In quiver rep,is it$\mathrm{Ext}^i_{\mathrm{rep}}(\mathcal{X},\mathcal{R})=0 ‎\leftrightarrow‎ \forall v \mathrm{Ext}^i_R(\mathcal{X}_v,R)=0$?

Question

Let $\mathcal{Q}$ be a finite acyclic quiver, and $R$ be a ring Let $\mathcal{X}$ be a representation in $\mathrm{Rep}(\mathcal{Q},R)$. Let $\mathcal{R}$ represent the image of $R\mathcal{Q}$ under the equivalence $\mathrm{Mod}-R\mathcal{Q} \cong \mathrm{rep}(\mathcal{Q},R)$. How can I prove that for every $i>0$, $\mathrm{Ext}^i_{\mathrm{rep}}(\mathcal{X},\mathcal{R})=0$ iff for every vertex $v$ we have $\mathrm{Ext}^i_{R}(\mathcal{X}_v,R)=0$, where $\mathcal{X}_v$ is the $R$-module associated with the vertex $v$?

Answer 1

This statement is false. Take $R=k$ a field, and $\mathcal{X}$ any finite-dimensional representation of $\mathcal{Q}$ over $k$. Then for $i>0$ and $v\in\mathcal{Q}_0$, there is some $n\geq0$ such that

$$\operatorname{Ext}^i_R(\mathcal{X}_v,R)=\operatorname{Ext}^i_k(k^n,k)=0$$

since $k$ is a field. However, if $\mathcal{X}$ is not projective, one has $\operatorname{Ext}^1_{\text{rep}}(\mathcal{X},\mathcal{R})\ne0$, since $\mathcal{R}=k\mathcal{Q}$ is a hereditary algebra. See for example the simple module $S_1$ for the quiver $1\to 2$, which has an extension on top of the summand $S_2$ of $\mathcal{R}=k\mathcal{Q}$ when $k$ is a field.