I have written a proof for the following and would like you to correct me if I made any mistakes, thanks in advance:

claim: $X = \sqcup_i X_i$ then $H_q (X) \cong \oplus_i H_q (X_i)$

proof: Proof of case $X = A \sqcup B$, the general case follows by induction.

By the Mayer-Vietoris theorem the following sequence is exact:

$$ \dots \xrightarrow{k_\ast} H_{n+1}(X) \xrightarrow{\partial_\ast} H_n(A \cap B) \xrightarrow{(i_\ast, j_\ast)} H_n(A) \oplus H_n(B) \xrightarrow{k_\ast} H_n(X)\xrightarrow{\partial_\ast} \dots$$

Then $A \cap B = \emptyset \implies H_n(A \cap B) = 0 \implies \partial_\ast = 0$

Then $k_\ast$ is injective because $ker k_\ast = im \partial_\ast = 0$ and $k_\ast $ is surjective because $im k_\ast = ker \partial_\ast = H_n(X)$

  • 4
    $\begingroup$ Well, yes, if you take Mayer-Vietoris for granted. However, you should probably try and prove this directly from the definition of homology... Suggestion: go through Hatcher's proof of Mayer-Vietoris and extract a direct argument. This would probably be more illuminating than what you did. $\endgroup$ – t.b. Aug 8 '11 at 11:39
  • 3
    $\begingroup$ If your collection is not countable, you will need transfinite induction. $\endgroup$ – gary Aug 8 '11 at 11:42
  • 1
    $\begingroup$ Maybe you can argue that if a cycle bounds in one of the $A_i$'s, then it will bound in the disjoint union, and, conversely, a trivial cycle will also be trivial in the union. $\endgroup$ – gary Aug 8 '11 at 12:09
  • $\begingroup$ @Theo, ok, I'll do that. $\endgroup$ – Rudy the Reindeer Aug 8 '11 at 12:11
  • 1
    $\begingroup$ You could also make an analogous observation about the singular (or simplicial, or $\Delta$) chain complex for the disjoint union and work from there. $\endgroup$ – wckronholm Aug 8 '11 at 15:54

Wouldn't be easier to argue like this?

Since $\Delta^n$ is connected, then the image of every continuous map $\sigma : \Delta^n \longrightarrow \bigsqcup_{\alpha \in J} X_\alpha$ must be contained in some $X_\alpha$: $\sigma (\Delta^n) \subset X_\alpha$.

I didn't check the details, but I think that this would say that you can find an inverse to the universal map

$$ \bigoplus_{\alpha \in J} H_p(X_\alpha) \longrightarrow H_p(\bigsqcup_{\alpha \in J} X_\alpha) $$

induced by the inclusions $X_\alpha \longrightarrow \bigsqcup_{\alpha \in J} X_\alpha$. And this would be true for any set of indices $J$.

| cite | improve this answer | |
  • $\begingroup$ I am confused; did the OP specify he is working with singular or simplicial homology? $\endgroup$ – user641 Aug 8 '11 at 19:24
  • $\begingroup$ The OP didn't specify, but it's really the same argument either way. $\endgroup$ – Aaron Mazel-Gee Aug 8 '11 at 20:12

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.