Previously, I ask for comments/suggestions on setting up the calculation in Quotient space, homogeneous space, and higher homotopy groups. There, however, I was looking for whatever methods and tools you recommend.

Being more down to earth now, I had decided to do the explicit calculation simply based on the exact sequence and see whatever clues I can find. I like to present my results and see whether this leads to a complete answer.

The goal is the same. But since this new post contains more information than the previous ones, I believe that these will be useful to post as a sequel question suitable for other people who are interested in this too.

Question: For this precise quotient space $ G/H, $ (Can we think it of as some more familiar manifold?)

- What is the homotopy group? $$ \pi_j(G/H)=? $$ for $j=1,2,3,4,5,6$.

Here we still take

$${G}=\frac{SU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2} \times U(1)}{(\mathbb{Z}_N)^2}\equiv \frac{G_0}{G_1},$$ $$=\frac{PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2}}{(\mathbb{Z}_N)}\equiv \frac{G_0'}{G_1'} $$ and the subgroup $H$ of the $G$ as: $$H=\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2=PSU(N)_{A,B} \times \mathbb{Z}_2.$$ Note $H$ is not a normal subgroup of $G$. So generally $G/H$ is not a nice Lie group but just a quotient space. $SU(N)$ is the special unitary group written in terms of rank-$N$ matrix, and $PSU(N)$ is the projective special unitary group written in terms of rank-$N$ matrix. See more details here. The $\mathbb{Z}_N \equiv \mathbb{Z}/(N\mathbb{Z})$ is a cyclcic elementary abelian group of order $N$.

Here $N$ is an integer, we can take $N=3$ an odd integer throughout this post.

I like to determine the homotopy groups from the long exact sequence $$ ... \to \pi_n(H) \to \pi_n(G) \to \pi_n(G/H) \to \pi_{n-1}(H) \to \pi_{n-1}(G) \to \pi_{n-1}(G/H) \to .... \tag{(!!)} $$

So I start from computing the homotopy group of $H$: We can see that $$\pi_{0}(H)=\pi_{0}(\mathbb{Z}_2)=\mathbb{Z}_2,$$ $$\pi_{1}(H)=\pi_{1}(PSU(N))=\mathbb{Z}_N,$$ $$\pi_{2}(H)=\pi_{2}(PSU(N))=0,$$ $$\pi_{3}(H)=\pi_{3}(PSU(N))=\mathbb{Z},$$ $$\pi_{4}(H)=\pi_{4}(PSU(N))=0,$$ $$\pi_{5}(H)=\pi_{5}(PSU(N))=\mathbb{Z},$$ $$\pi_{6}(H)=\pi_{6}(PSU(N))=\mathbb{Z}_6, \text{ if } N=3$$

Then I compute the homotopy group of $G$: We can see that $$\pi_{0}(G)=0,$$ $$\pi_{1}(G)=\mathbb{Z}\times \mathbb{Z}_N,$$ Here $\pi_{1}(G)$ is related to the comment here given by @YCor and @Dan Ramras, who says that "the fundamental group is isomorphic to the preimage $\tilde{G}_1$ of $G_1$ in the universal covering of $G_0$." I have another way to look at it, and I agree that $\pi_{1}(G)=\mathbb{Z}\times \mathbb{Z}_N$ seems correct (Further criticism is welcome?).

$$\pi_{2}(G)=\pi_{2}(PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2})=0,$$ $$\pi_{3}(G)=\pi_{3}(PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2})=\mathbb{Z}^3,$$ $$\pi_{4}(G)=\pi_{4}(PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2})=0,$$ $$\pi_{5}(G)=\pi_{5}(PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2})=\mathbb{Z}^3,$$ $$\pi_{6}(G)=\pi_{6}(PSU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2})=\mathbb{Z}_6^3, \text{ if } N=3$$

By the method in Eq.(!!), we can find that: $$ \pi_2(G/H)=0 \to \pi_1(H)=\mathbb{Z}_N \to \pi_1(G)=\mathbb{Z} \times \mathbb{Z}_N \to \pi_1(G/H) \to \pi_{0}(H)=\mathbb{Z}_2 \to \pi_{0}(G)=0 \to \pi_{0}(G/H)=0, \tag{(a-1)} $$ so $$ 0 \to \mathbb{Z}_N \to \mathbb{Z} \times \mathbb{Z}_N \to \pi_1(G/H) \to \mathbb{Z}_2 \to 0 \to 0, \tag{(a-2)} $$ Notice Eq.(a-2) boils down to $$ 0 \to 0 \to \mathbb{Z} \to \pi_1(G/H) \to \mathbb{Z}_2 \to 0 \to 0, \tag{(a-3)}, $$ Eq.(a-3) is related to classified the $H^2(B\mathbb{Z}_2,\mathbb{Z})$. There are 3 choices, $\pi_1(G/H)=\mathbb{Z}\times \mathbb{Z}_2, \mathbb{Z}, \mathbb{Z}\rtimes \mathbb{Z}_2$. These 3 cases are mentioned in a nice post https://math.stackexchange.com/questions/1040450/extensions-of-mathbbz-by-mathbbz-2. I don't have a definite way to determine the unique answer, so this I require whatever advice you suggested.

Then we also have $$ \pi_7(G/H)=0 \to \pi_6(H)=\mathbb{Z}_6 \to \pi_6(G)=\mathbb{Z}_6^3 \to \pi_6(G/H)$$ $$\to \pi_5(H)=\mathbb{Z} \to \pi_5(G)=\mathbb{Z}^3 \to \pi_5(G/H) \to \pi_{4}(H)=0 \to \pi_{4}(G)=0 \to$$ $$ \pi_4(G/H)=0 \to \pi_3(H)=\mathbb{Z} \to \pi_3(G)=\mathbb{Z}^3 \to \pi_3(G/H) \to \pi_{2}(H)=0 \to \pi_{2}(G)=0 \to \pi_{2}(G/H)=0, \tag{(b-1)} $$ We can determine $\pi_3(G/H)$, $\pi_5(G/H)$ and $\pi_6(G/H)$ from $$ 0 \to \mathbb{Z} \to \mathbb{Z}^3 \to \pi_3(G/H) \to 0 \tag{(b-2)} $$ $$ 0 \to \mathbb{Z}_6 \to \mathbb{Z}_6^3 \to \pi_6(G/H) \to \mathbb{Z} \to \mathbb{Z}^3 \to \pi_5(G/H) \to 0 \tag{(b-3)} $$

My claim/tentative answer is that: $$ \pi_{0}(G/H)=0 $$ $$ \pi_{1}(G/H)=\mathbb{Z} \times \mathbb{Z}_2 (?) $$ $$ \pi_{2}(G/H)=0 $$ $$ \pi_{3}(G/H)=\mathbb{Z}^2 $$ $$ \pi_{4}(G/H)=0 $$ $$ \pi_{5}(G/H)=\mathbb{Z}^2 (?) $$ $$ \pi_{6}(G/H)=\mathbb{Z}_6^2 (?) $$

In particular, I am now so sure $\pi_{5}(G/H)$ and $\pi_{6}(G/H)$, am I too innocent about my answers?

P.S. If you are not familiar with the homotopy groups of unitary Lie group, you can find the info here.