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Let $\mathcal{A}$ be a commutative weakly amenable Banach algebra and $\mathcal{B}$ be a Banach algebra, let $\theta:\mathcal{A} \to \mathcal{B}$ be a continuous homomorphism with dense range; then it is known that $\mathcal{B}$ is weakly amenable (and commutative).

Now my question

Let $\mathcal{A}$ be a weakly amenable Banach algebra and $\mathcal{B}$ be a Banach algebra, let $\theta:\mathcal{A} \to \mathcal{B}$ be a continuous homomorphism with the dense range. Is $\mathcal{B}$ a weakly amenable Banach algebra?

I am grateful for any help.

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  • $\begingroup$ Didn't we have some version of this question on MO before? In any case, the answer is no: you can have noncommutative WA Banach algebras which quotient onto non-WA Banach algebras $\endgroup$
    – Yemon Choi
    Commented May 28, 2019 at 18:32
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    $\begingroup$ However, this is not an obvious fact unless one knows the right sources of examples - I will have to refresh my memory and get back to you. I think that one source of counter-examples arises by taking $A= E\hat\otimes E^*$ for certain Banach spaces $E$, with multiplication defined by $(a\otimes\phi)\cdot (b\otimes\psi)= \phi(b) a\otimes\psi$, but I must confess I've forgotten the details at the moment $\endgroup$
    – Yemon Choi
    Commented May 28, 2019 at 18:37
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    $\begingroup$ Ah yes, this (very natural and good!) question was asked independently mathoverflow.net/questions/260170/… $\endgroup$
    – Yemon Choi
    Commented May 28, 2019 at 18:38
  • $\begingroup$ @Yemon Choi thank you $\endgroup$
    – user62498
    Commented May 28, 2019 at 18:50

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