Let $G$ be a group with normal subgroup $N \triangleleft G$. Let $\pi: G \to G/N$ be the quotient map.
Is the induced map on the group cohomology with coefficients in $U(1)$, i.e., $\pi^*: H^n(G/N,U(1)) \to H^n(G,U(1))$, injective?
(If this is not true in general, is it perhaps true for finite groups or Lie groups?)
On
Max already gave a counter-example, but here is an even simpler one:
Let $G$ be the quaternion group (i.e., the group of order eight generated by $i$, $j$ and $k$ with $i^2 = j^2 = k^2 =-1$). It has no non-trivial projective rep: $H^2(G,U(1))= 0$. However, it has a normal subgroup $N$ generated by $-1$ such that $G/N \cong \mathbb Z_2 \times \mathbb Z_2$. Since $H^2(\mathbb Z_2 \times \mathbb Z_2,U(1)) = \mathbb Z_2$, it is clear that there exists no injective mapping $$\mathbb Z_2 = H^2(G/N,U(1)) \to H^2(G,U(1)) = 0. $$
It's definitely not true, even for finite groups. In all that follows, all coefficients will be trivial $G$-modules, and $G$ will be finite.
Firsr of all, note that as groups, $S^1\simeq \mathbb{Q/Z}\oplus V$ where $V$ is some $\mathbb{Q}$-vector space. In particular, since $G$ is a finite group, $H^*(G,V)=0$ for $*>0$, so $H^*(G,S^1)\simeq H^*(G,\mathbb{Q/Z})$ for $*>0$.
Then we have a short exact sequence $0\to \mathbb{Z\to Q\to Q/Z}\to 0$ and $H^*(G,\mathbb Q)=0$ for $*>0$ so that $H^*(G,\mathbb{Q/Z}) \simeq H^{*+1}(G,\mathbb Z)$ for $*>0$, all of this being natural in (the finite ) $G$.
So your question becomes (for finite groups) : is $H^*(G/N, \mathbb Z)\to H^*(G,\mathbb Z)$ injective for $*>1$ (knowing that for $*=0$ it trivially is, and for $*=1$ both are $0$ so it is injective too) ? The answer to that is well-known to be no. As per the comments, here's an example :
Consider $G= C_4\times C_2$ ($C_n$ the cyclic group of order $n$) and $G/N = C_2\times C_2$. Then we know that $H^2(C_2,\mathbb Z)\to H^2(C_4,\mathbb Z)$ is multiplication by $2$ from $C_2\to C_4$ and so $H^2(C_2,\mathbb Z)\otimes H^2(C_2,\mathbb Z)\to H^2(C_4,\mathbb Z)\otimes H^2(C_2,\mathbb Z)$ sends $1= 1\otimes 1$ to $2\otimes 1 = 1\otimes 2 = 0$ and in particular isn't injective. By the Künneth formula, it follows that $H^4(C_2\times C_2,\mathbb Z)\to H^4(C_4\times C_2,\mathbb Z)$ is not injective.