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Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.

EDIT: For non-central groups one can still use the basic gerbe $\mathcal{G^1}$, but I don't know if the obstruction classes are then still accessible for calculations.

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis (Wayback Machine).

EDIT: For non-central groups one can still use the basic gerbe $\mathcal{G^1}$, but I don't know if the obstruction classes are then still accessible for calculations.

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.

EDIT: For non-central groups one can still use the basic gerbe $\mathcal{G^1}$, but I don't know if the obstruction classes are then still accessible for calculations.

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|Z|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.

Here is a solution for $G$ a compact, simple, connected, simply-connected Lie group and $\Gamma$ a subgroup of the center of $G$.

The group $H^3(G,Z)$ classifies $U(1)$-gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$-equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$.

The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lie-theoretical construction in the framework of bundle gerbes. The existence of $\Gamma$-equivariant structures can then be checked by inspection of a certain obstruction class.

This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $|\Gamma|$ is odd, or $\frac{n}{|\Gamma|}$ is even. That's how I came up with the comment to the question.

All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 28-55. An overview is in Table 5.1 on page 143 of my phd thesis.