In corollary 2 of the article "From triangulated categories to cluster algebras II" by Caldero and Keller, it is stated that a seed $(u,Q)$ (where $u$ is a cluster and $Q$ a quiver) of an acyclic cluster algebra $\mathcal{C}$ is determined by $u$. (i.e., two different clusters can't be associated with the same quiver in the exchange graph.) The authors say that this is an immediate consequence of the correspondence (Theorem 4 of the same article) between cluster variables/clusters in the cluster algebra and indecomposable objects/cluster-tilting objects of the cluster category.
I assume this may follow from the fact that cluster-tilted algebras $A_1=\text{End}_{\mathcal{C}}(T_1)$, $A_2=\text{End}_{\mathcal{C}}(T_1)$ derived from two different cluster-tilting objects $T_1,T_2$ in the cluster category, will have two different associated quivers $Q_{A_1}$, $Q_{A_2}$. Is this statement true? and if so, how can it be shown? If it is not, then why is the above corollary an immediate consequence of the theorem?
This is not the correct implication: for example, taking $T_2$ to be the shift of $T_1$ will make $A_1$ and $A_2$ isomorphic, even though $T_1$ and $T_2$ will be different.
Instead, the reasoning is as follows: let $(u,Q)$ be a seed, with $u = (u_1, \ldots, u_n)$ a cluster. Then by Theorem 4, for each $i=1,\ldots, n$, there is a unique object $M_i$ without self-extension in the cluster category (up to isomorphism) such that $X_{M_i} = u_i$. The quiver $Q$ is then the Gabriel quiver of the endomorphism algebra of $M_1 \oplus \cdots \oplus M_n$. Thus $Q$ is determined by $u$.