## Do “surjective” degree zero maps exist?

Is there a map $f\colon X \to Y$ of closed, connected, smooth and orientable $n$-dimensional manifolds such that the degree of $f$ is 0 but $f$ is not homotopic to a non-surjective map?

Added: The motivation is: There is a "mild version" of the Nearby Langrangian conjecture stating: any exact Lagrangian manifold $X \to T^*Y$ has non-zero degree when composed with the projection $T^*Y \to Y$. It is known that the map is always surjective. I am looking at a possible inbetween stating that the map cannot be homotoped to a non-surjective map.

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I seems like it will be very very hard to prove that a given map is not homotopic to a non-surjective map. – Chris Schommer-Pries Apr 8 2010 at 12:42
Partial answer: If $Y=S^n$, it follows by the Theorem of Hopf that the degree determines the homotopy class. It's on the last page before the exercises in Milnor's Topology from a Differentiable Viewpoint. This gives a negative answer for spheres, but I don't know about the general case. Also, by closed, do you mean closed as a submanifold of euclidean space? – Harry Gindi Apr 8 2010 at 13:44
"Closed" is standard terminology for a compact manifold without boundary. – Tyler Lawson Apr 8 2010 at 13:57
Ah, I've never heard of that before. – Harry Gindi Apr 8 2010 at 14:42
If you're getting into this "nearby Lagrangian" stuff, make sure you're up to date! You need to know the theorem of Fukaya-Seidel-Smith/Nadler about Maslov-zero exact Lagrangians in simply connected cotangent bundles, and the recent work of Abouzaid about cotangent fibres, relevant to relaxing the simple connectedness assumption. – Tim Perutz Apr 8 2010 at 16:29

It is a theorem of H. Hopf that a map between connected, closed, orientable n-manifolds of degree 0 is homotopic to a map that misses a point, when n > 2. See D. B. A. Epstein, The degree of a map. Proc. London Math. Soc. (3) 16 1966 369--383, for a "modern" discussion including the analogous situation in the non-orientable case. The same result holds for n = 2, but is more difficult and is due to Kneser. See Richard Skora, The degree of a map between surfaces. Math. Ann. 276 (1987), no. 3, 415--423, for a thorough discussion of the non-orientable case in dimension 2.

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 Great - Thanks! I am assuming that you forgot the connected assumption. – Thomas Kragh Apr 8 2010 at 20:03 Right. "All manifolds are connected unless otherwise stated." I'll fix it for the record. – Allan Edmonds Apr 9 2010 at 1:24

Think of $F$ as being embedded into $\mathbb{R}^3$ symmetrically, with a handle in the middle. Mirroring wrt the symmetry plane of this embedding perpendicular to the direction of the handle maps $F$ to a torus minus a disc. Is your map not homotopic to this factor map? – Thorny Apr 8 2010 at 14:56