The problem that I am working on says to either prove or disprove that there is a smooth map f : $S^3$ to $S^1$x$S^2$ such that the differential is an isomorphism at each point.
It seems like the easiest way would be to say that $S^3$ is simply connected, hence no such f exists, but I cannot figure out whether this works or not.
For reference, the closest thing I could find is the fact that local diffeomorphisms from compact manifolds are a stable homotopy type, which (I think) is something different from preserving homotopy class. Well-known theorems are considered fair game - the problem is from a recent qualifying exam (UMD Topology, August 2018).
Given a manifold $X$, there always exists a simply connected manifold $\tilde{X}$ and a local diffeomorphism $\tilde{X} \to X$: it's the universal cover of $X$. In this case, $\mathbb{R} \times S^2$. So this approach won't work.
Here's something that will. A continuous map from a compact space to a Hausdorff space is proper. So the purported local diffeomorphism $f : S^3 \to S^1 \times S^2$ would be proper. A proper local homeomorphism is a covering map. So $f$ would be a covering map. Since $S^3$ is simply connected, it should be the universal cover. But the universal cover of $S^1 \times S^2$ is actually $\mathbb{R} \times S^2$. So we reach a contradiction.