When is a $*$-homomorphism between multiplier algebras strictly continuous? The strict topology on the multiplier algebra $M(A)$ of a C*-algebra $A$ is that generated by the seminorms
$$ x\mapsto \| ax \|\qquad x\mapsto\| xa \| \qquad (x\in M(A), a\in A) $$
Whereas a $*$-homomorphism $\phi : M(A)\to M(B)$ between two multiplier algebras is necessarily norm-continuous, if I understand things correctly it will not always be continuous with respect to the strict topologies on either side.  Does anyone have a good reference for this?
On the other hand an easily-proven theorem states that $\phi$ is strictly continuous if the image of $\phi$ contains $B$.  This is not necessary, however; take $\phi : \mathcal{B}(\ell^2)\to \mathcal{B}(\ell^2)$ to be the map $x\mapsto sxs^*$ where $s$ is the unilateral shift.  This is strictly continuous even though its image doesn't contain $\mathcal{K}(\ell^2)$.  Are there other conditions which guarantee $\phi$ to be strictly continuous?
I'm particularly interested in the case where $\phi$ maps $A$ into $B$, and both are nonunital.  Is this enough to show that $\phi$ is strictly continuous?
 A: I am posting here an answer to the last question I had, since that did not come up at the MO question.  (Readers should look there for the answers to the other questions, though.)  It is essentially a slight modification of a comment by Farah in his book "Analytic Quotients."  (See Example 3.2.3.)
Recall that $\ell^\infty$ is the multiplier algebra of $c_0$.  I'll construct a $*$-homomorphism $\phi : \ell^\infty \to \ell^\infty$ which sends $c_0$ into $c_0$ but is not strictly continuous.
Let $\mathcal{U}_n$, $n\in\mathbb{N}$, be a sequence of nonprincipal ultrafilters.  Consider the $*$-homomorphism $\phi : \ell^\infty \to \ell^\infty$ defined by
$$\phi(x)(n) = \lim_{k\to\mathcal{U}_n} x(k)$$
Since each $\mathcal{U}_n$ is nonprincipal, it follows that $\phi$ takes $c_0$ into $c_0$; indeed, $\phi$ takes $c_0$ to $0$!  But it is obvious that $\phi$ is not strictly continuous; to see this, let $e_k\in\ell^\infty$ be the sequence which begins with $k$-many $1$'s and ends with $0$'s.  Then $e_k$ converges in the strict topology to the sequence $e$ with constant value $1$.  However, the images $\phi(e_k)$ are all just the constant sequence with value $0$, whereas $\phi(e) = e$.
