Functorial isomorphism involving tensor products Let $R$ be a commutative ring and $E', E, F', F$ be free, f.g. $R$-modules of equal rank. 
For $f\in L(E',E):={\rm Hom}_R(E',E)$ and $g\in L(F',F)$, let $T(f.g)\in L(E'\otimes_R F', E\otimes_R F)$ be the unique linear map sending each generator $x'\otimes y'$ of $E'\otimes F'$ to $f(x')\otimes g(y')$. 
Lang's 'Algebra' p.627 says that the map $$L(E',E)\otimes_R L(F',F)\to L(E'\otimes_R F', E\otimes_R F)$$ defined by $f\otimes g \mapsto T(f,g)$ is functorial. 
To show this do I need to prove that the following diagram commutes (for a variable $E'$ and fixed $E$, $F'$ and $F$ say)
$$\begin{matrix}
L(E',E)\otimes L(F',F)&\longrightarrow&L(E'\otimes F',E\otimes F)\\
\downarrow{\Lambda(\varphi)}&&\downarrow{\Omega(\varphi)}\\
L(E'',E)\otimes L(F',F)&\longrightarrow&L(E''\otimes F',E\otimes F)
\end{matrix}
$$
where $\Lambda$ and $\Omega$ are the obvious contravariant functors and $\varphi: E''\to E'$ is any $R$-map and then do the same for the other cases? 
If so, how do I define the maps $\Lambda(\varphi)$ and $\Omega(\varphi)$? 
Many thanks!
 A: First, $L( \;\cdot\;, E) = \operatorname{Hom}_R( \;\cdot\;, E)$ is a contravariant functor; i.e., given $\varphi:  E'' \to E'$, we have
$$
\begin{align}
L(\varphi,E): L(E', E) &\to L(E'', E) \\
\alpha &\mapsto \alpha \circ \varphi.
\end{align}
$$
In order to define $\Lambda(\varphi)$, an $R$-linear map on a tensor product, we have to first define an $R$-bilinear map $\Lambda'(\varphi)$ on the cartesian product:
$$
\begin{align}
\Lambda'(\varphi): L(E',E) \times L(F',F) &\to L(E'',E) \otimes L(F',F) \\
\alpha \times \beta &\mapsto (\alpha \circ \phi) \otimes \beta
\end{align}.
$$
After verifying that $\Lambda'(\varphi)$ is balanced, i.e.,
$$
\Lambda'(\varphi)(r \alpha, \beta) = \Lambda'(\varphi)(\alpha, r \beta),
$$
the universal property for tensor products guarantees the well-defined map
$$
\begin{align}
\Lambda(\varphi): L(E',E) \otimes L(F',F) &\to L(E'',E) \otimes L(F',F) \\
\alpha \otimes \beta &\mapsto (\alpha \circ \phi) \otimes \beta.
\end{align}
$$
By a similar routine, we can define
$$
\begin{align}
\Omega(\varphi): L(E' \otimes F', E \otimes F) &\to L(E'' \otimes F', E \otimes F) \\
\alpha \otimes \beta &\mapsto (\alpha \circ \phi) \otimes \beta.
\end{align}
$$
(It's the same formula!  The issue is not so much what the formula should look like but rather whether the maps are well-defined.)
