It's a well-known fact that $H^*(\bigsqcup_{\alpha}X_{\alpha};R)\xrightarrow{\cong}\prod_{\alpha}H^*(X_{\alpha};R)$ that is also in Hatcher Chapter 3, example 3.13. My attempt by definition is that

$H^*(\bigsqcup_{\alpha}X_{\alpha};R) = \bigoplus_{k\geq 0}H^k(\bigsqcup_{\alpha}X_{\alpha};R)=\bigoplus_{k\geq 0}\prod_{\alpha}H^k(X_{\alpha};R)$


$\prod_{\alpha}H^*(X_{\alpha};R)=\prod_{\alpha}\bigoplus_{k\geq 0}H^k(X_{\alpha};R).$

However, is it true that $\bigoplus_{k\geq 0}\prod_{\alpha}H^k(X_{\alpha};R)=\prod_{\alpha}\bigoplus_{k\geq 0}H^k(X_{\alpha};R)$? I remembered from my algebra course that direct sum and direct product can not be interchanged, but I have trouble constructing a good example. Would anyone like to point out my mistake here or a good example for failing to interchange $\prod$ and $\bigoplus$? If it fails, how could we prove this fact?

Appreciate your help!

  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/1003299/… and math.stackexchange.com/questions/4019359/…. Basically, the correct definition is to define the graded ring $H^*$ as just consisting of the sequence of $H^k$'s (together with the relevant operations), rather than their direct sum (or their direct product). $\endgroup$ Mar 8, 2022 at 4:01
  • 1
    $\begingroup$ But, if you define the graded ring $H^*$ as a direct sum the way Hatcher does, then his assertion in Example 3.13 is simply wrong. $\endgroup$ Mar 8, 2022 at 4:11
  • $\begingroup$ Note that if you take $H^*$ to be a sequence of abelian groups, then it’s not closed under addition, and some people might be hesitant to call such an object a ring. $\endgroup$ Mar 8, 2022 at 7:49
  • $\begingroup$ I think it doesn't really matter whether we think of the graded ring as a sequence of $H^k$ or their direct sum as long as we understand that product to be taken in the category of graded rings. $\endgroup$
    – Thorgott
    Mar 9, 2022 at 17:48


You must log in to answer this question.