And sorry if this is naive, I'm also having trouble proving point 1 rigorously, clearly we have the commutative diagram $$\require{AMScd} \begin{CD} Y @>{i}>> X;\\ @V{q_2}VV@VV{q_1} V\\ Y/\sim @>{f}>>X/\sim \end{CD},$$ where $i$ is inclusion, $q_1,q_2$ are quotient maps, and $f$ is the induced map, it is easy to prove f is bijective and continuous, but I failed to prove it is a homeomorphism. If I can prove $q_1\circ i$ is a quotient map , then it should be done, but my point-set topology failed me.

Ah, thanks. For $U(2)/D\cong S^2$, I guess an explicit construction would be $A\mapsto A/\sqrt{\det A}\in SU(2)$,then followed by a Hopf map?

Hi Marco, I just tried to follow your argument in detail, and I had difficulty verifying the claim $Y/\sim\cong (U(2)/D)\times U(2)$. For $Y/\sim$ we are identifying $(A,B)\sim(A\Lambda, B\Lambda)$ where $\Lambda$ is any 2 by 2 diagonal unitray matrix, while it seems for $(U(2)/D)\times U(2)$ we are identifying $(A,B)\sim(A\Lambda, B)$, how do I see there is a homeomorphism between them? Is there an explicit construction?

@FernandoMuro: Done. Just curious, what is popping up?

And just to confirm, this would mean $\pi_1=Z$ and $\pi_2=Z$, right?

Thanks and +1. I'll need to digest this and come back to it after I learn some more homotopy theory

@MarcoGolla: It'd be nice if you can expand it into an answer, it is a shame that I'm quite ignorant about the machinery related.

@MarcoGolla: This looks like a promising simplification, but I just started peeking into some general homotopy theory so I'm not able to tell if an answer is immediate using this simplification, would you care to elaborate?

@FernandoMuro: You might be right. I'm really not sure if it is a subspace, let me edit the post.

@JohannesNordström: I meant N has that form and N is unitary at the same time. In other words, $AA^\dagger=BB^\dagger=\frac{1}{2} I$