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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    $\begingroup$ I seems like it will be very very hard to prove that a given map is not homotopic to a non-surjective map. $\endgroup$ Apr 8, 2010 at 12:42
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    $\begingroup$ 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? $\endgroup$ Apr 8, 2010 at 13:44
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    $\begingroup$ "Closed" is standard terminology for a compact manifold without boundary. $\endgroup$ Apr 8, 2010 at 13:57
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    $\begingroup$ Ah, I've never heard of that before. $\endgroup$ Apr 8, 2010 at 14:42
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    $\begingroup$ @Bogdan: You don't need to remove smoothness for that, and that is not the question. You may have missed the part "homotopic to". $\endgroup$ Jul 22, 2014 at 12:02

1 Answer 1


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.

  • $\begingroup$ Great - Thanks! I am assuming that you forgot the connected assumption. $\endgroup$ Apr 8, 2010 at 20:03
  • $\begingroup$ Right. "All manifolds are connected unless otherwise stated." I'll fix it for the record. $\endgroup$ Apr 9, 2010 at 1:24

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