Is the conformal group $\mathrm{Conf}(p,q) ≃ \mathrm{O}(p + 1, q + 1) / \mathbb{Z_2} $ connected or not? In a pseudo-Euclidean space $\mathbf{E}^{p,q}$, the conformal group is $\mathrm{Conf}(p, q) ≃ \mathrm{O}(p + 1, q + 1) / \mathbb{Z_2}$. See this. Or some notes  say that $\mathrm{Conf}(p, q) ≃ \mathrm{SO}(p+1,q+1)$.
But we know $\mathrm{O}(p,q)$ has $4$ connected components and $\mathrm{SO}(p,q)$ has $2$ connected components for $p,q\ge1$. See this.
So if this notation $\mathrm{Conf}(p, q) ≃ \mathrm{O}(p + 1, q + 1) / \mathbb{Z_2}≃\mathrm{SO}(p+1,q+1)$ is correct, then $\mathrm{Conf}(p,q)$ should have $2$ connected components.
However from every textbook of CFT, they firstly derived the Lie algebra (commutation relation) of conformal symmetries. It's isomorphic to $\mathfrak{so}(p+1,1+1)$ or $\mathfrak{o}(p+1,q+1)$ (It doesn't matter because at Lie algebra level they are same.) Then they will say the conformal group is $\mathrm{SO}(p+1,q+1)$ or $\mathrm{O}(p+1,q+1)/\mathbb{Z}_2$. (Francesco's CFT p98)
It puzzles me. Since if they only want to talk about the connected component of conformal group, then neither $\mathrm{SO}(p+1,q+1)$ nor $\mathrm{O}(p+1,q+1)/\mathbb{Z}_2$ is connected. If they want to talk about the full conformal symmetry including the discrete conformal symmetry $P,T$ and inversion $x^\mu \rightarrow x^\mu/x^2$,  then I think the full conformal group should be $\mathrm{O}(p+1,q+1)$. Is it right?
 A: Firstly, the answer is that $\mathrm{Conf}(p,q)$ is connected. By definition, the conformal group $\mathrm{Conf}(p,q)$ is the connected component that contains the identity in the group of conformal diffeomorphisms of the conformal compactification of $\mathbb{R}^{p,q}$.
To my feeling, the confusion comes from the definition of these groups. Let's recall that $\mathrm{SO}(p+1,q+1)$ is defined as the connected component of the identity in group $\mathrm{O}(p+1,q+1)$. Notice that: -1 may lie in this group. And $\mathrm{SO}(p+1,q+1)$ is a subgroup of 
$$\{\Lambda\in \mathrm{Conf}(p+1,q+1) : \det \Lambda =1\}.$$
These two facts may against your instincts obtained from usual linear algebra with the metric $(+,\cdots,+)$, but they are true.
What you have written $\mathrm{O}(p+1,q+1)/\{\pm 1\}$ is actually the group of all conformal transformations of the conformal compactification of $\mathbb{R}^{p,q}$. The conformal group is the connected subgroup containing $1$ in this group: (i) if $-1$ is not in the connected component containing 1 of $\mathrm{O}(p+1,q+1)$, we have $\mathrm{Conf}(p,q)\simeq \mathrm{SO}(p+1,q+1)$, (ii) if $-1$ is contained in the connected component containing 1 of $\mathrm{O}(p+1,q+1)$, we have $$\mathrm{Conf}(p,q)\simeq \mathrm{SO}(p+1,q+1)/\{\pm 1\}.$$
