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I am troubled by the following fact:

If $f,g : X\to Y$ are homotopic and both transversal to $Z$ then the mod.$2$ intersection numbers are equal $I_2(f,Z)=I_2(g(Z)$. (the book by Guillemin and Pollack, Differential Topology, page 78).

If $f,g : [-2,2]\to[-2,2]\times\mathbb{R}$, with $f(x)=(x,x^2)$ and $g(x)=(x,x^3)$ then $f$ and $g$ are homotopic (by $H(t,x)=(x,tx^3+(1-t)x^2)$) to $\Delta=\{(x,x),\ x\in[-2,2]\}$ but $I_2(f,\Delta)=2[2]=0[2]$ and $I_2(g,\Delta)=3[2]=1[2]$.

I don't know what I am missing here! Thanks for any help!

user102980
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  • I think there might be a "closedness" (i.e. compact and no boundary) issue here but I'm not sure. – Tim kinsella Nov 23 '13 at 20:52
  • ooh this is actually really interesting. so the proof of the statement in GP (which is definitely for closed manifolds) takes a homotopy between f and g which is transversal to Z. Then if you take the preimage of Z under the homotopy you get a compact 1-manifold whose boundary lies in the boundary of $X \times I$. But if $X\times I$ has boundary points not in $X\times 0 \cup X \times 1$, the 1-manifold could have boundary points on the side of $X\times I$ instead of just the top and bottom allowing $f$ and $g$ to have different mod 2 intersection numbers. – Tim kinsella Nov 23 '13 at 21:05
  • Sorry if this seems like mad ravings. If its not clear I'll try to clean it up into an answer in a bit. – Tim kinsella Nov 23 '13 at 21:05
  • Thanks Tim, I wasn't able to reply quickly! So in the Theorem we must take a closed manifold $X$ in order to get the boundary of $X\times I$ is $X\times0\cup X\times1$. Now ts clear! – user102980 Nov 26 '13 at 17:54

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