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This is a question related to this page.

https://ncatlab.org/nlab/show/restricted+product .

Let $I$ be a directed set. Let $X_i(i\in I)$ be a group. Let $\prod'_{i\in I}(X_i,Y_i)$ be a restricted product with respect to $Y_i$. That is , it is a set of tuples $(a_1,,a_2,\cdots )\in \prod_{i\in I} X_{i}$ such that almost all $a_i$ belongs to $Y_i$.

In the above linked page, $\prod'_{i\in I}(X_i,Y_i)$ is defined as $\varinjlim_{S\subset I \text{ runs finite subset of} I} (\prod_{i\in S} X_{i}\times \prod_{v\in I-S}Y_i)$.

My question is, is these two definiton equivalent ? I cannot see the isomorphism between them.

This is the end of my question.

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The following background is related to the theory of elliptic curves, but this question is purely algebraic, so you don't need to worry about it. I am only writing it in case someone who is also knowledgeable about elliptic curves happens to read it and provide comments. It is sufficient to read just this question itself.

Background : https://mathoverflow.net/questions/471560/bad-prime-of-torsor-and-original-elliptic-curve-definition-of-tate-shafarevich

In the above linked page, $X_i=H^1(G_{K_i},E)[2]$ where $E/K$ is an elliptic curve over number field $K$ and $G_{K_i}$ is the absolute galois group of $K_i$ and $Y_i$ is not explicitly written(may be unramified cohomology $H^1_{nr}(G_{K_i},E)[2]$). $S$ runs all finite set of places of $K$ and contains fixed bad primes of $E/K$ and infinite places.

Last Professor Wuthrich's comment says 'the sum is no longer a direct sum, but a restricted product'. But if above is correct, $Y_i=0$, $\bigoplus_{i\in I}X_i=\varinjlim_{S\subset I \text{ runs finite subset of} I} \oplus_{i\in S} X_{i}$ and contradicts Professor Wuthrich's comment because $\bigoplus_{i\in I}X_i\neq \prod'_{i\in I}X_i$.

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    $\begingroup$ This limit is just a union, since all the maps are inclusions and everything is contained in some ambient set (namely the product of the $X_i$) which makes the equivalence of thetwo definitions clear. For your interpretation of the mathoverflow answer, I do not understand why you are saying $Y_i = 0$. $\endgroup$
    – hunter
    Commented Jun 24 at 13:24
  • $\begingroup$ Why are you saying that $I$ is directed? That seems to be irrelevant. The set that is directed is the partially ordered set of finite subsets of $I$, ordered by inclusion. $\endgroup$
    – Arturo Magidin
    Commented Jun 24 at 17:16
  • $\begingroup$ @Arturo Magidin. You're right. I need not to be directed set here. We take limit by inclusion $I\subset I'\implies \bigoplus_I\subset \bigoplus_{I'}$. $\endgroup$
    – Poitou-Tate
    Commented Jun 24 at 22:47
  • $\begingroup$ Thank you very much. Wuthrich's comment is saying $ \varinjlim_S \prod_{v\in S} H^1(G_{K_v},E)[2]=\prod’ H^1(G_{K_v},E)[2]$・・・① where $S$ runs all finite set counting but primes and infinite places of $K$ with some restricted product $\prod’$. Let say this is a restricted product with respect to $Y_i$, ① does not hold unless $Y_i=0$ because $H^1(G_{K_v},E)[2]$ does not vanish for infinitely many $v$. But when $Y_i=0$, ① just means $ \varinjlim_S \prod_{v\in S} H^1(G_{K_v},E)[2]=\bigoplus_{v} H^1(G_{K_v},E)[2]$. But he says ‘the sum is no longer direct sum, but restricted product’. $\endgroup$
    – Poitou-Tate
    Commented Jun 24 at 22:49
  • $\begingroup$ In short, I’m having difficulty to understand why left hand side of $ \varinjlim_S \prod_{v\in S} H^1(G_{K_v},E)[2]=\prod’ H^1(G_{K_v},E)[2]$・・・① is not just $\bigoplus H^1(G_{K_v},E)[2]$.@hunter $\endgroup$
    – Poitou-Tate
    Commented Jun 24 at 22:50

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