4
$\begingroup$

In Bott & Tu's "Differential forms", Example 10.1 states:

$\textbf{Example 10.1}$ Let $\pi: E \to M$ be a fiber bundle with fiber $F$. Define a presheaf on $M$ by $\mathcal F(U) = H^q(\pi^{-1}(U))$. For $U$ contractible, $H^q(\pi^{-1} U) \cong H^{q}(F)$ by the Kunneth formula, and if $V \subseteq U$ with $V$ connected, then the restriction $\rho^U_V: H^q(\pi^{-1}(U)) \to H^q(\pi^{-1} (V))$ is the identity. Therefore $\mathcal F$ is a locally constant presheaf on $M$.

However, consider the trivial rank $1$ bundle on $\mathbb R^2$. For a connected subset $V \cong S^1 \times \mathbb R \subseteq D^2$, we have $\mathcal F(V) = H^q(S^1) \ncong \mathcal F(U) = H^q(\text{point})$.

What am I missing here?

Improve this question
$\endgroup$
6
  • $\begingroup$ If I understand the question correctly, it seems that the OP's doubt is that a contractible open set $U$ can contain a non-contractible open set $V$. For instance, a disk in $\mathbb{R}^2$ contains an open annulus, whose cohomology is non-trivial. $\endgroup$
    – Francesco Polizzi
    Commented May 21 at 9:01
  • 3
    $\begingroup$ Actually, using the notation in the book for the trivial line bundle on $\mathbb{R}^2$, if $U=D^2$ and $V=D^2-\{(0, \, 0)\}$, I do not see how the pullback ("restriction") map $$H^1( U \times \mathbb{R}) \to H^1(V \times \mathbb{R})$$ can be the identity, since $$H^1(U \times \mathbb{R})= 0, \; H^1(V \times \mathbb{R})= \mathbb{R}$$ $\endgroup$
    – Francesco Polizzi
    Commented May 21 at 10:29
  • 3
    $\begingroup$ This is a good question. As your example shows, the map $\rho_V^U$ need not be an isomorphisms. That seems to be a mistake. I think it is still true that $\mathcal{F}$ is a locally constant presheaf, but this might depend on the definition that Bott and Tu are using. Is it enough for the sheafification to be locally constant? $\endgroup$
    – Chris Schommer-Pries
    Commented May 21 at 14:40
  • 2
    $\begingroup$ As GregFriedman highlights, different editions of the book may be more/less accurate -- I hope your version is older? (My version is "revised third printing" which agrees with Friedman's post.) $\endgroup$
    – Chris Gerig
    Commented May 22 at 5:59
  • $\begingroup$ @ChrisSchommer-Pries The sheaf associated with the given presheaf is locally constant, but I do not understand the significance of that sheaf. $\endgroup$
    – Jaehwan Kim
    Commented May 23 at 5:10

1 Answer 1

Reset to default
9
$\begingroup$

In my version of Bott and Tu, the example reads (emphasis mine):

"Moreover, if $V \subset U$ is an inclusion of CONTRACTIBLE open sets, then $\rho^U_V: H^q(\pi^{-1} U) \to H^q( \pi^{-1} V)$ is an isomorphism."

It goes on to say that this "is an example of a locally constant presheaf on a good cover, to be defined in Section 13." The idea then is that in this context one only allows certain open sets as inputs.

Looks like this is the original edition, copyright 1982, 4th printing.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Oh, my version is outdated. It seems fixed. Thank you :) $\endgroup$
    – Jaehwan Kim
    Commented May 23 at 5:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.