Unitary representations of $SO(3)$ and $so(3)$

Question

According to my skript:
Quantum mechanic states $ψ ∈ \mathcal H$ changes under a rotation $R ∈ \text{SO(3)}, \vec{x} \rightarrow R\vec{x}$ according to $ψ \rightarrow U(R)ψ$, whereas $U(R)$ is a unitary representation of $\text{SO(3)}$ is, that means: $$U: \text{SO(3)} \rightarrow \mathcal L(\mathcal H) = \{\text{linear transformation } \mathcal H \rightarrow \mathcal H\} = \text{GL}(\mathcal H,\mathcal H)$$ $$R \longmapsto U(R) $$
is a homomorphism, i.e. $U(R_1)U(R_2) = U(R_1R_2), U(1) = \mathbb I$ which is unitary $U(R)^{-1} = U(R)^*$.

Infinitesimal rotations are elements $Ω$ of the tangent space $T_{\mathbb I}SO(3) = \{\dot{γ}(0)|γ:[-ε,ε] \rightarrow \text{SO(3)}, γ(0) = \mathbb I\}$, where $γ(ε) = e^{εΩ} ∈ \text{SO(3)},γ(0) = e^{0Ω} = \mathbb I$, on $\text{SO(3)}$ at the point $\mathbb I$: $$Ω = \frac{d}{dt}R(t)\bigg|_{t=0},$$
whereas $t \longmapsto R(t)$ is a differentiable curve in $\text{SO(3)}$ through $R(0) = \mathbb I$.

Every Lie group representation of $\text{SO(3)}$ on $\mathcal H$ corresponds to a Lie algebra representation of $\text{so(3)}$ (but not vice versa):
$$U(Ω):= \frac{d}{dt}U(R(t))\bigg|_{t=0}.$$

The transformation $Ω \longmapsto U(Ω)$ is a homomorphism of $\text{so(3)}$ $(\alpha_1,\alpha_2 ∈ ℝ)$: $$U(\alpha_1Ω_1 + \alpha_2Ω_2) = \alpha_1U(Ω_1) + \alpha_2U(Ω_2)$$ $$U([Ω_1,Ω_2]) = [U(Ω_1),U(Ω_2)],$$ whereas the last follows from $U(RΩR^{-1}) = U(R)U(Ω)U(R)^{-1} \quad (R ∈ \text{SO(3)}).$

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I want to check the last statement, i.e. that $Ω \longmapsto U(Ω)$ is a homomorphism.
Calculation Questions:
1) If I just consider $\alpha_1U(Ω_1)$, is it then correct to state:
$$\alpha_1U(Ω_1) = \alpha_1 U(\frac{d}{dt}R_1(t)\bigg|_{t=0})$$ $$\alpha_1U(Ω_1) = \alpha_1\frac{d}{dt}U(R_1(t))\bigg|_{t=0},$$
by simply plugging in the definitions?
2) If this holds, can I then conclude: $$U(\alpha_1Ω_1 + \alpha_2Ω_2) = U(\frac{d}{dt}(R_1(\alpha_1t)R_2(\alpha_2t))\bigg|_{t=0})\\ \quad= \frac{d}{dt}U(R_1(\alpha_1t)R_2(\alpha_2t))\bigg|_{t=0} = \frac{d}{dt}(U(R_1(\alpha_1t))U(R_2(\alpha_2t)))\bigg|_{t=0} =\alpha_1Ω_1 + \alpha_2Ω_2.$$
3) I ask this, because someone else wrote down that:
$$U(Ω_1+\alphaΩ_2) := \frac{d}{dt}U(R_1(t)R_2(\alpha t)\bigg|_{t=0}.$$
4) So in general I am a bit confused about the notation, maybe one could clarify this a bit.

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General Questions & Remarks:
I still don't get the main concept behind representations.
Let me place a few words into this room: Quantum Mechanics; Born's rule; Symmetries; Projective Representations; Wigner's Theorem; Irreducible Representations; Eigenstates; Composite Systems & Clebsch-Gordan-Coefficients; Wigner-Eckart-Theorem.
It would be great if one could show me his Idea in a few lines using the words given above. I myself will later try (so far I have only a long version in another language which I can poste later on). Thank you in advance! :)

Answer 1

The answer is yes to all your questions. In general one assumes the representation to be strongly continuous (i.e. continuous in the strong operator topology) so expressions make sense when evaluated against a vector from the Hilbert space. The existence of the "derivative" of a 1-parameter subgroup like $U(R_1(t))$ is given by Stone theorem which says that there exists a self-adjoint operator $H_1$ such that $U(R_1(t)) = e^{iH_1t}$ (through functional calculus) and $H_1$ is precisely the "derivative" of $U(R_1(t))$ at $t=0$.