# Lifting problem of a map $g:M\to B^2\mathbb{Z}_n$ to $f:M\to BPSU(n)$

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!

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