Let $Q :Z \rightarrow Y$ be a quotient operator and let $I$ be the identity operator on X.Show that the tensor product operator $I \otimes Q : X \hat{{\otimes}}_\epsilon Z \rightarrow X \hat{{\otimes}}_\epsilon Y$ is a quotient operator if the following "lifting" condition is satisfied: for every operator $S: X^* \rightarrow Y$ and every $\epsilon >0$, there exists an operator $T :X^* \rightarrow Z$ such that $QT = S$ and $||T|| \leq ||S|| + \epsilon$

My try:

Let $Q : Z \longrightarrow Y$ is a quotient operator implies $Q$ is surjective and $||y|| = inf \{||z||: z \in Z , Q(z)= y\}$ {definition on Page 18 of Ryan}

$I \otimes Q : X \hat{{\otimes}}_\epsilon Z \longrightarrow X \hat{{\otimes}}_\epsilon Y$.

$ I \otimes Q$ is clearly surjective.

let $u = \sum_{i=1}^{n} x_i\otimes y_i$

Claim: $\epsilon(u) = inf \{\epsilon(w) : w \in X \hat{{\otimes}}_\epsilon Z \ and\ (I\otimes Q)(w) = u\}$ $ = inf \{\sup_{\phi \in B_{X*}}||\sum_{i=1}^{n} \phi(x_i) z_i|| : z_i \in Z \ and \ Q(z_i) =y \}$

Define, $S_u : X* \longrightarrow Y$ as $S_u(\phi) = \sum_{i=1}^{n} \phi (x_i) y_i$ $\Rightarrow \exists$ $T_u : X* \longrightarrow Z$ such that $QT_u = S_u$ and $||T_u|| \leq ||S_u|| + \epsilon$. $\epsilon(u) = sup_{\phi \in B_{X*}} ||\sum_{i=1}^{n} \phi(x_i) y_i||_Y$ =$\sup_{\phi \in B_{X*}}||S_u(\phi)||_Y$ =$||S_u||$ = $||QT_u||$ \ = $\sup_{\phi \in B_{X*}}||QT_u(\phi)||$ = $\sup_{\phi \in B_{X*}} inf{||z|| : z \in Z , Qz = QT_u (\phi)}$


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