Is taking inverse automatically well-defined? By the usual definition, Lie group is a manifold $G$ with a group structure on it such that the multiplication $m\colon G\times G\to G$ and taking inverse $i\colon G\to G$ are both smooth maps. But it is not difficult to proove that actually we need only the smoothness of the multiplication map $m$. The proof uses the fact that the preimage $m^{-1}(1)$ of $1\in G$ is a smooth submanifold of $G\times G$, and the inverse function theorem.
Now I am wondering, is that true for algebraic groups? It might be obvious, but it's not obvious to me, since in this case we can't use methods from the smooth case. More precisely, suppose $G$ is a variety with a group structure such that $m\colon G\times G\to G$ is a regular map. Is it true that it automatically guarantees regularity of the inversion map $i\colon G\to G$ ?
Thank you very much for your help!
 A: I think that the answer is yes.
In my argument I will use the following fact: a morphism of varieties which is a bijection on points and whose target is normal (e.g. smooth) is necessarily an isomorphism of varieties.  (This follows from Zariski's main theorem.)
I will first prove that $G$ is necessarily smooth.  Since $G$ is a variety it is generically smooth, i.e. it contains a non-empty Zariski open subset $U$ consisting of smooth points.  Let $g\in G$ be an arbitrary point, and choose some $h \in G$ such that $gh \in U$ (possible because $G$ is a group under multiplication and $U$ is non-empty).
Then $m^{-1}(U) \cap (G \times h)$ is a Zariski open subset of $G \times h$,
say $V \times h$, such that $g \in V$.  Then right multiplication by $h$
gives a regular map $V \to U$ which is a bijection on points, with $U$ being
smooth.  Thus this map must be an isomorphism of varieties, and so $V$ is also smooth.  Thus every point of $G$ is smooth.
Now consider the preimage $\Gamma$ of $1$ under $m$; this is a subvariety of $G \times G$,
which set-theoretically consists of the graph of $i$.  In particular, each of the projection maps is a set-theoretic bijection $\Gamma \to G$.
Since $G$ is smooth (by what we proved above), the morphism $\Gamma \to G$ is a morphism of varieties which is a set-theoretic bijection with smooth target: it is thus necessarily an isomorphism.

Added: the fact about bijections requires a char. zero assumption as stated, since, in the application of ZMT that underlies it, it assumes that the quasi-finite morphism in question if generically etale (so that we can go from set-theoretic bijection to birational morphism).  This is automatic in char. zero, but can fail in positive char.
My guess is that with a little more care the above argument should actually work in char. $p$ too; on the one hand mult. by $g$ induces a morphism $G \to G$ which is set-theoretically a bijection, so on function fields it gives a purely inseparable embedding $k(G) \hookrightarrow k(G)$ of some degree $p^{n_g}$.  Now $n_g$ should be generically a constant, but on the other hand, composing mult. by $g$ with mult. by $h$ shows that $n_{gh} = n_g n_h$.  Taking $g$ and $h$ to be generic gives that $n_g = 1$ for generic $g$, hence mult. by $g$ is generically etale of degree $1$, i.e. generically an isomorphism.  This should give the generic etaleness needed for the ZMT argument to go through.  I didn't think this through carefully though.
