Question about the connectivity of the ordinary quiver of a connected algebra. I am reading the book Elements of Representation Theory of Associative Algebras, Volume 1. I have some questions about  the connectivity of the ordinary quiver of a connected algebra. On page 61, Lemma 3.3, it is said that $\{x_{\alpha} + \text{rad}^2A \mid \alpha: i \to j \}$ is a basis of $e_i(\text{rad} A\backslash \text{rad}^2A) e_j$ (this follows from Lemma 3.2(b) on page 60). I think that $\{x_{\alpha} + \text{rad}^2A \mid \alpha: i \to j \}$ should be $\{x_{\alpha} + e_i( \text{rad}^2A )e_j \mid \alpha: i \to j \}$.
On line -5 of page 61, why $$x = \Sigma_{\alpha: a\to b} x_{\alpha} \lambda_{\alpha} + \cdots \quad \quad \mod e_a(\text{rad}^3 A)e_b?$$ 
Why here is $ \mod e_a(\text{rad}^3 A)e_b $ but not $\mod e_a(\text{rad}^4 A)e_b$? Thank you very much. I attached page 61 of the book. 
 A: The answers to your two questions:
(1) It is maybe a bit unprecise, but it doesn't matter whether you take residue classes with respect to $\operatorname{rad}^2 A$ or $e_i\operatorname{rad}^2Ae_j$ for arrows $i\to j$. Indeed if $x,y:i\to j$, then $x+\operatorname{rad}^2=y+\operatorname{rad}^2$ iff $x+e_i\operatorname{rad}^2e_j=y+e_i\operatorname{rad}^2e_j$ (one direction follows since one is a subset of the other, the other implication follows by multiplying with $e_i$ and $e_j$ from the right and left.
(2) For your second question, I think this is their reasoning:
\begin{align*}
x&=x'+\sum x_\alpha \lambda_\alpha=\sum x_\alpha\lambda_\alpha+\sum x_c'y_c'\\
&=\sum x_\alpha\lambda_\alpha+\sum x_\beta y_c'\lambda_\beta+\sum r_{c,2}y_c'\\
&\equiv \sum x_\alpha \lambda_\alpha+\sum x_\beta y_c'\lambda_\beta\mod\operatorname{rad}^3\\
&=\sum x_\alpha\lambda_\alpha+\sum x_\beta y_\gamma\lambda_\beta\lambda_\gamma+\sum x_\beta s_{c,2}\lambda_\beta\\
&\equiv \sum x_\alpha\lambda_\alpha+\sum x_\beta y_\gamma\lambda_\beta\lambda_\gamma\mod\operatorname{rad}^3
\end{align*}
where $r_{c,2}$ and $s_{c,2}$ are elements of $\operatorname{rad}^3 A$.
