Free product of two algebras and actions of algebras. Let $A, B$ be two algebras. Suppose that $A$, $B$ acts on $V$. Then we have two maps
$$
\delta_1: A \otimes V \to V, \\
\delta_2: B \otimes V \to V,
$$
which satisfy the axioms of actions. 
Do we have the free product $A * B$ of $A, B$ acts on $V$? 
It seems that the $\delta_1 \circ (1 \otimes \delta_2): A \otimes B \otimes V \to V$ is not an action. That is, it is possible that $\delta_1 \circ (1 \otimes \delta_2)$ does not satisfy the axioms of actions.
Thank you very much.
 A: An action of $A$ on $V$ is a simply a homomorphism $A \to \operatorname{End}(V)$, and similarly for $B$. By the universal property of the free product, the homomorphisms $A, B\to \operatorname{End}(V)$ lift (uniquely) to the required action $A*B \to \operatorname{End}(V)$.
A: If we have $a_1, a_2 \in A$, $b_1, b_2 \in B$, then
\begin{align}
& ((a_1 \otimes b_1)(a_2 \otimes b_2))(v) \\
& = (a_1a_2 \otimes b_1 b_2)(v) \\
& = a_1(a_2(b_1(b_2(v)))). \\
& (a_1 \otimes b_1)((a_2 \otimes b_2)(v)) \\
& =  a_1(b_1(a_2(b_2(v)))).
\end{align}
If the actions of $A, B$ do not commute, then $\delta_1 \circ (1 \otimes \delta_2)$ is not a coaction.
For $a_1 b_1 a_2 b_2 \cdots a_m b_m, a_1' b_1' a_2' b_2' \cdots a_{m'}' b_{m'}' \in A*B$, we have
\begin{align}
& ((a_1 b_1 a_2 b_2 \cdots a_m b_m )*(a_1' b_1' a_2' b_2' \cdots a_{m'}' b_{m'}'))(v) \\
& = a_1 (b_1(a_2 (b_2 \cdots (a_m (b_m (a_1' (b_1' (a_2' (b_2' \cdots (a_{m'}' (b_{m'}'(v)))\cdots) \\
& = (a_1 b_1 a_2 b_2 \cdots a_m b_m a_1' b_1' a_2' b_2' \cdots a_{m'}' b_{m'}')(v).
\end{align}
Therefore $A*B$ acts on $V$.
