# 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 '10 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 '10 at 13:44
"Closed" is standard terminology for a compact manifold without boundary. –  Tyler Lawson Apr 8 '10 at 13:57
Ah, I've never heard of that before. –  Harry Gindi Apr 8 '10 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 '10 at 16:29