Let $A$ and $B$ be Banach algebras and $\varphi$ a continuous epimorphism from $A$ onto $B$. Suppose $D:B\rightarrow X$ is a continuous derivation where $X$ is both a Banach B-bimodule and Banach A-bimodule. That is for $m,n\in B$, we have that $D(mn)=D(m)n+mD(n).$
What happens to the composition $D\circ\varphi$? Will it be a continuous derivation also?
I understand that the composition will definitely be continuous since the composition of two continuous maps is continuous. Although I perceive that the composition $D\circ \varphi$ will be a derivation but I cant show how.
I know that for $a,b\in A$ \begin{eqnarray*} (D\circ\varphi)(ab)&=&D(\varphi(a)\varphi(b))\\&=&D(\varphi(a))\varphi(b)+\varphi(a)D(\varphi(b))\\&=&(D\circ\varphi)(a)b+a(D\circ\varphi)(b). \end{eqnarray*}
but my question is this: what makes $\varphi(b)=b\,\,\,$ and $\varphi(a)=a\,\,\,$ from the last two lines above.
Thanks.
Your epimorphism $\varphi: A \to B$ gives $B$ a structure of $A$-algebra whose scalar multiplication is $$\begin{matrix} A \times B & \longrightarrow & B \\ (a,b) & \longmapsto & a \cdot b &= \varphi(a) b \end{matrix}$$
In the hypothesis that $X$ is a $A$-$B$-bimodule, left scalar multiplication by $A$ and left scalar multiplication by $B$ are compatible, i.e. $$\begin{matrix} A \times X & \longrightarrow & X \\ (a,x) & \longmapsto & a \cdot_A x &= \varphi(a) \cdot_B x \end{matrix}$$