Showing $[f^{-1}(y)]$ is Poincare dual to $f^*(\operatorname{vol})$.

Let $f: N^n \to M^m$ be a smooth map between closed oriented manifolds. Then I'm trying to show that for almost all $y \in M$, the homology class $[f^{-1}(y)] \in H_{n-m}(N)$ is Poincare dual to $f^* \operatorname{vol}_M$, where $\operatorname{vol}_M$ is the volume form on $M$ (here I'm using the fact that for generic $y$, $f^{-1}(y)$ is a submanifold of dimension $n-m$).

Unwinding the definitions, I need to show that if $\phi \in \Omega^{n-m}(N)$ is closed then $$\int_{f^{-1}(y)} i^* \phi = \int_N \phi \wedge f^* vol_M$$ where $i: f^{-1}(y) \to N$ is the inclusion. Now this is easy to see if, for example $N = M \times F$ and the map $f$ is just the projection, but I am having trouble proving this in general. My issue is that the left hand side appears dependent on $y$ while the right hand side doesn't. But I believe all such $f^{-1}(y)$ are homologous so the left hand side is really independent of $y$ but I don't know how to show it's equal to the right hand side.

I'm sure this is really standard thing in differential topology but for some reason I haven't found it in my literature search.

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1 Answer

Look at the Griffiths-Harris “Principles of Algebraic geometry” middle of the page 59, for the state and proof of following general and important result.

General result: With the same assumption, one can show that, if $f$ is non-singular over a cycle $C \subset M,$ then with the proper orientation, the cycle $f^{-1}(C) \subset N$ is Poincare dual to the pull-back via $f$ of the Poincare dual of $C.$

Now, all you have to show is that the Poincare dual of a (generic) point $y \in M$ is the volume form. The rest, follows from the Poincare duality and pairing of closed differential forms of complementary degree in de Rham cohomology.

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