& h(b)^d(a_d,\dots,a_1) = b^{d+1}(e_{Y_0},a_d,\dots,a_1).
\end{aligned}$$

In other words, the Yoneda embedding is cohomologically full and faithful. Moreover, it canonically extends to a cohomologically full and faithful functor $A^{\mathit{tw}} \rightarrow A^{\mathit{mod}}$. The easiest way to see this is to think of it as the composition $$\label{eq:yoneda-restriction}
A^{\mathit{tw}} \longrightarrow (A^{\mathit{tw}})^{\mathit{mod}} \longrightarrow A^{\mathit{mod}}.$$ where the first arrow is the Yoneda embedding for $A^{\mathit{tw}}$, and the second is restriction of modules from $A^{\mathit{tw}}$ to $A$. Since objects of $A$ generate $A^{\mathit{tw}}$ by definition, the restriction functor is a quasi-equivalence, so is again cohomologically full and faithful.

Idempotent splittings\[subsec:splittings\]
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Suppose that we have an object $Y$ of $A$ together with an endomorphism in the category $H^0(A)$ which is idempotent. One can always lift it to a [*homotopy idempotent*]{}, which is a sequence $p = \{p^d\}$ of elements $p^d \in \mathit{hom}_A(Y,Y)^{1-d}$ ($d \geq 1$) satisfying the equations $$\label{eq:homotopy-idempotent}
\sum_r \sum_{k_1 + \cdots + k_r = d} \mu_A^r(p^{k_r},\dots,p^{k_1}) =
\begin{cases} p^{d-1} & \text{$d$ even,} \\ 0 & \text{$d$ odd} \end{cases}$$ for any $d \geq 1$, and such that $p^1$ represents our original idempotent. This is proved in [@seidel04 Section 4], but it is maybe useful to summarize the argument in more elementary language. The choice of 