unital $*$-homomorphism $ \mathcal{O}_\infty \to ( \mathcal{O}_\infty)_{\omega}\cap \mathcal{O}_\infty'$ Consider the Cuntz algebra $\mathcal{O}_\infty$. To prove that this $C^*$-algebra is strongly selfabsorbing, one way is to construct a unital $*$-homomorphism $ \mathcal{O}_\infty \to ( \mathcal{O}_\infty)_{\omega}\cap  \mathcal{O}_\infty'$ for some free ultra filter $\omega$. Howeverm I don't know how to construct it and I'm interested in how to do this. ($( \mathcal{O}_\infty)_{\omega}$ means the ultrapower $C^*$-algebra of $\mathcal{O}_\infty$, and $\mathcal{O}_\infty'$ is the commutant of $\mathcal{O}_\infty$).
For $\mathcal{O}_2$, one can use that there is a sequence of unital *-endomorphisms $\rho_n: \mathcal{O}_2\to  \mathcal{O}_2$ with $\|a\rho_n(b)-\rho_n(b)a\|\to 0$ for $n\to\infty$, for all $a,b\in \mathcal{O}_2$. Then you can define a unital $*$-homomorphism $\rho: \mathcal{O}_2 \to ( \mathcal{O}_2)_{\omega}\cap  \mathcal{O}_2'$ as follows $$\rho(d)=[(\rho_n(d))_{n\in\mathbb{N}}].$$
Is there an analogous result as for $\mathcal{O}_2$ with a sequence $\rho_n: \mathcal{O}_\infty\to  \mathcal{O}_\infty$ with $\|a\rho_n(b)-\rho_n(b)a\|\to 0$ for $n\to\infty$? 
If not, how to construct a unital $*$-homomorphism $ \mathcal{O}_\infty \to ( \mathcal{O}_\infty)_{\omega}\cap  \mathcal{O}_\infty'$ for some free ultra filter $\omega$?
 A: All references made belong to Rørdam's book Classification of Nuclear C-algebras. $\DeclareMathOperator{\O}{\mathcal O_\infty}$
Define $A := (\O)_\omega \cap \O'$, where $\omega$ is a fixed free ultrafilter on $\mathbb N$.
We want to apply 7.2.2. to show that $\O \otimes \O \cong \O$. 
I: Show that there exists a unital embedding $\O \hookrightarrow A$.
Proof: Since $\O$ is purely infinite we know by 7.1.1. that $A$ is purely infinite and by 4.2.3. we thus get a unital embedding $\O \hookrightarrow A$.
II: Show that $\O$ has approximate inner half flip
Proof: By 7.2.5. (and 4.1.10.) we see that the $*$-homomorphisms  $\O \to \O \otimes \O$ given by $x \mapsto x \otimes 1$ and $x \mapsto 1 \otimes x$ are approximately unitarily equivalent. 

Now, we know that
$$
 \bigotimes_{i=1}^n \O \cong \O \qquad (n \in \mathbb N).
$$
Applying the same argument as in 5.1.2. one has that 
$$
 \bigotimes_{i=1}^\infty \O \cong \O.
$$
By 1.10. of the paper Strongly self-absorbing $C^*$-algebras by Toms and Winter, you get that $\O$ is strongly self-absorbing.
