Explicit isomorphism of $\mathbb G_m/\{\pm 1 \}$ with $\mathbb G_m$ Let $G$ be a linear algebraic group over an algebraically closed field, and let $H$ be a closed subgroup.  Identify all groups with their closed points.  General theory guarantees the structure of a quasiprojective variety on the space of cosets $G/H$ in the quotient topology.  
This is done by constructing a rational representation $\pi: G \rightarrow \textrm{GL}(V)$, together with an element $0 \neq v \in V$, such that $H$ is the stabilizer in $G$ of the line $[v]$ through $v$, and $\mathfrak h$ is the stabilizer in $\mathfrak g$ the same line.  The representation induces an action of $G$ on $\mathbb{P}(V)$, and $G/H$ is given its variety structure via the resulting bijection with the orbit $G[v]$.  
Let $G = \mathbb{G}_m$, and let $H = \{ \pm 1 \}$.  The quotient $G/H$ is a one dimensional algebraic group consisting of semisimple elements, so it should be isomorphic to $\mathbb{G}_m$.  I'm wondering how one can construct an explicit isomorphism of $G/H$ with $\mathbb{G}_m$.
 A: Assume we aren't in characteristic two.  Let $\lambda$ be a generator of $X(G)$.  Then $2\lambda$  is a generator of $X(G/H)$.
Define an isomorphism of abelian groups $\phi:X(\mathbb{G}_m) \rightarrow X(G/H)$ by $\phi(\chi) = 2\lambda$, where $\chi$ is a generator of $X(\mathbb{G}_m)$.  The morphism of algebraic groups $G \rightarrow \mathbb{G}_m, x \mapsto x^2$ induces a morphism of algebraic groups $f: G/H \rightarrow \mathbb{G}_m$.
We have $\phi(2\lambda) \circ f(xH) =\chi \circ f(xH) =  x^2 = 2\lambda(xH)$.  This shows that $f$ is the morphism of tori corresponding to $\phi$.  Since $\phi$ is an isomorphism, the equivalence of categories between tori and free abelian groups tells us that $f$ is an isomorphism as well.
This is the weird thing: the inverse of $f$ does not look like a morphism of varieties.  At least to me, it does not appear to be continuous.  But it must be.  Explicitly, one chooses for each $x \in \mathbb{G}_m$ a square root $\sqrt{x}$, and defines $f^{-1}(x) = \sqrt{x}H$.
