# Injective tensor product (Ryan 3.3)

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