I'm currently reading proof of Corollary $2.7$ from the paper The ideal structure of the Haagerup tensor product of $C^{\ast}$-algebras. Let $X_1$ and $X_2$ be operator spaces and $Y_i \subseteq X_i (i=1, 2)$ be operator spaces. Let $\pi_i : X_i \to X_i/Y_i$ be quotient maps of operator spaces and $E$ is a closed subspace of $X_1 \otimes^h X_2$ containing $X_1 \otimes^h Y_2+ Y_1 \otimes^h X_2$ then in the linked paper following is stated without proof.
$(\pi_1 \otimes \pi_2)^{-1}((\pi_1 \otimes \pi_2)(E))=E$
I know that $\pi_1 \otimes \pi_2$ is a quotient map. Also, It is obvious that $E \subseteq (\pi_1 \otimes \pi_2)^{-1}((\pi_1 \otimes \pi_2)(E))$. Can someone explain the reverse inclusion?
As stated very clearly in the paper, this uses Corollary 2.6, which says that
$$ (\pi_1\otimes\pi_2)^{-1}(0)=X_1 \otimes^h Y_2+ Y_1 \otimes^h X_2. $$ So if $x\in E$ and $(\pi_1\otimes\pi_2)(x)=(\pi_1\otimes\pi_2)(z)$, then $$x-z\in(\pi_1\otimes\pi_2)^{-1}(0)=X_1 \otimes^h Y_2+ Y_1 \otimes^h X_2\subset E.$$ So $z\in E$.