Prove isomorphism between universal enveloping algebras I have the following problem I can't solve. I am following a couse in Lie Algebras, and one of the exercises is prove the following isomorphism: Given a semisimple lie algebra $\mathfrak{g}$ over a field $\mathfrak{k}$ of characateristic 0, and let $\mathfrak{h}$ be a toral maximal subalgebra, then by the cartan-chevalley decomposition, $\mathfrak{g}$ splits as $\mathfrak{g}=\mathfrak{n}^- \oplus \mathfrak{h} \oplus \mathfrak{n}$. Let $\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}$. Then one has that the map:
$$
f:\mathfrak{U}(\mathfrak{n}^-)\otimes\mathfrak{U}(\mathfrak{b})
\rightarrow
\mathfrak{U}(\mathfrak{g})
$$
$$
x\otimes y\longrightarrow
xy
$$
The exercise is to prove that this map is in fact an isomorphism. Where $\mathfrak{U}(\mathfrak{g})$ is the universal enveloping algebra of the $\mathfrak{g}$ and respectively for the rest.
I am very confused, because if you remove the $\mathfrak{U}$ part this map is no longer injective, since it's easy to find $x\in\mathfrak{n},y\in\mathfrak{b}$ so that $xy=0$,
but I don't understand how that is possible without violating the injectivity of $f$.
Any help is appreciated. Thanks to everyone.
P.D: For clarification, the Cartan-Chevalley decomposition establishes than
$\mathfrak{g}=
\mathfrak{h}
\oplus
\bigoplus_{\alpha\in\Phi}
\mathfrak{g}_\alpha
$
where $\mathfrak{h}$ is a maximal abelian subalgebra of ad-diagonalizable elements. (This just means that the linear map $ad(x)$ is diagonalizable).
Furthermore
$\mathfrak{g}_\alpha=\{
x\in\mathfrak{g} : [h,x]=\alpha(h)x
\}
$. Where $\Phi\subset\mathfrak{h}^*$ and $\Phi$ can
be split into $\Phi=\Phi^+\cup\Phi^-$. Then we define
$n
=
\bigoplus_{\alpha\in\Phi^+}
\mathfrak{g}_\alpha
$
and
$n^-
=
\bigoplus_{\alpha\in\Phi^-}
\mathfrak{g}_\alpha
$.
 A: In fact, if $L$ is any Lie algebra over any field $k$, and $L_1, L_2$ are subalgebras such that $L = L_1 \oplus L_2$ as $k$-vector spaces, then the $k$-linear map $f$ given by
$$U(L_1) \otimes_k U(L_2) \rightarrow U(L)$$
$$x \otimes y \mapsto xy$$
is an isomorphism of $k$-vector spaces (and hence of $(U(L_1),U(L_2))$-bimodules).
As noticed in the first bullet point in this question, this follows from the Poincaré-Birkhoff-Witt theorem, compare e.g. Bourbaki, Lie Groups and Algebras ch. 1 §7. Indeed, it is a corollary of PBW (in Bourbaki, Corollary 3) that if $(e_1, ..., e_n)$ is an ordered $k$-basis of a Lie algebra $\mathfrak g$, then the monomials $e_1^{r_1} \cdot ... \cdot e_n^{r_n}$ are a $k$-basis of $U(\mathfrak g)$. From this the assertion (which is stated in greater generality in Bourbaki's Corollary 6) follows. Namely, if now $(x_1, ..., x_m)$ is an ordered $k$-basis of $L_1$, and $y_1, ..., y_n$ one of $L_2$, then on the one hand PBW and tensor products over fields tell us that the set of all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \otimes y_1^{l_1} \cdot ... \cdot y_n^{l_n}$ is a $k$-basis of $U(L_1) \otimes U(L_2)$, on the other hand the images of these elements under $f$, namely, all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \cdot y_1^{l_1} \cdot ... \cdot y_n^{l_n}$, again by PBW are a $k$-basis of $U(L)$, because by assumption $(x_1, ..., x_m, y_1, ..., y_m)$ is an ordered $k$-basis of $L$.
Now apply this to $L=\mathfrak g, L_1 = \mathfrak{n}^-, L_2= \mathfrak b$.

To see where your supposed counterargument goes wrong (copied from comments and discussion):
Your computation of $xy=0$ for certain $x, y \neq 0$ seems to be happening in $M_3(\mathfrak k)$ (or $End_{\mathfrak k} ({\mathfrak k}^3)$). That matrix ring is an enveloping associative algebra of $\mathfrak{sl}_3$, and maybe likewise would be certain associative subalgebras $B:= \pmatrix{*&*&*\\0&*&*\\0&0&*}$, $N^-:=\pmatrix{0&0&0\\*&0&0\\*&*&0}$ corresponding to $\mathfrak b, \mathfrak n^-$. These algebras are quotients of the respective universal enveloping algebras (by the latter's universal property, although surjectivity might be more subtle), but of course a product being zero in a quotient tells us little about the product being zero in the original ring.
What this computation shows, then, is that even though there are canonical surjections of associative algebras
$$U(\mathfrak n^-) \twoheadrightarrow N^-, U(\mathfrak b) \twoheadrightarrow B, U(\mathfrak g) \twoheadrightarrow M_3(k)$$
the map $$N^- \otimes_{\mathfrak k} B \rightarrow M_3(k)$$
induced via those projections from the given map $U(n^-) \otimes U(b) \simeq U(g)$, is no longer an isomorphism. But viewed this way, that is maybe not as surprising.
Full disclosure, I have made the mistake of mixing up standard enveloping matrix algebras with the universal enveloping algebra before, and I still sometimes do calculations in those matrix rings to get a first feeling for the universal enveloping one, but it's important to remember that the matrix algebras (like the above $M_3(\mathfrak k), B, N^-$) are just quotients -- and for that matter, with a lot factored out -- of the "much, much bigger" universal enveloping algebras. In particular, note that the above PBW corollaries also imply that $x^my^n \neq 0$ in $U(\mathfrak g)$ for all $m,n \in \mathbb N$ and any $x \neq 0 \neq y \in \mathfrak g$.
