Let me recall the following fact:

If $A$ is a $C^*$-algebra and $\pi: A \to \mathcal{B}(\mathcal{H})$ is a faithful non-degenerate representation, then we can explicitely realise the multiplier algebra $M(A)$ as bounded operators on $H$. More precisely, $M(A) \cong M_\pi(A)$ where $$M_\pi(A) = \{T \in \mathcal{B}(\mathcal{H})\mid T\pi(A) \subseteq \pi(A), \ \pi(A) T \subseteq \pi(A)\}.$$
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Let $A$ be a $C^*$-algebra and $\mathcal{H}$ be a Hilbert space. The canonical $*$-morphism
$$\mathcal{B}_0(\mathcal{H}) \otimes A \to \mathcal{B}_0(\mathcal{H}) \otimes A \otimes A: x \otimes a \mapsto x \otimes a \otimes 1$$
is non-degenerate, so extends uniquely to a strict $*$-morphism between the multiplier algebras. Let $\pi_u$ be a faithful representation of $A$ on the Hilbert space $\mathcal{K}$. Let $\text{id}$ be the identity representation of $\mathcal{B}_0(\mathcal{H})$. I want to show that the following diagram commutes:


[![enter image description here][1]][1]

I can see that we have two maps
$$\mathcal{M}(\mathcal{B}_0(\mathcal{H}\otimes A) \to \mathcal{M}(\mathcal{B}_0(\mathcal{H}\otimes A \otimes A)$$
such that $\mathcal{B}_0(\mathcal{H})\otimes A \ni x \otimes a \mapsto x \otimes a \otimes 1$. Somehow I want to invoke uniqueness of the extension that these maps must agree everywhere but for that I need to show that the composition along the two isomorphisms and the map $T \mapsto T \otimes 1$ (first down, then go right, then go up) is strictly continuous, and I can't see why this holds.

**Question**: What is the quickest way to see that this diagram commutes?

  [1]: https://i.sstatic.net/F6cWt.png