Compute a map on a tensor product The context is the following:

2.3. The canonical $r$-matrix.
  Let $(\mathfrak{d}, \mathfrak{g}, \mathfrak{h})$ be a Manin quasi-triple.
  We identify $\mathfrak{d}$ with $\mathfrak{g} \oplus \mathfrak{g}^*$ using the isomorphism $j^{-1}$ of $\mathfrak{h}$ onto $\mathfrak{g}^*$.
  The map $r_{\mathfrak{d}} \colon \mathfrak{d}^* \to \mathfrak{d}$ defined by $r_{\mathfrak{d}} \colon (\xi, x) \mapsto (0, \xi)$ for $x \in \mathfrak{g}$, $\xi \in \mathfrak{g}^*$ defines an element $r_{\mathfrak{d}} \in \mathfrak{d} \otimes \mathfrak{d}$, called the canonical $r$-matrix.
  Let $(e_i)$, $i = 1, \dotsc, n$, be a basis of $\mathfrak{g}$ and $(\varepsilon^i)$, $i = 1, \dotsc, n$, be the dual basis of $\mathfrak{g}^*$.
  Then
  \begin{equation}
  \tag{2.3.1}
    r_{\mathfrak{d}}
  = \sum_{i=1}^n e_i \otimes \varepsilon^i.
\end{equation}
(Original scan here.)

(This is an excerpt from 'Manin pairs and moment maps' by A. Alekseev and Y. Kosmann-Scharzbach p5)
What I don't understand is how the operator defined in (2.3.1) realizes the promised action, i.e. sending a pair $(\xi, x)$ to $(0,\xi)$ with $x$ in $\frak g$ and $\xi \in \frak g^*$.
 A: $\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\g}{\mathfrak{g}}
\newcommand{\d}{\mathfrak{d}}
$Recall that for finite dimensional vector spaces $U$ and $V$,
$$
    U^* \otimes V \cong \Hom(U,V)
$$
The forward map is a pretty natural one: Given $\mu\in U^*$ and $v\in V$, we can define $\phi(\mu \otimes v)$ to be the map which takes $u \in U$ to $\mu(u)v$.  To go in the other direction, choose a basis $u_1,\dots,u_n$ of $U$, a dual basis $\mu^1,\dots,\mu^n$ of $U^*$, and   given $T \in \Hom(U,V)$, let $$\psi(T) = \sum_i\mu^i \otimes T(u_i)$$
The authors are using a map like $\psi$ but suppressing the notation.
Here $V = \d \cong \g \oplus \g^*$, and $U = \d^* \cong \g^* \oplus \g$.  We have $r_{\d}\in\Hom(\d^*,\d)$; the canonical $r$-matrix is $\psi(r_{\d}) \in \d \otimes \d$.  If $e_1,\dots,e_n$ is a basis of $\g$, and $\varepsilon^1,\dots,\varepsilon^n$ the dual basis of $\g^*$, then $e_1,\dots,e_n,\varepsilon^1,\dots,\varepsilon^n$ is a basis of $\d$ and $\varepsilon^1,\dots,\varepsilon^n,e_1,\dots,e_n$ is the corresponding dual basis of $\d^*$.  We have $r_\d(e_i) = 0$ and $r_\d(\varepsilon^i) = \varepsilon^i$.
So
$$
    \psi(r_{\d}) = \sum_{i=1}^n \varepsilon^i \otimes r_\d(e_i)  + \sum_{i=1}^n e_i \otimes r_\d(\varepsilon^i) = \sum_{i=1}^n e_i \otimes \varepsilon^i
$$
