= Solution
Every $A\in O(1,3)$ preserves the quadratic form $x^TMx$, so it permutes the two connected components of
$$
\{x:x^TMx=-1\}.
$$
The set $X$ is its future component. Since $e_0=(1,0,0,0)\in X$, a Lorentz transformation preserves $X$ exactly when $Ae_0$ has positive zeroth coordinate, equivalently
$$
A_{00}>0.
$$
Thus
$$
O^+(1,3)=\{A\in O(1,3):A_{00}>0\}.
$$
If $A$ and $B$ preserve $X$, then so does $AB$. Moreover, $A$ maps the connected component $X$ onto a connected component; the assumption $A(X)\subseteq X$ forces $A(X)=X$, so $A^{-1}$ also preserves $X$. The identity does as well. Hence $O^+(1,3)$ is a subgroup.
The inequality $A_{00}>0$ defines an open subset of the six-dimensional manifold $O(1,3)$. Every open subset of a manifold is a submanifold of the same dimension. Consequently the <Orthochronous Lorentz group> is
$$
\boxed{\text{a subgroup and a full six-dimensional submanifold of }O(1,3)}.
$$
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