Hilbert-C*-Modules and interior tensor products Let $A$ and $B$ be $C^*$-Algebras, $E$ a Hilbert-$A$-module, $F$ a Hilber-$B$-Module and 
$$\pi: A \rightarrow \mathcal{L}_B(F)$$
a $*$-homorphisms of C*-Algebras, where $\mathcal{L}_B(F)$ denotes the C*-Algebra of adjointable operators on $F$. Now, let $x \in E$ and $y \in F$ such that
$$\langle y, \pi(\langle x,x \rangle)(y) \rangle=0.$$
I want to show that for any $T \in \mathcal{L}_A(E)$ we have 
$$\langle y, \pi(\langle T(x),T(x) \rangle)(y) \rangle=0.$$
This would prove that the homomorphism $$F \otimes 1: E\otimes_\pi F \rightarrow E\otimes_\pi F $$ is well defined.
Thank you
 A: Have a look in Chapter 4 of Lance's book (Hilbert $C^*$-modules, a toolkit for operator algebraists), in particular page 42 where he shows that $T\otimes 1$ is well defined. 
Lance explains that you actually need to show something more general to get the well-definedness i.e. what you want to show is that, for any algebraic tensor $\sum_i x_i\otimes y_i$ in $E\otimes_\pi F$, $\|\sum_i Tx_i\otimes y_i\|^2\leq \|T\|^2 \|\sum_i x_i\otimes y_i\|^2$. By definition of $E\otimes_\pi F$, this amounts to showing that $$\|\sum_{i,j} \langle y_i , \pi(\langle Tx_i, Tx_j\rangle) y_j\rangle\|\leq  \| T\|^2 \|\sum_{i,j} \langle y_i , \pi(\langle x_i, x_j\rangle) y_j\rangle\| \quad\quad\qquad(1)$$ 
How to prove this? Let $X=(\langle x_i, x_j\rangle)_{i,j} \in M_n(A)$ and let $W=(\langle Tx_i, Tx_j\rangle)_{i,j} \in M_n(A)$. Lemma 4.2 (page 32/33) states that $W\leq \|T\|^2 X$. 
By complete positivity of $\pi$, we have $\pi^{(n)}(W)\leq\|T\|^2 \pi^{(n)}(X)$ and hence $0\leq \langle y,\pi^{(n)}(W)y\rangle \leq \|T\|^2\langle y,\pi^{(n)}(X)y\rangle$ where $y=(y_1,\cdots,y_n)\in F^n$- e.g. see proof of Prop 4.5 (page 40). Hence $\|\langle y,\pi^{(n)}(W)y\rangle\| \leq \|T\|^2\|\langle y,\pi^{(n)}(X)y\rangle\|$. This is exactly equivalent to equation $(1)$, so the proof is complete.
A: To answer what you asked exactly first observe that, $$\langle Tx,Tx\rangle\leq \Vert T\Vert^2 \langle x,x\rangle$$
Now use this as follows, 
\begin{align*}
\Vert Tx \otimes y\Vert^2 &= \Vert\langle  Tx \otimes y , Tx \otimes y \rangle\Vert \\
&= \Vert \langle y, \pi (\langle Tx,Tx \rangle )y\rangle \Vert \\
&\leq \Vert T\Vert^2 \Vert \langle y, \pi(\langle x,x\rangle)y\rangle\Vert \\
&= \Vert T \Vert^2 \Vert \langle x\otimes y, x\otimes y\rangle\Vert \\
&= \Vert T\Vert^2 \Vert x\otimes y\Vert^2
\end{align*}
So, $ \Vert x\otimes y\Vert=0 \Rightarrow \Vert Tx \otimes y\Vert=0$ 
