I am working through the book Quiver Representations by Schiffler and I am confused about some notation regarding pullbacks. The following is example 2.13 (pp. 65):
Let $k$ be a field and $Q$ the quiver $1==>2$ with two arrows $\alpha$ and $\beta$ going from 1 to 2. Let $g\colon E\to S(1)$ and $g'\colon E'\to S(1)$ be morphisms of representations where $S(1)$ is the simple representation $k==>0$ (both arrows are $0$), $E\colon E_1==>E_2$ is the representation $k==>k$ with top arrow $\varphi_\alpha =1$ and bottom arrow $\varphi_\beta =0$, and $E'\colon E_1'==>E_2'$ is the representation $k==>k$ with top arrow $\varphi'_\alpha =0$ and bottom arrow $\varphi'_\beta =1$.
We also denote the elements of $E$ and $E'$ as ordered pairs $(e_1,e_2)$ and $(e_1',e_2')$.
Then the pullback with respect to $g$ and $g'$ is the set $E''=\{((e_1,e_2),(e_1,e_2'))\in E\times E'\}$ and we want to figure out what are the corresponding maps in the representation $E''\colon E_1''==>E_2''$.
This is where I get confused by the notation: the corresponding linear maps are $\varphi_\alpha''$ for the top arrow, and $\varphi_\beta''$ for the bottom arrow, and we have
$$\varphi_\alpha''((e_1,e_2),(e_1,e_2'))=(\varphi_\alpha(e_1,e_2),\varphi_\alpha'(e_1,e_2'))=((0,e_1),(0,0)),$$ $$\varphi_\beta''((e_1,e_2),(e_1,e_2'))=(\varphi_\beta(e_1,e_2),\varphi_\beta'(e_1,e_2'))=((0,0),(0,e_1)).$$
I am confused about what $\varphi_\alpha(e_1,e_2)$ means and why does it map to $(0,e_1)$, and similarly for the other three mappings? Recall that $\varphi_\alpha$ is a map from $E_1=k$ to $E_2=k$ and it doesn't make sense to me for it to take $(e_1,e_2)$ as an argument.
The only way that I can make half-sense out of this is that we have $\varphi_\alpha(e_1,e_2)=(0,e_1)$ because $e_1\in E_1$ from the first bracket gets mapped to $e_1 \in E_2$ in the second bracket (since $\varphi_\alpha=1$), and the $0$ is there because we don't have a map $E_2\to E_1$.