Irreducible representations of SO(3) I have just read that all irreducible representations of $SO(3)$ are in bijective correspondance with the irreducible representations $V_n$ of $SU(2)$ in which $-E$ acts as the identity.
Can someone help me how to understand that this indeed holds.
As far as I know, there is only an epimorphism $\pi : SU(2) \to SO(3)$, having kernel {$E,-E$} (diagonal matrices.) 
But not sure how this helps to get bijection between their irreducible representations nor what exactly is meant when saying "in which $-E$ acts as the identity".
Many thanks for any help
 A: This a consequence of the universal property of quotients and the first isomorphism theorem.
A representation of $SU(2)$ is the same thing as a continuous group homomorphism
$$\rho :SU(2)\to  GL(V),$$
(with $V$ a vector space) and representations with $-E$ acting as the identity are those where $\rho(-E)$ is the identity on $V$, i.e. the homomorphisms $\rho$ whose kernel contains the normal subgroup $\{E,-E\}$. By the universal property of quotient groups, these are the same thing as continuous homomorphisms
$$SU(2)/\{E,-E\}\to  GL(V)$$
(where $SU(2)/\{E,-E\}$ is endowed with the quotient topology). Now $SU(2)/\{E,-E\}$ is isomorphic to $SO(3)$ by the first isomorphism theorem, and the isomorphism is continuous, hence an homeormorphism (for a more detailed explanation about this isomorphism, see this question). Thus the continuous homomorphisms $SU(2)/\{E,-E\}\to  GL(V)$ are exactly the representations of $SO(3)$.
This shows that representations of $SO(3)$ are in bijection with those of $SU(2)$ such that $-E$ acts trivially. Now irreducible representations are those where $V$ has no proper invariant subspace. Since this only depends on the image of the homomorphism $\rho $ in $GL(V)$, the bijection described above restricts to a bijection between irreducible representations.
