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HHH
HHH
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Let $M$ be a representation of a quiver $Q=(V, E)$ by $R$-modules. By $M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\to M^{++}$ of representations of $Q$. Is it a pure monomorphism? That is, for any finitely presented representation $F$, can we prove that $\mathrm{Hom}(F, M^{++})\to \mathrm{Hom}(F, M)$$ \mathrm{Hom}(F, M^{++})\rightarrow {\rm Hom}(F, \frac{M^{++}}{M})$ is a pure epimorphisman pimorphism?

Let $M$ be a representation of a quiver $Q=(V, E)$ by $R$-modules. By $M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\to M^{++}$ of representations of $Q$. Is it a pure monomorphism? That is, for any finitely presented representation $F$, can we prove that $\mathrm{Hom}(F, M^{++})\to \mathrm{Hom}(F, M)$ is a pure epimorphism?

Let $M$ be a representation of a quiver $Q=(V, E)$ by $R$-modules. By $M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\to M^{++}$ of representations of $Q$. Is it a pure monomorphism? That is, for any finitely presented representation $F$, can we prove that $ \mathrm{Hom}(F, M^{++})\rightarrow {\rm Hom}(F, \frac{M^{++}}{M})$ is an pimorphism?

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Andrej Bauer
Andrej Bauer
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Let $M$ be a representation of a quiver Q=(V, E)$Q=(V, E)$ by R$R$-modules. By $M^+$$M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)={\rm Hom}(M(v), \frac{Q}{Z})$$M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\rightarrow M^{++}$$M\to M^{++}$ of representations of $Q$. Is it a pure monomorphism? i.e.That is, for any finitely presented representation $F$, can we prove that ${\rm Hom}(F, M^{++})\rightarrow {\rm Hom}(F, M)$$\mathrm{Hom}(F, M^{++})\to \mathrm{Hom}(F, M)$ is a pure epimorphism?

Let $M$ be a representation of a quiver Q=(V, E) by R-modules. By $M^+$ we mean a representation of $Q^{op}$ with $M^{+}(v)={\rm Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\rightarrow M^{++}$ of representations of $Q$. Is it a pure monomorphism? i.e. for any finitely presented representation $F$, can we prove that ${\rm Hom}(F, M^{++})\rightarrow {\rm Hom}(F, M)$ is a pure epimorphism?

Let $M$ be a representation of a quiver $Q=(V, E)$ by $R$-modules. By $M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\to M^{++}$ of representations of $Q$. Is it a pure monomorphism? That is, for any finitely presented representation $F$, can we prove that $\mathrm{Hom}(F, M^{++})\to \mathrm{Hom}(F, M)$ is a pure epimorphism?

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HHH
HHH
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Is this a pure monomorphism?

Let $M$ be a representation of a quiver Q=(V, E) by R-modules. By $M^+$ we mean a representation of $Q^{op}$ with $M^{+}(v)={\rm Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is anatural monomorphism $M\rightarrow M^{++}$ of representations of $Q$. Is it a pure monomorphism? i.e. for any finitely presented representation $F$, can we prove that ${\rm Hom}(F, M^{++})\rightarrow {\rm Hom}(F, M)$ is a pure epimorphism?