I am under the impression that if I had a product $X = A \times B$ the induced fibration $A \rightarrow X \rightarrow B$ would have a global section, namely $\sigma: B \rightarrow X$ taking $b$ to $(b, a_0)$ for any $a_0 \in A$. Furthermore, I have seen (e.g. here) that $SO(4) \cong SO(3) \times S^3$ as topological spaces. However, it seems that the fiber bundle $SO(3) \rightarrow SO(4) \rightarrow S^3$ doesn't have a global section (see e.g. here; the second answer shows that there are local sections, which is sort of where I got this impression from).
I am not sure what to make of this contradiction. My best guess at what's going on is that the "canonical" fiber bundle in the question linked above doesn't have a section, though one could write down a separate, perhaps less natural bundle $SO(3) \rightarrow SO(4) \rightarrow S^3$ that does have a section. In this case, what is this bundle, and what is the section? I am interested in using a global section to show that $SO(4)$ is homeomorphic to $SO(3) \times S^3$.
EDIT: It just occurs to me that perhaps in the general case the fiber bundle $SO(n) \rightarrow SO(n+1) \rightarrow S^n$ does not have a section (which would be consistent with the fact that in general the product does not seem to hold), but when $n = 3$ it happens to. In this case, can someone write down an explicit section $S^3 \rightarrow SO(4)$?