How to calculate the first and second homotopy groups of the following space constructed from $U(4)$ In solving a physics problem, I came across a weird topological space constructed from $U(4)$, the group of $4\times4$ unitary matrices. I want to know the first two homotopy groups of it. Here is how it is defined:
Consider the set $X$ of matrices in $U(4)$
$$X=\{F\in U(4):F=N\Lambda\},$$
where $\Lambda$ is any diagonal unitary matrix, and $N$ has the block form
$$N=\begin{bmatrix}A&A\\B&-B\end{bmatrix},$$
where $A$ and $B$ are $2\times2$ matrices. This, of course, means $N$ must be also unitary, and $A,B$ must satisfy $AA^\dagger=BB^\dagger=\frac{1}{2} I$. 
We can form the quotient space $M=X/\sim$, where $F_1\sim F_2$ if $F_1=F_2\Lambda$, $\Lambda$ being any diagonal unitary matrix. 
Is there a way to calculate $\pi_1(M)$ and $\pi_2(M)$? 
 A: Yes, you can compute them, and $\pi_1(M) = \mathbb{Z}$, $\pi_2(M) = \mathbb{Z}$.
The way I see it, there are three steps in the proof. For convenience, let me call $Y = U(2)\times U(2)$ the set of $N$s of the form above.


*

*Instead of looking at $X/\sim$, let's look at $Y/\sim$ (where I denote with the same symbol the relation on $X$ and its restriction on $Y$). Topologically, this is exactly the same, so the result is still $M$.

*Notice that $Y/\sim$ is simply $U(2)\times U(2)$ quotiented by the diagonal action of the diagonal subgroup $D$ of $U(2)$. From this it follows that $M$ is $(U(2)/D)\times U(2)$ (this does not hold at the group level, if you're quotienting by a normal subgroup, but it holds at the topological level).

*$D = U(1)\times U(1)$, and since $\mathbb{CP}^n = U(n+1)/(U(1)\times U(n))$, so $U(2)/D = S^2$.
From this, since $\pi_1(S^2)$ and $\pi_2(U(2))$ are both trivial (the latter is a general fact for Lie groups, see this question), and $\pi_2(S^2)$ and $\pi_1(U(2))$ are both infinite cyclic.

The first two steps are more or less just general topology, while the third is apparently well-known.
You can also convince yourself that 3. is true since you can see a "fibration" of $U(2)/D$ over $S^1$ such that the generic fiber is $S^1$ and there's one exceptional fiber which is $S^0$ (though this is not a proof, of course).
