# Why is the isomorphism of Artin-Wedderburn a ring isomorphism?

I'm somewhat confused on the following calculation in the proof of Artin-Wedderburn: let $R$ be a semisimple ring such that (since $R$ is f.g. over itself) we have $R \simeq \oplus_{i = 1}^r S_i^{n_i}$. Now,

$$R^{op} \stackrel{(1)}\simeq End_R(R) \simeq End_R(\oplus_{i = 1}^r S_i^{n_i}) \stackrel{(2)}\simeq \bigoplus_{i = i}^rEnd_R(S_i)^{n_i \times n_i}$$

Now, I know that $(1)$ is a ring isomorphism between $R^{op}$ and the left $R$-module morphisms from $R$ to istelf, but why is $(2)$ a ring isomorphism? I know we have $Z(R)$-module isomorphism between $Hom_R(M \oplus N,P)$ and $Hom_R(M,P) \oplus Hom_R(N,P)$, so $(2)$ could be stated as a module isomorphism, but however $(1)$ is a ring homomorphism so it would not make sense in this context.

$End(M\times N)\cong \begin{bmatrix}Hom(M, M)& Hom(N,M)\\Hom(M,N)& Hom (N,N)\end{bmatrix}$ as rings where the multiplication is matrix multiplication. Take a look at how the elements from the various entries compose with each other and generalize it to your situation.
The final piece of the puzzle is to notice that when $M$ and $N$ are different Wedderburn components of $R$, there is no nonzero homomorphism between them. In our toy example above, if $Hom(M,N)=Hom(N,M)=\{0\}$, then you're looking at the diagonal matrix ring
$\begin{bmatrix}Hom(M, M)& 0\\0& Hom (N,N)\end{bmatrix}\cong Hom(M,M)\times Hom(N,N)=End(M)\times End(N)$, all as ring homomorphisms.