2
$\begingroup$

If a map $f:X\to Y$ of CW-complexes induces an isomorphism on homotopy groups it is a homotopy equivalence by the Whitehead theorem. In particular it also induces isomorphisms on homology and cohomology.

If now $f$ is just an $n$-equivalence (induces isomorphisms on homotopy up to degree $n$), intuitively I would say it still induces isomorphisms on (co)homology in lower degrees. But does it? Or is there a similar statement?

$\endgroup$
2
$\begingroup$

If $f$ is a $CW$-map, then $f$ induces an equivalence between $X[n]\rightarrow Y[n]$ where $X[n]$ is the $n$-truncature of $X$, you deduce that $H^*(X[n],.)\rightarrow H^*(Y[n],.)$ is an isomorphism by Whitehead. $H^*(X[n],.)=H^*(X,.)$ for $*<n$. To see this, use cellular (co)homolgy since all the maps above are natural, you deduce that the morphism $H^*(X,.)\rightarrow H^*(Y,.)$ is an isomorphism for $*<n$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

It does. See again Switzer's book (Theorem 10.28). Note that all versions of the Whitehead theorem require connected CW-complexes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah, now that I see it, I remember! Thank you! $\endgroup$ – Pepe Jun 29 '18 at 17:43

Your Answer

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

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