Why is a Lie group homomorphism from $SO(3)$ to $SU(2)$ always trivial? The Lie group $SU(2)$ is a double cover of $SO(3)$. $SU(2)$ is simply-connected as a manifold, and $SO(3)$ is $\mathbb{R}\mathbb{P}^3$. But why must a Lie group homomorphism from $SO(3)$ to $SU(2)$ be trivial? That is, the homomorphism carries any element of $SO(3)$ to the unit of $SU(2)$. How to prove that?   
 A: $\text{SO}(3)$ is simple, I assume we know this.
So any non-trivial homomorphism would be injective and by compactness, a homeomorphism onto its image. The image would be a 3-dim compact submmanifold of the connected manifold $\text{SU}(2)$, hence everything. But the two manifolds are no homeomorphic, e.g $\text{SO}(3)$ is not simply connected (we know a double cover of it, $\text{SU}(2)$, right?).
A: You haven't made clear what's assumable, so I'll use *s to point out where I'm assuming a fact. The basic principle: representation theory is easier than group theory, so try to argue from representations where possible.
Let $\pi:SU(2)\to SO(3)$ be the known* double cover, quotienting by the $Z_2$ center. Then every irrep of $SO(3)$ extends to one of $SU(2)$, on which $SU(2)$'s center acts trivially. The irreps of $SU(2)$ are known* to be $\{ Sym^k \mathbb C^2\ :\ k\in \mathbb N\}$, and only the odd-dimensional ($k$ even) ones have that $Z_2$ acting trivially.
A map $SO(3)\to SU(2)$ gives a $2$-d rep of $SO(3)$. That has to be a sum of irreps of $SO(3)$, but the only irrep of dimension $\leq 2$ is the trivial one.
