# Homology of disjoint union is direct sum of homologies

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 $im k_\ast = ker \partial_\ast = H_n(X)$

• 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. – t.b. Aug 8 '11 at 11:39
• If your collection is not countable, you will need transfinite induction. – gary Aug 8 '11 at 11:42
• 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. – gary Aug 8 '11 at 12:09
• @Theo, ok, I'll do that. – Rudy the Reindeer Aug 8 '11 at 12:11
• You could also make an analogous observation about the singular (or simplicial, or $\Delta$) chain complex for the disjoint union and work from there. – wckronholm Aug 8 '11 at 15:54

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 $X_\alpha$: $\sigma (\Delta^n) \subset X_\alpha$.
$$\bigoplus_{\alpha \in J} H_p(X_\alpha) \longrightarrow H_p(\bigsqcup_{\alpha \in J} X_\alpha)$$
induced by the inclusions $X_\alpha \longrightarrow \bigsqcup_{\alpha \in J} X_\alpha$. And this would be true for any set of indices $J$.