Given a (compact) Lie group $G$, persumably disconnected, there exists a short exact sequence
$$1\rightarrow G\_c\rightarrow G\rightarrow G/G\_c\rightarrow 1$$
where $G\_c$ is the normal subgroup which contains all elements in the same connected component as the identity element, and $G/G\_c$ can be thought of as the "finite part" of $G$.
Suppose A is a finite $G/G\_c$ module (as well as a $G$ module).


The question is: Is the cohomology map
$$H^3(G/G\_c, A)\rightarrow H^3(G, A)$$ induced by the projection $p:G\rightarrow G/G\_c$ always injective?


The background is as follow: given any finite group T, a homomorphism $G\rightarrow T$ always factors through $G/G\_c$. I wish to prove (or disprove) the similar statement for $\mathcal{T}$ a "finite" 2-group.



Injectivity of the cohomology map induced by some projection map
Ok, I will follow Fernando's advice and post an answer. I learned the computation below from the beginning of [Pin(2)-equivariant Seiberg--Witten Floer homology and the triangulation conjecture](https://arxiv.org/abs/1303.2354).


The group $\text{Pin}(2) = S^1 \cup jS^1 \subset S^3$ is a subgroup of the unit quaternions. The quotient is the topological space $S^3/\text{Pin}(2) \cong \Bbb{RP}^2$. This ultimately leads one to the fiber sequence


$\Bbb{RP}^2 \to B\text{Pin}(2) \to BS^3 = \Bbb{HP}^\infty$.


Then we have a spectral sequence $H^\*(\Bbb{HP}^\infty; H^\*(\Bbb{RP}^2;\Bbb F\_2)) \implies H^\*(B\text{Pin}(2);\Bbb F\_2)$.


The $E\_2$ page is given by $$\Bbb F\_2[U] \otimes \Bbb F\_2[V]/(V^3) = \Bbb F\_2[U,V]/(V^3).$$ More precisely, when $j \le 2$ we have that $E\_2^{4i,j}$ is 1-dimensional with nonzero generator $U^i V^j$ and otherwise $E\_2^{i,j} = 0$.


Now the only possibly nonzero higher differentials are those of bidegree $(r,1-r)$ where (a) $r$ is divisible by $4$ and (b) $-2 \le 1-r \le 2$. These imply $r = 0$, so there are no nontrivial higher differentials.


Thus $H^\*(B\text{Pin}(2); \Bbb F\_2) = \Bbb F\_2[U,V]/(V^3)$ and in particular $H^3$ is trivial. But $\pi\_0 \text{Pin}(2) = \Bbb Z/2$ has nontrivial group cohomology in all degrees.


*Remark.* Here is why you might think this example is profitable. If $G \to \pi\_0 G$ admits a section, your homomorphism actually is injective. However, $\text{Pin}(2) \to \Bbb Z/2$ admits no section: $(jz)^2 = -1$ regardless of what $z \in S^1$ is. The non-identity component consists entirely of elements of degree 4.


