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Let $Q$ be a finite quiver and denote the stationary parts of $Q$ by $e_{i}$. Suppose we have two arrows $f,g$ such that their composition $f \circ g$ is equal to $e_{i}$. Does this always implies that $f=g=e_{i}$?

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When you say arrow, I assume $f,g\in Q_1$. Then the answer is no. Note that the trivial path (which I assume here is the stationary path you wrote) is defined to have path length zero, and we usually do not think of $e_i$ as elements in $Q_1$, they really are elements of $Q_0$.

Be careful if you are asking a question about representations of path algebra of quiver, then $fg=e_i$ is not in any admissible ideal since path length of $e_i$ is zero. The path algebra of a quiver arise from $kQ/I$ where $I$ needs to be admissible, i.e. contains elements of path length $\geq 2$. Representation-theoretically, $fg=e_i$ means you go two radical layer down the projective (indecomposable) module corresponding to $i$ and you ended up in the simple top, this does not make any sense (this is also the reason why path algebra are defined using admissible ideals).

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