Artin - Wedderburn theorem corollary 
Corollary of Artin Wedderburn: (p18, Introudction to finite group theory, Webb) Let $A$ be a finite dimensional semsimple algebra over an algebrically closed field $k$. In any decomposition 
  $$ _AA = S_1^{n_1} \oplus \cdots \oplus S_r^{n_r} \cong M_{n_1}(D_1) \oplus \cdots \oplus M_{n_r}(D_r)$$ 
  where $S_i$ are pairwise nonisomoprhic simple modules, and $D_i=End_A(S_i)^{op}$. We have $n_i = \dim_k S_i$. 

For the proof the author begins with

$M_{n_i}(k) \cong S_i^{n_i}$ as left $A$-modules, since term on left is isomorphic to the quotient of $A$ by the left submodules consisting of elements that the summand $M_{n_i}(k)$ annihilates by right multiplication, and the term on right is an image of this quotient and we have $\dim_k(A) = \sum_i \dim_k S_i^{n_i}$ they must be isomorphic. 

I am really confused, what are the explicit maps involved here? 
 A: Why you don't just act $A=\bigoplus_{i=1}^rS_i^{n_i}$ by $\mathrm{End}_A$ for this corollary? If so, one has
$$
\mathrm{End}_A(A)=\bigoplus_{i=1}^r\mathrm{End}_A(S_i^{n_i})\cong\bigoplus_{i=1}^rM_{n_i}(D_i^{\mathrm{op}}).
$$
Moreover, $A^\mathrm{op}\cong\mathrm{End}_A( _AA)$ is given by $a\mapsto(x\mapsto xa)$, and hence
$$
A^\mathrm{op}\cong\bigoplus_{i=1}^rM_{n_i}(D_i^{\mathrm{op}})\cong\bigoplus_{i=1}^rM_{n_i}(D_i)^{\mathrm{op}},
$$
which implies
$$
A\cong\bigoplus_{i=1}^rM_{n_i}(D_i).
$$
I don't understand why in your corollary $A$ is assumed to be over an algebraically closed field $k$, since in this case, $D_i$'s are finitely dimensional division algebras over $k$ and hence all equal to $k$.

In allusion to your question, as is said above, $k\cong D_i^\mathrm{op}\cong \mathrm{End}_A(S_i)$, and hence $M_{n_i}(k)\cong\mathrm{End}_k(S_i)$. Then the Jacobson density theorem provides an epimorphism
\begin{align}
\varphi\colon A&\twoheadrightarrow \mathrm{End}_k(S_i)\\
a&\mapsto(x\mapsto xa).
\end{align}
Pick a $k$-basis $\{e_1,\cdots,e_{n_i}\}$ of $S_i$. Define
\begin{align}
\Phi\colon\mathrm{End}_k(S_i)&\twoheadrightarrow S_i^{n_i}\\
f&\mapsto (f(e_1),\cdots,f(e_{n_i}))
\end{align}
and $\Phi$ is also epimorphic. These are the explicit maps in your questioned part.
