Algebraic $K\_0$ group for an algebra $A$ may be defined in terms of stable isomorphism classes of idempotents in $M\_n(A)$ or equivalently in terms of isomorphism clasess of finitely generated projective modules. If $f:A \to B$ is a morphism of algebras and $e=(e\_{ij})\_{i,j}$ is an idempotent in $M\_n(A)$ then we define $f\_\*([e]):=[(f(e\_{ij}))\_{i,j}]$. In the "module picture" given $P$-projective finitely generated $A$ module we form $f\_\*([P]):=[P \otimes\_A B]$ where the right $B$-module structure on $P \otimes\_A B$ is via $(x \otimes b)b'=x \otimes bb'$. You can wonder where the map $f$ is involved: well, it is in the relation $xa \otimes b=x \otimes f(a)b$.  

Now one can show that *any* finitely generated projective $A$ module is of the form $eM\_n(A)$ for some idempotent $e=(e\_{ij})\_{i,j}$. So my question is 



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> Why these two description are equivalent, i.e. why is $(f(e\_{ij}))\_{i,j}M\_n(B) \cong P \otimes\_A B$ where $P=eM\_n(A)$ is finitely generated projective $A$ module? 
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Please note that I posted a similar question on stck exchange however I didn;t received a satisfactory answer. 



Morphisms in K-theory: comparison of two pictures
It seems that by $M\_n(A)$ you denote two different objects: actual $M\_n(A)$ -- which is the same as $\text{Hom}(A^n, A^n)$ -- and $A^n$ itself.


One can show that any finitely generated projective $A$-module $P$ can be written in the form $P = eA^n$ for some idempotent $e = (e\_{i,j}) \in \text{Hom}(A^n,A^n)$. And, of course, $P \otimes\_A B = (eA^n) \otimes\_A B \simeq f(e)B^n$. Indeed, we have an isomorphism $\alpha : A^n \otimes\_A B \longrightarrow B^n$ given by $(a\_j)\otimes 1 \mapsto (f(a\_j))$. Take now a vector $(a\_j) \in eA^n$, so $(e\_{i,j})(a\_j) = (a\_j)$. Its image in $B^n$ is preserved by $(f(e\_{i,j}))$: $(f(e\_{i,j})) (f(a\_j)) = f ((e\_{i,j}) (a\_j)) = f((a\_j))$ since $f$ is a morphism of rings. On the other hand, every object in $f(e)B^n$ can be written as $f(e)f((a\_j)) \cdot b$ for some $(a\_j) \in A^n$ and $b \in B$; if $(f(e\_{i,j})) (f(a\_j) \cdot b) = f ((e\_{i,j}) (a\_j)) \cdot b$ does not equal $f((a\_j)) \cdot b$, then $(a\_j) \notin eA^n$. So the image of $(eA^n) \otimes\_A B$ under the isomorphism $\alpha$ is exactly $f(e)B^n$.


