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Since there is a short exact sequence of groups: $$1\to\mathbb{Z}_n\to SU(n)\to PSU(n)\to1,$$ we have a fiber sequence: $$B\mathbb{Z}_n\to BSU(n)\to BPSU(n)\stackrel{\iota}{\to} B^2\mathbb{Z}_n\to\cdots.$$

My question: If we are given a 3-manifold $M$ and a map $g:M\to B^2\mathbb{Z}_n$, can we find a map $f:M\to BPSU(n)$ such that $g$ is homotopic to $\iota\circ f$?

Edit: Thanks to Tyrone's useful comment, the lifting problem is true even for any 4-manifold $M$ since $\pi_k BSU(n)=0$ for $k\le3$.

The general statement is: If $F\to E\stackrel{p}{\to} B$ is a fibration and $\pi_kF=0$ for $k\le n-1$, we consider the lifting problem of a map $g:M\to B$ to $f:M\to E$ such that $p\circ f=g$, then a lift over the $n$-skeleton of $M$ always exists.

Thank you!

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    $\begingroup$ $BSU_n$ is $3$-connected, so you see that the map $BPU_n\rightarrow K(\mathbb{Z}_n,2)$ is actually $4$-connected. $\endgroup$ – Tyrone Nov 4 '18 at 13:24
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    $\begingroup$ You can also use elementary obstruction theory to see what you want quite easily. $\endgroup$ – Tyrone Nov 4 '18 at 13:24

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