Does the interior semidirect product of Lie groups $G = N \rtimes H$ respect the projection $G \to H$? Suppose we have some matrix Lie groups $N$ and $H$ both subsets of the $n  \times n$ matrices, and a matrix Lie group $G$ for which we know $G = N H$, $N$ is a normal subgroup of $G$, and $N \cap H = \{ I \}$.
(This is just the definition of the inner semidirect product of ordinary groups.)
In this situation, we know that every element $g \in G$ is uniquely a product $g = n h$ for $n \in N$ and $h \in H$. Do we automatically know that the map $G \to N \times H$ given by this decomposition is smooth?
It is certainly a bijection, but I am worried that it could possibly be pathological in some way.
 A: The answer is yes if everything is finite dimensional, but I do not know an elementary argument and would be happy to accept one instead of this (or a proof of the general case). 
The argument
For $K$ any Lie group realise the Lie algerba $\text{Lie}(K)$ of $K$ as those matrices $X$ for which $e^{t X} \in K$ for all $t \in \mathbb{R}$. Then there is a natural map of underlying vector spaces from $\text{Lie}(N) \times \text{Lie}(H) \to \text{Lie}(G)$ defined by $(X, Y) \mapsto X + Y$; the fact that $X + Y \in \text{Lie}(G)$ follows directly from the Lie product formula. We can turn this map into a Lie algebra homomorphism by upgrading $\text{Lie}(N) \times \text{Lie}(H)$ into the semidirect product $\text{Lie}(N) \rtimes \text{Lie}(H)$ over the $\text{ad} : H \to \text{Der}(N)$ map (the semidirect product of Lie algebras has underlying vector space the Cartesian product of underlying vector spaces).
Call the resulting Lie algebra homomorphism $\iota : \text{Lie}(N) \rtimes \text{Lie}(H) \to \text{Lie}(G)$, which is still defined by $(X, Y) \mapsto X + Y$ (it is easy to check that this map is a morphism of Lie algebras). Moreover, it is easy to see that this map is injective because $N$ and $H$ have trivial intersection.
At this point, we now upgrade $G$, $N$, and $H$, from Linear Lie groups into Lie groups, i.e. smooth manifolds with are groups and which have a smooth multiplication. Because $\text{exp} : \text{Lie}(K) \to K$ is a local homeomorphism at the identity for any Lie group $K$, we have $\dim \text{Lie}(K) = \dim K$ for all such. In order to establish $\dim(N) + \dim(H) = \dim(G)$ it now sufficies to show that $\dim(N) + \dim(H) \geq \dim(G)$ (the other direction of the inequality follows since $\iota$ is an injective map of finite dimensional vector spaces implies that $\dim \text{Lie}(N) + \dim \text{Lie}(H) \leq \dim \text{Lie}(G)$). But the multiplication $m : G \times G \to G$ restricts to a map $\overline{m} : N \times H \to G$ which is surjective by the definition of $G$, so $G$ is the image of a smooth manifold of dimension $\dim(N \times H) = \dim(N) + \dim(M)$, so our desired inequality follows from Sard's theorem.
This all shows that $\dim(G) = \dim(N) + \dim(H)$, from which it follows that the map $\iota$ is an isomorphism. Turn $N \times H$ into a group via taking the outer semidirect product (using the conjugation action). Then via the isomorphisms from the respective Lie algebras (by our definition) of $N \rtimes H$ and $G$ we see that the map of tangent spaces induced by $\overline{m}$ at the identity is an isomorphism (this just follows because $\iota$ is).
The inverse function theorem now applies, showing that the map $p : G \to N \rtimes H$ inverse to $\overline{m}$ is smooth in a neighbourhood of the identity. By translating this neighbhourhood we see that $p$ is smooth everywhere, and forgetting about the group structure on $N \rtimes H$ we get the decomposition map $p : G \to N \times H$ of the question statement, as desired.

Appendix on Sard's theorem
Here we just elaborate slightly the piece of the argument where we used Sard's theorem.
Lemma. Let $f : M \to N$ be a surjective smooth map of smooth manifolds. Then $\dim N \leq \dim M$.
Proof. Let $X$ be the subset of points of $M$ where $f$ has rank strictly less than $\dim N$. Sard's theorem for manifolds then exactly says that $f(X)$ has measure zero in $N$. As a consequence, $f(X)$ has empty interior in $N$. So long as $N$ is not discrete we can therefore find a point in $n \in N$ and a preimage $m \in M$ (i.e. with $f(m) = n$) such that the derivative of $f$ at $m$ has rank $\dim M$. A consequence of this in particular is that both manifolds $M$ and $N$ must have dimension at least $\dim M$ (their tangent spaces at each point have the same dimension, and the rank of a linear map is bounded by the minimum of the dimensions of its domain and codomain), as claimed.
A: First of all, you do not need the assumption that your groups are matrix groups, just the fact that they are Lie groups. Next, consider the topological product $N\times H$ and the continuous map $\mu: N\times H\to G$ sending $(n,h)$ to $nh$. By the assumption, this map is a continuous bijection (continuity comes from the continuity of the multiplication in $G$. Since $N, H, G$ are topological manifolds, $\mu$ is a homeomorphism (by the invariance of domain theorem). In particular, projections $G\to H, G\to N$ given by the composition of $\mu^{-1}$ with the projections $N\times H$ to its factors, are also continuous. Verifying smoothness takes more work.
