Let $\xi:Y\to X$ be a vector bundle.
Spherization of this bundle is the quotient space $(X\backslash \Theta) / \sim $ (here $\Theta$ is a graph of the zero section of the bundle) with equivalence relation $\sim$: $u\sim v$ if $\xi(u)=\xi(v)$ and $u=\lambda v$, where $\lambda>0$.
How to prove that $SO(3)$ diffeomorphiс to a spherization of tangent bundle of two-dimensional sphere?
Let $Q$ be the spherization of the tangent bundle of the standard unit 2-sphere $S^2$ in $\mathbb R^3$. Define a map $\phi: SO(3)\to Q$ as follows. An element $g\in SO(3)$ is a $3\times 3 $ matrix, whose columns are a positively oriented orthonormal basis $(v_1, v_2, v_3)$ of $\mathbb R^3$ (with respect to the standard inner product and orientation on $\mathbb R^3$). In particular, $v_1\in S^2$. The tangent space $T_{v_1}S^2=(v_1)^\perp$, hence $v_2\in T_{v_1}S^2$. Let $[v_2]=\mathbb R^+ v_2\in Q$ be the corresponding equivalence class. Finally define $\phi(g):=[v_2]$. To show that $\phi$ is a diffeomorphism we construct an inverse $\psi:Q\to SO(3)$ as follows. Given an element $[v]\in Q$, where $v\in T_{v_1}S^2$, $v\neq 0$, let $v_2=v/\|v\|$ and $v_3=v_1\times v_2$ (vector product). Then the matrix $g$ whose columns are $(v_1,v_2,v_3)$ is in $SO(3)$ so we can define $\psi([v])=g.$ It is quite easy to show that $\phi,\psi$ are inverse to each other hence $Q$ is diffeomorphic to $SO(3)$.