I am reading the book “elements of the representation theory of associative algebras” and I have a problem about the proof of the theorem 6.8 in chapter 1.
Let $A$ be a finite dimensional $k$-algebra, $e\in A$ be a non-zero idempotent and $B=eAe$.
The functor $T_e$, $L _e:mod~B \rightarrow mod~A$ and $res _e: mod~A \rightarrow mod~B$ are defined by
$$res_e(-)=(-)e, \quad T_e(-)=-\otimes eA, \quad L_e(-)=Home_B(Ae,-)$$
where $Mod~B$ is the category of all finitely generated right $B$-modules and $Mod~A$ is the category of all finitely generated right $A$-modules
If we choose $Y$ is a right $B$-module, the author says that there are isomorphisms $$(Y\otimes_B eA)e\cong Y\otimes_B eAe\quad(1),$$ $$(Hom_B(Ae,Y))e\cong Y_B\quad (2).$$
The first isomorphism seems natural, but I don't know how to prove it strictly.
The second isomorphism I don't seem to have any ideas about.
Any help and references are greatly appreciated.
Thanks!
Here are the ideas. I leave to you the verification of each claim below.
For the first isomorphism, note that the map $$ Y\times eA\to Y\otimes_B B,\quad (y,ex)\mapsto y\otimes exe $$ is $B$-balanced, so it induces a homomorphism of $k$-vector spaces $$ \alpha_Y':Y\otimes_B eA\to Y\otimes_B B, \quad y\otimes ex\mapsto y\otimes exe. $$ Then define $\alpha_Y:(Y\otimes_B eA)e\to Y\otimes_B B$ by $$ \alpha_Y((y\otimes ex)e) = \alpha_Y'(y\otimes ex), $$ this map is a well defined $B$-module homomorphism. We can construct an inverse similarly: The map $$ Y\times B\to (Y\otimes_B eA)e,\quad (y,exe)\mapsto (y\otimes ex)e $$ is a well defined $B$-balanced map, and the induced $k$-linear homomorphism $$ \beta_Y:Y\otimes_B B\to (Y\otimes_B eA)e $$ is an inverse for $\alpha_Y$.
For the second isomorphism note that $Ae$ is a $(A,B)$-bimodule, which induces a structure of right $A$-module on $L_e(Y_B)=\mathrm{Hom}_B(Ae,Y_B)$ for all right $B$-module $Y_B$. This structure is given by $$ (f\cdot a)(x)=f(ax) $$ for all $x\in Ae$, $a\in A$ and $f\in \mathrm{Hom}_B(Ae,Y_B)$. Now consider the map $$ \varphi_Y:\mathrm{Hom}_B(Ae,Y_B)e\to Y_B, \quad f\cdot e\mapsto f(e). $$ This map is well defined and is a $B$-module homomorphism. We can construct an explicit inverse for $\varphi_Y$, namely $$ \psi_Y:Y_B\to \mathrm{Hom}_B(Ae,Y_B)e, $$ where for $y\in Y_B$ and $xe\in Ae$ (with $x\in A$), we define $$ \psi_Y(y)(xe) = yexe, $$ which shows that $\varphi_Y$ is an isomorphism of $B$-bmodules.