intersection of a multiplier algebra with a commutant of a $C^*$-algebra

I have a question about multiplier algebras and commutants of $C^*$-algebras in general.

First of all, the question is related to this structure theorem about completely positive order zero maps (you can find the theorem in Completely positive maps of order zero", by Winter and zacharias, Theorem 2.3 ):

Theorem: Let $A$ and $B$ $C^*-$algebras and $\phi:A\to B$ be a completely postive map of order zero. Set $C:=C^*(\Phi(A))\subset B$.
Then there is a positive element $h\in M(C)\cap C'$ with $\|h\|=\|\phi\|$ and a $*-$homomorphism $$\pi_{\phi}:A\to M(C)\cap \{h\}'$$ such that $$\phi(a)=\pi_{\phi}(a)h$$ for $a\in A$.

If $A$ is unital, then $\phi(1_A)=h\in C$.

Ma Problem: I have problems to understand why the intersection $M(C)\cap C'$, similar $M(C)\cap \{h\}'$, does make sense.

$M(C)$ is the multiplier algebra of $C$ and the elements are maps $m:C\to C$ such that there is a map $m^*:C\to C$ with $(m(a))^*b=a^*m^*(b)$ for all $a,b\in C$, so called multipliers of $C$. $C'$ is the commutant of $C$: $C'=\{y\in L(H); xy=xy\; \text{ for all$x\in$C}\}$, this means that this is a subset of $L(H)$ for a suitable Hilbert space $H$.

Why it is possible to intersect this sets? They seem very different. And is $H$ here the universal Hilbert space with $C$ acts nondegenerately on $H$? I appreciate your help. Regards

If you represent $C$ in $B (H)$ faithfully and nondegenerately, then it is well-known that $$M (C)\simeq\{x\in B (H):\ xc\in C,\ cx\in C,\ \forall c\in C\}.$$
• $C$ should be represented faithfully and nondegenerately for this to hold. – Mike F Oct 5 '15 at 16:52
• Indeed. $\ \ \$ – Martin Argerami Oct 5 '15 at 17:47