Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)\cong \mathbb{K})$ of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.

Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)\cong \mathbb{K})$ which is a general representation of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.

Edit: The answer of @wcb is a perfectly good counter-example of the question that I had asked before editing my question. I apologize for not posing my question properly.

Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)= \mathbb{K})$ of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.

Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)\cong \mathbb{K})$ of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.

Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

My question is:

If I consider quivers with relations, but without oriented cycles, is the forward direction of the above result still true? Assuming $\beta\not=0, \alpha$

Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)= \mathbb{K})$ of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.