Surjective map in a proof of Morita's theorem

Question

I'm trying to prove the following theorem : Two rings $R$ and $S$ are Morita-equivalent iff there's a progenerator $P$ of Mod-$R$ s.t. $$ S \cong End_R(P)$$
In $(\Leftarrow)$ direction we define a left $S$-module structure on $P$ by $$End_R(P) \times P \rightarrow P$$ $$ (f,x) \mapsto f(x)$$ and the isomorphism $S \cong End_R(P)$ so that $P$ has a structure of $S-R$ bimodule.
Let $Q:= Hom_R(P,R)$. By $$ Q \times S \rightarrow Q$$ $$ (q,s) \mapsto q \circ \phi(s)$$ where $\phi : S \cong End_R(P)$ and by $$ R \times Q \rightarrow Q$$ $$ (r,q) \mapsto \psi(r) \circ q$$ where $ \psi : R \cong End_R(R)$, $Q$ has a $R-S$ bimodule structure.
We need to prove that $P \otimes_R Q \cong S$ e $Q \otimes_S P \cong R$ ( because there is a characterization of the morita equivalence in terms of bimodules).
Define $$ \beta : P \otimes_R Q \rightarrow End_R(P)$$ on generators by $$ p \otimes f \mapsto pf$$ How can I prove that $\beta$ is surjective? In some lecture notes the author said "$\beta$ is onto since $P$ is a direct summand of $R^n$ and so any endomorphism of $P$ is a sum of endomorphisms factoring through $R$"
I don't understand what the author mean here. Any help?

Answer 1

You have a natural morphism $A\otimes B^* \to \hom(B,A)$ for $A,B\in \mathrm{Mod}-R$, where $B^* = \hom(B,R)$, and it is defined by $x\otimes f \mapsto (y\mapsto xf(y))$

$B$ fixed, if $A=R^n$, $n$ an integer, this natural morphism is an isomorphism : the converse is defined by : $g\mapsto \sum_i e_i \otimes (\pi_i\circ g)$ where $e_i = (0,...,1,0,...)$ with a $1$ in $i$th position and $\pi:R^n\to R$ is the $i$th projection. It is easy to check that it is a converse.

Now if $P$ is a summand of $R^n$, say $P\oplus S = R^n$, then this morphism is (by naturality) $P\otimes B^* \oplus S \otimes B^* \to \hom(B, P)\oplus \hom(B,S)$ the sum of the two natural morphisms for $P,S$. Therefore it is also an isomorphism for $P$ : $P\otimes B^* \overset{\sim}\to \hom(B,P)$.

Apply this with $B=P$ to get $P\otimes Q\overset{\sim}\to \mathrm{End}(P)$