Here are statements that I attempt to make sure I am doing things on the right track.
Below we take $N=3$ for all $N$.
- Consider the group:
$$G=\frac{SU(N)_A \times SU(N)_B \times U(1)}{(\mathbb{Z}_N)^2},$$
this group can be understood as a triplet $$(g_A, g_B, e^{i \theta}) \in SU(N)_A \times SU(N)_B \times U(1),$$ such that $(e^{i \frac{2\pi}{N}}, 1, e^{-i \frac{2\pi}{N}})$ is the first $\mathbb{Z}_N$ generator mod out in $G$, while $(1,e^{i \frac{2\pi}{N}}, e^{-i \frac{2\pi}{N}})$ is the second $\mathbb{Z}_N$ generator mod out in $G$. Namely (1) the center of $SU(N)_A$, (2) the center of $SU(N)_B$ and (3) the $e^{i \frac{2\pi}{N}}\in U(1)$ overlap, thus we only mod out the twice redundant $(\mathbb{Z}_N)^2$.
Now consider the subgroup $H$ of the $G$ as:
$$H=\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2,$$
where this group can be understood as a doublet $(g_{A,B}, g_1) \in SU(N)_{A,B}\times \mathbb{Z}_2= (SU(N)_{A,B}, \{\pm 1 \}) =H \subset G,$ where there is a one-to-one correspondence between $$(g_{A,B}, g_1) \in H$$ and $$(g_{A,B},g_{A,B}^*, g_1) \in G$$ with $g_{A,B}=g_A=g_B^*$, where $g_B^*$ means the complex conjugation (without the transpose $T$) of $g_B $.
Finally, we need $\frac{SU(N)_{A,B}}{\mathbb{Z}_N}$ to mod out ${\mathbb{Z}_N}$, where we consider $(g_{A,B}',g_{A,B}'^*)=(e^{i \frac{2\pi}{N}},e^{-i \frac{2\pi}{N}})\mathbb{I} \in {\mathbb{Z}_N}$; this particular rank-$N$ diagonal matrix is the ${\mathbb{Z}_N}$ we modded out .
We have that $$\frac{SU(N)_{A,B}}{\mathbb{Z}_N} \subset SU(N)_A \times SU(N)_B, $$ $$ \mathbb{Z}_2 = \{\pm 1 \} \subset U(1). $$
So this explains how $H$ is embedded as a subgroup in $G$.
Question:
What is the precise topological space of the set of left coests $$ G/H=\frac{\frac{SU(N)_A \times SU(N)_B \times U(1)}{(\mathbb{Z}_N)^2}}{\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2}? $$ Is this certain smooth homogeneous space like a sphere or a complex/real projective space?
What is the homotopy group? $$ \pi_j(G/H)=?, $$ for $j=1,2,3,4,5$.
Some background info that I made attempt and showed for your free use:
(1). $\pi_i(U(N))=\pi_i(\frac{SU(N)\times U(1)}{\mathbb{Z}_N})$: $$\pi_m(U(N))=\pi_m(SU(N)), \text{ for } m \geq 2$$ $$\pi_1(U(N))=\mathbb{Z}, \;\;\pi_1(SU(N))=0,$$
(2). $\pi_1(\mathbb{Z}_N)=\mathbb{Z}_N$ and $\pi_i(\mathbb{Z}_N)=0$ for $i\geq 2$.
This is not a complete answer, since on the one hand, I think there is a small error in your question and second the various quotients by centers got too complicated for me. (They only lead to some finite coverings and thus quotients in $\pi_1$ and isomorphisms in the higher $\pi_i$'a anyway.) What I think is an error in your question is that you can view $$\{(g,g^*):g\in SU(N)\}$$ as a subgroup in $SU(N)\times SU(N)$ since $$(gh)^*=h^*g^*$$ but not $g^*h^*$.
But for any Lie group you can take $G\times G$ and naturally view $G$ as the diagonal subgroup $\Delta:=\{(g,g):g\in G\}\subset G\times G$. As one may expect naively, the quotient $(G\times G)/\Delta$ is diffeomorphic to $G$ itself. (This is the standard way to view a Lie group as a symmetric space.) Explicitly, this diffeomorphism can be seen by considering the left action of $G\times G$ on $G$ defined by $(g,h)\cdot\tilde g:=g\tilde gh^{-1}$. Denoting by $e$ the neutral element of $G$, you see that $(g,e)\cdot e=g$, so the action is transitive and that $(g,h)\cdot e=e$ if and only if $gh^{-1}=e$ and hence $g=h$. Thus the isotropy group is $\Delta$ and hence $G\cong (G\times G)/\Delta$.
I am sure that a small adaption of this construction can be used to identify the homogeneous space you are interested in with something that is close to $U(N)$ or $SU(N)$, which should also allow identifying the homotopy groups.