Lorentz group is the group of linear transformations that leave the quadratic form $q(x_0, x_1, x_2, x_3) = -x_0^2 + x_1^2 + x_2^2 + x_3^2$ invariant. Linear transformations $A$ with $\det A = +1$ and $A_{00} > 0$ are called orthochronous proper transformations.
If we call the original group $O(1,3)$ and the restricted group $SO^+(1,3)$, I want to show that $SO^+(1,3)$ is a normal subgroup of $O(1,3)$ and the quotient group is isomorphic to the Klein 4-group.
I'm trying to do so by proving three little lemmas first:
Lemma 0: $SO^+(1,3)$ is connected I have no idea why... It is sufficient to prove that $SO^+1(1,3)$ is path-connected. So, for any orthochornous proper transformation, I should somehow deform it in a way that it ends up being identity. But I don't know how.
Lemma 1: In a Lie group G, the connected component containing identity element is a subgroup.
Proof: Suppose $H$ is the connected component of $G$ containing the identity element of the group. Then $h^{-1}H$ is a clopen connected set (because inversion and multiplication are homeomorphisms/diffeomorphisms as maps). Therefore, $h^{-1}H$ is connected and contains identity, i.e. $H \cap h^{-1}H$ is not empty and hence $h^{-1}H=H$, i.e. $h^{-1} \in H$
Now suppose $h_1$ is any element in $H$ and $h_2$ is any other element in $H$. Consider $h_2H$. Since $h_2$ is in $H$, so is $h_2^{-1}$. Therefore $h_2H$ contains identity and hence, like above, $h_2H=H$. Therefore, for any $h_1 \in H$, there exists $h' \in H$ such that $h_2h_1 = h'$, i.e. $H$ is a non-empty set closed under multiplication and inversion so it must be a group.
Lemma 2: The connected component containing identity in a lie group is a normal subgroup.
Proof: for any $g \in G$, consider $gHg^{-1}$. $gHg^{-1}$ is connected and it contains the identity element, hence, $gHg^{-1}=H$ (the argument is very similar to the above one)
Now, if it's proved that $SO^{+}(1,3)$ is connected, and my two other lemmas were true, it would imply the result I'm looking for. But I'm not sure about how I can prove $SO^+(1,3)$ is connected.