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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).

Pepper
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  • A proper local diffeomorphism is a covering map. – Michael Albanese Oct 02 '18 at 01:35
  • I think that map can't exist, and to prove it I would use the fact that @MichaelAlbanese stated, and use higher homotopy groups. E.g, check page 339 in Hatcher's Algebraic Topology for help with homotopy groups of spheres. Again, I haven't thought about it much, but I suspect they are not locally diffeomorphic (I have some reasonable geometric reasons), and that would be my approach to prove I'm right. – Laz Oct 02 '18 at 04:25
  • Another approach would be de Rham cohomology. – Laz Oct 02 '18 at 04:29
  • How about this? If they were locally diffeomorphic, then subtracting a point in both cases would give us a local diffeomorphism between $\mathbb{R}^3$ and $\mathbb{S}^1\times \mathbb{S}^2$ minus a point. – Laz Oct 02 '18 at 04:36

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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.

Najib Idrissi
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