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Let $M^n$ be a differentiable manifold and $\pi\colon E\to M$ is $n$-dimensional vector bundle over $M$.

We have a zero section $s\colon M\to E$ of $\pi$.

How can I make a section $s'$ which is trnasversal to $s$? (i.e., $s'$ vanishes $s$ finitely many times.)

(In some text, it seems even possible to make $s$ and $s'$ are isotopic.)

I need this to interpret the euler class of $\pi$, $\chi(\pi)$ as an algebraic intersection number of $s$ and $s'$.

Are there anybody who can give me any references?

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  • $\begingroup$ I'll answer the easy one: $s$ and $s'$ are always isotopic, since you can consider the isotopy $s_t = t\cdot s'$ (using the linear structure on each fiber, so $t\cdot s'$ is well-defined). $\endgroup$
    – Marco Golla
    Apr 25, 2011 at 14:50

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This is a straightforward application of transversality theorem, which roughly speaking states that we can make a map transverse to a submanifold with an arbitrary small perturbation. It is a consequence of Morse-Sard theorem.

The statement you need is the following:

Theorem. Let $A$, $B$ be $\mathcal{C}^r$-submanifolds of $M$, $1 \leq r \leq \infty$. Then every neighborhood of the inclusion $i_B \colon B \to M$ in $\mathcal{C}^r(B, M)$ contains an embedding which is transverse to $A$.

For a proof, see [Hirsch, Differential Topology, Thm. 2.4 pag. 78].

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  • $\begingroup$ As for the second part of the question, it doesn't look like Hirsch covers the theorem "Euler number = self-intersection of the zero-section". That's covered in Bott and Tu's, Differential forms in algebraic topology, Theorem 11.17. That requires lots of technology, though (at least, the book does). $\endgroup$
    – Marco Golla
    Apr 25, 2011 at 14:56

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