A coordinate free proof of the identity $Tr(A \otimes  B)= Tr(A) \otimes Tr(B)$ I was going over some of the definitions of trace and was trying to find a way to prove one of the facts stated about the trace listed in the referenced Wikipedia article.
Let $M, N$ be finitely generated projective $R$-modules over a commutative ring $R$ with identity. 
In particular I was wondering if there was a slick proof using the coordinate free language of the trace for the following fact. 

How do we show that for any $f \in End_R(M), v \in End_R(N)$ then $ Tr( f \otimes_R g) = Tr(f) \otimes_R Tr(g)$?

 A: Warning: There will be some cheating below, in the sense that the phrase "the obvious map from ... to ..." will often be used. Thank you in advance for letting me know if I should be more explicit. 
References will be made to H. Cartan and S. Eilenberg's classic book Homological algebra, legally available at Internet Archive. 

LEMMA. Let $A$ be a ring (that is, an associative ring with $1$), let $\mathcal A$ be the category of $A$-modules, let $\mathcal B$ be the category opposite to $\mathcal A$, let $F$ and $G$ be additive functors from $\mathcal B^p\times\mathcal A^q$ to the category of abelian groups, and let $\varphi$ be a functorial morphism from $F$ to $G$. Assume that 
  $$
\varphi(A,\dots,A):F(A,\dots,A)\to G(A,\dots,A) 
$$ 
  is an isomorphism, and that $P_1,\dots,P_{p+q}$ are finitely generated projective $A$-modules. Then 
  $$
\varphi(P_1,\dots,P_{p+q}):F(P_1,\dots,P_{p+q})\to G(P_1,\dots,P_{p+q}) 
$$ 
  is an isomorphism.

The statement results immediately from the following two facts: 


*

*Additive functors are compatible with finite direct sums in the sense of Proposition II.1.1 of Cartan-Eilenberg.

*Let $f_1,\dots,f_n$ be $A$-linear maps. Then $\bigoplus f_i$ is bijective if and only if each $f_i$ is. 
Assume that $A$ is commutative. The letters $U,V,X,Y$ will refer to "variable" $A$-modules. The letters $P,Q$ will refer to "variable" projective $A$-modules. Write respectively 
$$
\otimes,\quad\quad\text{Hom},\quad\quad\text{End},\quad\quad X^*,\quad\quad\text{map}
$$ 
for 
$$
\otimes_A,\quad\text{Hom}_A,\quad\text{End}_A,\quad \text{Hom}_A(X,A),\quad\text{$A$-linear map}.
$$ 
Associativity and commutativity of the tensor product will be tacitly used. 
Recall that the trace of an endomorphism $f$ of $P$ is defined as follows. On applying the Lemma to the obvious functorial morphism 
$$
U^*\otimes X\to\text{Hom}(U,X),
$$ 
we get a canonical isomorphism 
$$
P^*\otimes P\ \overset{\sim}{\to}\ \text{End}(P).
$$ 
Then the trace of $f$ is defined by transporting to $\text{End}(P)$ the obvious map from $P^*\otimes P$ to $A$. 
On applying the Lemma to the obvious functorial morphism 
$$
\text{Hom}(U,X)\otimes\text{Hom}(V,Y)\to\text{Hom}(U\otimes V,X\otimes Y),
$$ 
we get canonical isomorphisms 
$$
\text{End}(P)\otimes\text{End}(Q)\ \overset{\sim}{\to}\ \text{End}(P\otimes Q),\quad 
P^*\otimes Q^*\ \overset{\sim}{\to}\ (P\otimes Q)^*.
$$
Let $f$ be in $\text{End}(P)$ and $g$ be in $\text{End}(Q)$. To check 

$$\text{trace}(f\otimes g)=\text{trace}(f)\ \text{trace}(g),$$ 

it suffices to verify that the compositions of canonical isomorphisms 
$$
P^*\otimes P\otimes Q^*\otimes Q\ \overset{\sim}{\to}\ 
\text{End}(P)\otimes\text{End}(Q)\ \overset{\sim}{\to}\ 
\text{End}(P\otimes Q)
$$ 
and 
$$
P^*\otimes P\otimes Q^*\otimes Q\ \overset{\sim}{\to}\ 
(P\otimes Q)^*\otimes P\otimes Q\ \overset{\sim}{\to}\ 
\text{End}(P\otimes Q)
$$ 
coincide. But this is clear.
