Simple representations are projections(?) 
(p44, Webb, Finite group Theory) Let $\rho_1, \ldots, \rho_r$ be the simple representations of $G$ over $\mathbb{C}$ with degrees $d_1, \ldots, d_r$. The each $\rho_i : G \rightarrow M_{d_i}(\mathbb{C})$ extends linearly to a $\mathbb{C}$-algebra homomoprhism 
  $$ \rho_i: \mathbb{C}G = \bigoplus_{j=1}^r M_{d_j}(\mathbb{C}) \rightarrow M_{d_i}(\mathbb{C})$$
  the projection onto the $i$-th summand. 

 I know the equality is by Artin Wedderburn. Why is this map a projection?
EDIT: The comment pinpoints my problem
(i) Given the representations $\rho_1, \ldots , \rho_r$ why is $\mathbb{C}G = \bigoplus_1^r M_{d_j}(\mathbb{C})$?
 A: For your question (i), you need the

Maschke's Theorem. Let $G$ be a finite group and $F$ a field. If $\mathrm{char}(F)\not | |G|$, then the group algebra $F[G]$ is semi-simple.

The rest follows from the Wedderburn-Artin theorem, with $\rho_1,\cdots,\rho_r$ being all irreducible complex representations of $G$. Now having extended $\rho_i\colon G\to\mathrm{GL}_{d_i}(\mathbb C)$ linearly to a $\mathbb CG$-module epimorphism
$$
\widetilde\rho_i\colon \mathbb CG=\bigoplus_{j=1}^rM_{d_j}(\mathbb C)\twoheadrightarrow M_{d_i}(\mathbb C),
$$
we assert that $$\ker\widetilde\rho_i=\bigoplus_{j=1,j\neq i}^rM_{d_j}(\mathbb C).$$ This is obvious since otherwise $M_{d_i}(\mathbb C)$ would be isomorphic to a non-simple $\mathbb CG$-module, which is absurd. Hence it remains to show that $\widetilde\rho_i|_{M_{d_i}(\mathbb C)}$ is the identity map. We have known that $\widetilde\rho_i|_{M_{d_i}(\mathbb C)}\colon M_{d_i}(\mathbb C)\to M_{d_i}(\mathbb C)$ is an isomorphism (of $\mathbb CG$-modules and $\mathbb C$-algebras). By Skolem-Noether theorem, there is a $P\in\mathrm{GL}_{d_i}(\mathbb C)$ satisfying
$$
\widetilde\rho_i|_{M_{d_i}(\mathbb C)}(\cdot)=P^{-1}(\cdot)P.
$$
Then for each $M\in M_{d_i}(\mathbb C)$, it follows that
$$
M=MP^{-1}P=M\widetilde\rho_i|_{M_{d_i}(\mathbb C)}(I)=\widetilde\rho_i|_{M_{d_i}(\mathbb C)}(M)(=P^{-1}MP),
$$
where $I\in M_{d_i}(\mathbb C)$ is the identity matrix and thus $\widetilde\rho_i|_{M_{d_i}(\mathbb C)}=\mathrm{id}_{M_{d_i}(\mathbb C)}$. Therefore $\widetilde\rho_i$ is the projection onto the $i$-th summand.
