Fundamental group of $\{\text{diag}(A,B) \colon A,B \in SO(2)\} /\{I,-I\}$ Let $K=\{\text{diag}(A,B) \colon A,B \in SO(2)\}\subset SO(4)$, where 
$$ 
\text{diag}(A,B)=\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}\text{.}
$$
What is the fundamental group of $K/\{I,-I\}$? Does this space have a name? Any information of this space is welcome.
I know that $PSO(4)=SO(4)/\{I,-I\}$ has fundamental group $\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}$ and $\pi_{1}(PSO(2))=\mathbb{Z}$ so my first atempt was prove that $K/\{I,-I\}$ is homeomorphic to $PSO(2)\times PSO(2)$ but this is clearly not true.

It is well-known that:

Any compact, connected abelian Lie group is isomorphic to some $T^k$.

Now because $K$ is compact (why?), connected and abelian (why?), it must be isomorphic to $T^k$ for $k=\dim (K)= 1+1=2$. i.e. $K=\Bbb S^1\times \Bbb S^1$.
Like @JasonDeVito's argument in this post:

$\mathbb{Z}/2\mathbb{Z}\subseteq T^2$ generated by $\langle (\pi, \pi)\rangle$ (note that $-I\sim \tau(x,y):=(x+\pi,y+\pi)$) is normal (since $T^2$ is abelian, so we can form the quotient $K':=T^2/(\mathbb{Z}/2\mathbb{Z})$).  Being the continuous homomorphic image of $T^2$, $K'$ must be a compact abelian Lie group, so it must be isomorphic to $T^2$ as a Lie group.  In particular, $K'$ is diffeomorphic to $T^2$.

So $\pi_1(K')=\pi_1(T^2)$.
