Are $\text{SO}(3)$ and $S^2 \times S^1$ homotopy equivalent? There is a natural projection $\text{SO}(3) \to S^2$ which is the projection to the coset space of a subgroup $\text{Stab}(p)$ for any $p \in S^2$ under the standard action of the group. The preimage of any point in $S^2$ is a circle.
Is $\text{SO}(3)$ homotopy equivalent to $S^2 \times S^1$?
 A: The answer is NO!!
A different topological presentation of $SO(3)$ as $\mathbb RP^3$ would be helpful to complete this proof. So how to prove that this two spaces are homeomorphic? Observe that any matrix $A\in SO(3)$ is a orientation preservoing rotational matrix acts on $\mathbb R^3$. Since it is a $3\times 3$ matrix, so it has a real eigenvalue, and corresponding to this eigenvalue, the matrix fixes the eigen-vector in particular the line generated by the eigenvector. So we can think of any element of $SO(3)$ as a rotation of closed unit 3-disc with a fix axis. Now define a map $\phi :D^3\to SO(3)$ sends a non-zero vector $x$ to the rotation through angle of $|x|\pi$ about the axis joining origin and $x$.  Ans observe that antipodal points on the boundary are represent the same element, and in the interioi it is a $1-1$-map. So now if we consider the quotient map which quotient out the antipodal points of the boundary of the unit-disc, we get a homeomorphism $\bar{\phi}: \mathbb RP^3\to SO(3)$. 
Now from here you can easily see that $SO(3)$ and $S^2\times S^1$ are not homotopically equivalent, since their fundamental groups are different.
