2
$\begingroup$

Recently I came across a proposition:

Proposition. The projection $\pi:D^n\rightarrow D^n/S^{n-1}$ induces isomorphism of relative homology $\pi_*:H_p(D^n,S^{n-1})\rightarrow H_p(D^n/S^{n-1},pt)$.

By adapting the Mayer-Vietoris sequence, it is not hard to find that both groups are isomorphic to the reduced homology $\widetilde{H}_{p-1}\left(S^{n-1}\right)\cong\widetilde{H}_p\left(S^n\right)$, which are $\mathbb{Z}$ when $p=n$ and are trivial otherwise.

I’ve been trying to prove the proposition by showing that the homomorphism between those two groups is surjective (or epimorphic) since $\pi$ is itself surjective. However, I find it difficult to construct some kind of continuous “inverse” of the projection to bring the element of $H_p(D^n/S^{n-1},pt)$, i.e., a map $\Delta^p\rightarrow D^n/S^{n-1}$ to some $f:\Delta^p\rightarrow D^n$.

I believe my hypothesis is possibly correct, at least with respect to a certain class of “good” topological spaces, that an in/surjective continuous map between spaces will induce corresponding in/surjective homomorphism of homology.

So, my question is, whether or not my hypothesis is correct (or partly correct, which means it only holds when the topological spaces is “good” enough), and what properties of $D^n$ and $S^n$ does the proposition rely on.

Any help would be greatly appreciated.

$\endgroup$
1
$\begingroup$

This almost never happens in real life no matter how good the spaces and maps are.

For instance, the $n$-fold winding map $z\to z^n$ from $S^1$ to $S^1$ is surjective but not surjective on homology.

On the other hand the inclusion $S^1\to S^2$ is injective but not injective on homology.

$\endgroup$
  • $\begingroup$ yeah, I guess there could be a number of counterexamples. But I’m wondering why the projection here has the property. Or maybe I just ran into a completely wrong way? :( $\endgroup$ – J. Lizy Oct 25 '18 at 11:46
  • 1
    $\begingroup$ @J.Lizy If you want a reason why relative homology is isomorphic to the homology of the quotient, this is because the inclusion is a cofibration. See for example Hatcher pi.math.cornell.edu/~hatcher/AT/AT.pdf proposition 2.22. That is indeed very general but not equivalent to what you are asking in the title of the question. $\endgroup$ – Michal Adamaszek Oct 25 '18 at 12:17
  • $\begingroup$ Well, sometimes I just think too much... That proposition indeed helps me, thanks. $\endgroup$ – J. Lizy Oct 26 '18 at 4:57

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.