Tensoring with induced representation On J. Humphreys' book "Representations of Semisimple Lie Algebras in the BGG Category O", Theorem 3.6, a Tensor Identity is quoted:
$$ (U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} L) \otimes M\simeq  U(\mathfrak{g}) \otimes _{U(\mathfrak{b})} (L \otimes M)$$
where $L$ is a $U(\mathfrak{b})$-module and $M$ is a $U(\mathfrak{g})$-module.
While this isomorphism seems clear as a $k$-module isomorphism, it doesn't seem to be a $U(\mathfrak{g})$-module homomorphism.
Apparently, an element $x\in \mathfrak{g}$ should act on the left module by
$$ x\cdot (y\otimes a \otimes b) = xy\otimes a \otimes b + y\otimes a \otimes xb $$
while it acts on the right by
$$ x\cdot (y\otimes a \otimes b) = xy\otimes a \otimes b $$
Am I misunderstanding sometihng?
 A: Let $\Gamma$ be a sub-Hopf algebra of the Hopf algebra $\Lambda$.  Let $M$ be a $\Lambda$-module and $N$ a $\Gamma$-module.  Then 
$$ \Lambda \otimes _\Gamma (N \otimes M|_\Gamma) \cong (\Lambda \otimes_\Gamma N) \otimes M.$$
The isomorphism is given by
$$ \lambda \otimes_\Gamma (n\otimes m) \mapsto \sum (\lambda_{(1)} \otimes _\Gamma n) \otimes \lambda_{(2)} m.$$
I'm using Sweedler notation here: $\Delta(\lambda) = \sum \lambda_{(1)} \otimes \lambda_{(2)}$ where $\Delta$ is the comultiplication. 
Note that this does agree with Jyrki's map in the group case, for then $\Delta(g)=g\otimes g$ for $g$ a group element.
In your case, $\Delta(x) = x\otimes 1 + 1 \otimes x$ for $x \in \mathfrak{g}$.  You can use that to get an explicit expression for the isomorphism, for example it sends
$$ x \otimes (n \otimes m) \mapsto (x\otimes m) \otimes n + (1\otimes n) \otimes xm $$
for $x \in \mathfrak{g}$.
The inverse map is given by
$$ (\lambda \otimes_\Gamma n)\otimes m \mapsto \sum \lambda_{(1)} \otimes_\Gamma (n \otimes S(\lambda_{(2)}) m) $$
where $S$ is the antipode of $\Lambda$ (in your case, $S$ is the antihomomorphism defined by $S(x)=-x$ for $x \in \mathfrak{g}$).
A: Not an answer. Just explaining why we should not expect the obvious $k$-linear mapping to work. Hopefully this leads the OP or somebody else to dig out the correct isomorphism.

The analogous result at the level of finite groups is the following. If $H\le G$, $L$ is a $kH$-module, and $M$ a $kG$-module, then there is an isomorphism of $kG$-modules
$$
f:kG\otimes_{kH}(L\otimes M)\simeq (kG\otimes_{kH} L)\otimes M
$$
given at the level of elementary tensors by the recipe
$$
x\otimes (\ell\otimes m)\mapsto (x\otimes \ell)\otimes (xm)
$$
where $x\in G$, $\ell\in L$ and $m\in M$. This is a well-defined map, because for all $h\in H$ we have 
$$
(xh)\otimes(\ell\otimes m)\mapsto (xh\otimes\ell)\otimes (xhm),
$$
and also
$$
x\otimes (h\ell\otimes hm)\mapsto (x\otimes h\ell)\otimes (xhm)=(xh\otimes\ell)\otimes (xhm).
$$
This is absolutely needed, because in the range we have (the first tensor product is over $kH$)
$$
(xh)\otimes(\ell\otimes m)=x\otimes h(\ell\otimes m)=x\otimes(h\ell\otimes hm).
$$
Observe that the resulting map is an isomorphism of $kG$-modules as for all $g\in G$
$$
\begin{aligned}
f(g\cdot(x\otimes(\ell\otimes m)))&=f(gx\otimes(\ell\otimes m))\\
&=(gx\otimes \ell)\otimes gxm=g\cdot((x\otimes\ell)\otimes m))\\
&=g\cdot f(x\otimes(\ell\otimes m))
\end{aligned}
$$
by the definition of $f$ and the action of $G$ on the induced module as well as the tensor product.
The tensor identity that you want needs to reflect this somehow, and use something other than the obvious $k$-linear mapping. The action of $U({\mathfrak g})$ on the tensor product of two modules uses the coproduct. This we can sufficiently test (as you also indicated) at the level of ${\mathfrak g}$. But what about $f$?
