2
$\begingroup$

Let $f_1 : X_1 \to Y_1$ and $f_2 : X_2 \to Y_2$ be two continuous maps between topological spaces. Suppose that there exists a commutative diagram of singular cohomology groups (say with coefficients in $\mathbb{C}$):

$\require{AMScd}$ \begin{CD} H^*(Y_1) @>{f_1^*}>> H^*(X_1)\\ @VVV @VVV\\ H^*(Y_2) @>{f_2^*}>> H^*(X_2). \end{CD}

Is it enough to induce a map between the Leray spectral sequences of $f_1$ and $f_2$ ?

Note that I don't suppose that the above commutative diagram comes from topological maps, and therefore I have no way to define maps at the level of complexes preserving some filtration.

My question comes from the fact that, the $E_2$-term of the Leray spectral sequence of a map $f : X \to Y$ involves the presheaf $$U \to H^*(f^{-1}(U)),$$ and therefore only cohomological information on the map $f$ is required. Therefore, if we have a relation between two maps in cohomology, we should have a relation between their respective spectral sequences.

Thanks a lot !

$\endgroup$
  • 3
    $\begingroup$ No, absolutely not. A commutative diagram between cohomology groups contains pretty much no information at all. You need something like a map between filtered complexes. $\endgroup$ – Dan Petersen Jul 4 '18 at 16:48
  • $\begingroup$ Thanks for your answer @Dan. Then, maybe another question in this spirit: suppose that I have a map between the sheaves involved in the $E_2$-terms of the Leray sequences of these maps. Would I get a map between the spectral sequences ? $\endgroup$ – BrianT Jul 4 '18 at 16:55
  • $\begingroup$ What kind of map of sheaves do you have in mind? The sheaves live on different spaces, and you've assumed that there is no map between the spaces involved... Either way, you can not expect to get a map of spectral sequences just from a map between the individual sheaves $R^q f_\ast \mathbb Z$. It would be enough to construct a map in the derived category between the corresponding complexes $Rf_\ast \mathbb Z$. $\endgroup$ – Dan Petersen Jul 5 '18 at 5:37
  • $\begingroup$ Thanks @Dan. So if I have a map in the derived category between the complexes $R f_{1*} \mathbb{Z}$ and $R f_{2*} \mathbb{Z}$, it ensures me a map between the spectral sequences ? How can I prove this ? Would you have a good reference for this kind of results ? Thanks again $\endgroup$ – BrianT Jul 5 '18 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.