Throughout this chapter $A$ will denote a complexified semi-simple Lie-algebra, $H$ a Cartan sub-algebra, $R\sbe H$ the root system and $B$ a base for $R$.
For all $x\in\Z_p$, all $y\in\Z_q$ we have $p(x\otimes y)=(px)\otimes y=0$ and $q(x\otimes y)=x\otimes(qy)=0$. Since $\gcd(p,q)=1$ there are $a,b\in\Z$ such that $ap+bq=1$ and therefore $x\otimes y=(ap+bq)(x\otimes y)=ap(x\otimes y)+bq(x\otimes y)=0$.
$\proof$
Suppose $f\in\Hom(E,F;G)$, i.e. $f:E\times F\rar G$ is bi-linear. Then for each $x\in E$ the map $f_x:F\rar G$, $f_x(y)=f(x,y)$ is linear and $x\mapsto f_x$ is a linear map from $E$ into $\Hom(F;G)$. Hence $f\mapsto(x\mapsto f_x)$ is linear with inverse $g\mapsto((x,y)\mapsto g(x)(y)$.
2. For $f\in\Hom(E,F;G)$ there is a unique $\wh f\in\Hom(E\otimes F;G)$ such that for all $x\in E$ and all $y\in F$: $\wh f(x\otimes y)=f(x,y)$. Hence the linear map $f\mapsto\wh f$ is injective. On the other hand given a linear map $u\in \Hom(E\otimes F;G)$, the map $f(x,y)\colon=u(x\otimes y)$ is bi-linear and $\wh u=f$.
$\eofproof$
$\proof$
$m:R\times E\rar E$, $(r,x)\mapsto rx$, is bi-linear an thus there is a unique linear map $\wh m:R\otimes E\rar E$ such that $\wh m(r\otimes x)=rx$. On the other hand the map $l:E\rar R\otimes M$, $l(x)\colon=1\otimes x$ satisfies
$$
l\circ\wh m(r\otimes x)
=l(rx)
=1\otimes rx
=r(1\otimes x)
=r\otimes x
\quad\mbox{and}\quad
\wh m\circ l(x)=x,
$$
i.e. $l\circ\wh m|j(R\times E)=id|j(R\times E)$. By exam $l$ is an isomorphism with inverse $\wh m$.
$\eofproof$
1.,2.: For $p\in\Q$ and $q\in\Q$ or $q\in\R$ we get as im the previous proposition: $\wh m(p\otimes q)=pq$ and $l(p)=1\otimes x$
$$
l\circ\wh m(p\otimes q)
=l(pq)
=1\otimes pq
=p(1\otimes q)
=p\otimes q,
$$
3. $n\otimes q=n\otimes p(q/p)=p(n\otimes(q/p))=np\otimes(q/p)=0$.
$\proof$
Define $B:E\times(F\times G)\rar E\otimes F\times E\otimes G$ by
$$
B(x,(y,z))\colon=(x\otimes y,x\otimes z)~.
$$
Then $B$ is bi-linear and thus there is a unique linear map $\wh B:E\otimes(F\times G)\rar E\otimes F\times E\otimes G$ such that
$$
\wh B(x\otimes(y,z))=B(x,(y,z))\colon=(x\otimes y,x\otimes z)
$$
Put $J_2:E\times G\rar E\otimes(F\times G)$, $J_2(x,z)=x\otimes(0,z)$ and $J_1:E\times F\rar E\otimes(F\times G)$, $J_1(x,y)=x\otimes(y,0)$; both are bi-linear. Hence there are linear maps: $\wh J_2:E\otimes G\rar E\otimes(F\times G)$ and $\wh J_1:E\otimes F\rar E\otimes(F\times G)$ such that
$$
\wh J_1(x\otimes y)=J_1(x,y)=(x\otimes(y,0)),\quad
\wh J_2(x\otimes z)=J_2(x,z)=(x\otimes(0,z))~.
$$
Finally we define $\wh J:E\otimes F\times E\otimes G\rar E\otimes(F\times G)$ by
$$
\wh J(X,Y)\colon=\wh J_1(X)+\wh J_2(Y)~.
$$
Now for all $x,u\in E$, all $y\in F$ and all $z\in G$ we have
$$
\wh J\circ\wh B(x\otimes(y,z))
=\wh J(x\otimes y,x\otimes z)
=\wh J_1(x\otimes y)+\wh J_2(x\otimes z)
=x\otimes(y,0)+x\otimes(0,z)
=x\otimes(y,z)
$$
and
$$
\wh B\circ\wh J(x\otimes y,u\otimes z)
=\wh B(\wh J_1(x\otimes y)+\wh J_2(u\otimes z))
=\wh B(x\otimes(y,0)+u\otimes(0,z))
=(x\otimes y,0)+(0,u\otimes z)
=(x\otimes y,u\otimes z)~.
$$
By exam both maps $\wh J\circ\wh B$ and $\wh B\circ\wh J$ are the identities. Hence $\wh B$ is an isomorphism with inverse $\wh J$.
$\eofproof$
The assertion generalizes to finite products and to arbitrary direct sums, i.e.
$$
\Big(\bigoplus_j E_j\Big)\otimes E
\quad\mbox{is isomorphic to}\quad
\bigoplus_j(E_j\otimes E)~.
$$
Suppose $f_j\in\Hom(E_j;F_j)$, $j=1,\ldots,p$, then the map $\prod f_j:\prod E_j\rar\prod F_j\rar\bigotimes F_j$:
$$
\prod f_j(x_1,\ldots,x_p)\colon=f_1(x_1)\otimes\cdots\otimes f_p(x_p)
$$
is linear in each component. Hence there is a unique linear map $T(f_1,\ldots,f_p):\bigotimes E_j\rar\bigotimes F_j$ such that
$$
T(f_1,\ldots,f_p)(x_1\otimes\cdots\otimes x_p)=f_1(x_1)\otimes\cdots\otimes f_p(x_p)
$$
Now the map $T:\prod\Hom(E_j;F_j)\rar\Hom(\bigotimes E_j;\bigotimes F_j)$ is again linear in each component and thus there is a unique homomorphism
$$
\wh T:\bigotimes\Hom(E_j;F_j)\rar\Hom(\bigotimes E_j;\bigotimes F_j)
$$
such that $\wh T(f_1\otimes\cdots\otimes f_p)=T(f_1,\ldots,f_p)$. In abuse of notation the linear map $T(f_1,\ldots,f_p):\bigotimes E_j\rar\bigotimes F_j$ is also written $f_1\otimes\cdots\otimes f_p$. If all modules are finite dimensional and free, then $\wh T$ is an isomorphism.