I'm studying the book "quiver representations and quiver varieties" of Kirillov and I'm in Theorem 3.10:
The functor $\Phi_i^+$ is left exact. Moreover, $R^n\Phi_i^+(V)=0$ for all $n>1$, and $R^1\Phi_i^+(V)=0$ iff $V$ satisfies the following condition: \begin{align} \label{eq1} \left( \bigoplus_{h:k\rightarrow i} V_k\right) \rightarrow V_i \qquad \text{is surjective.} \end{align}
The left exact part I got it, on the next part, I wrote the injection resolution of a representation $V$:
So $I_0$ and $I_1$ are injective representations, I know that $I_0$ is the injective envelope for the representation $V$. And I know that $R^1\Phi_i^+(V)=\frac{\Phi_i^+(I_1)}{\text{Im}\Phi_i^+(d_0)}$, in the positions $j\neq i$ $(d_0)_j$ is surjective, so $(R^1\Phi_i^+(V))_j=0$ but in the position $i$ I can get the commutative diagram with exact lines and columns
Here $\overline{g}=\Phi_i^+(d_0)_i$, $Ker(\phi)=(\Phi_i^+(V))_i$, $Ker(\psi)=(\Phi_i^+(I_0))_i$ and $Ker(\eta)=(\Phi_i^+(I_1))_i$.
What I need to prove is that $\psi$ is surjective, then I can use the snake lemma and have a morphism between $V_i$ and $Coker(\overline{g})$ so then $Coker(\overline{g})=(R^1\Phi_i^+(V))_i$ and I have that $R^1\Phi_i^+(V)=0$ iff $\phi$ is surjective.
If someone can help me, or give me some other ideia for this prove, I will appreciate.
By assumption, $i$ is a sink. Note that the desired statement (that $\psi$ is surjective) is false if $i$ is an isolated vertex.
If $i$ does admit at least one incoming arrow, then the statement follows by considering each indecomposable injective summand of $I_0$. For each indecomposable injective module $I$, it's straightforward to check that $\psi:\bigoplus I_j \rightarrow I_i$ is surjective. That's because $I_i$ is actually only non-zero if $I$ is the indeomposable injective corresponding to vertex $i$, and in that case $\psi$ is clearly surjective.