connected? – Rasmus Oct 11 '11 at 19:55
A $K$-algebra $A$ is connected if $A$ is not a direct product of two algebras. It can be shown this is equivalent to saying that $0$ and $1$ are the only central idempotents of $A$. – user6495 Oct 11 '11 at 19:57

2. If the quiver is not acyclic, your algebra will be infinite dimensional. But as you point out, there is no problem with that. The aim of the book is not to provide the greatest generality. Your proof obviously also works for $R^m$ since this is also an admissible ideal.
Thanks, for $1$, is the map I wrote correct? – user6495 Oct 12 '11 at 8:56
It is correct if you normally write $\alpha\beta$ in this case for the concatenation of arrows. This should also be the convention of Assem, Simson, Skowronski. But there is also the differing convention of writing arrows as maps from right to left. – Julian Kuelshammer Oct 12 '11 at 11:23