The following selection is from Serre's FAC (Chapter 1, §3, n°22, page 26).

The relation `$\mathfrak{U}$ is finer than $\mathfrak{V}$' (which we denote hencforth by $\mathfrak{U} \prec \mathfrak{V}$) is a relation of a preorder$^{3}$ between coverings of $X$; moreover, this relation is filtered$^{4}$, since if $\mathfrak{U} = {U_i}_{i \in I}$ and $\mathfrak{V} = {V_j}_{j \in J}$ are two coverings, $\mathfrak{W} = {U_i \cap V_j}_{(i,j) \in I \times J}$ is a covering finer than $\mathfrak{U}$ and than $\mathfrak{V}$.

We say that two coverings $\mathfrak{U}$ and $\mathfrak{V}$ are equivalent if we have $\mathfrak{U} \prec \mathfrak{V}$ and $\mathfrak{V} \prec \mathfrak{U}$. Any covering $\mathfrak{U}$ is equivalent to a covering $\mathfrak{U'}$ whose set of indices is a subset of $\mathfrak{P}(X)$; in fact, we can take for $\mathfrak{U'}$ the set of open subsets of $X$ belonging to the family $\mathfrak{U}$. We can thus speak of the set of classes of coverings with respect to this equivalence relation; this is an ordered filtered set.$^{5}$

  • $^{3}$i.e. quasiorder
  • $^{4}$i.e. directed
  • $^{5}$i.e. To the contrary, we cannot speak of the `set' of coverings, because a covering is a family whose set of indices is arbitrary.

Please let me explain my interpretation of what is taking place here, and ask some related questions along the way.

At this point we have defined cohomology for an arbitrary open cover, but the information heavily dependend on the open cover chosen. We want a way to make the cohomology intrinsic to the space $X$ - so now in these two paragraphs Serre is setting the stage to take a direct limit of cohomology groups indexed by open covers ordered by refinement. Some questions about that:

  1. According to the wikipedia page, to take the direct limit we need the index set to be a directed preorder, but when I learned directed limits in Atiyah Macdonald (page 32) they require the index set be a directed partial order (thus also satisfy antisymmetry). What is the deal with this difference? After the equivalence relation Serre defines in paragraph 2 antisymmetry would hold (by definition), but is it required to take the limit in general?

  2. I don't understand footnote 5, why do open covers not form a set? He seems to say it is because the index set of the open cover is arbitrary, but by definition of open cover we have $\mathfrak{U} = \{U_i\}_{i \in I}$. Re-index this by some other index set $\mathfrak{U'} = \{U_j\}_{j \in J}$ but since each $U_i, U_j$ is a subset of the space $X$, if we considered the collection of open covers of $X$ wouldn't these two open covers be the same since they are equal as sets? They contain precisely the same elements, the same subsets of $X$. Am I being naive?

  3. Even once I understand why we have a set-theoretic issue, I am not sure I understand the solution. What is the solution Serre is proposing with the $\mathfrak{U'}$? -'Taking $\mathfrak{U'}$ the set of open subsets of $X$ belonging to the family $\mathfrak{U}$? Serre seems to do two things in paragraph 2, one is introduce the equivalence relation, and one is tell us how to choose a canonical index set? Can someone elaborate on what problems were presented and solved in paragraph 2?

  4. Is the equivalence relation he puts forth to help deal with the set-theoretic issue, or just to whittle down the number of open covers we are dealing with, since as shown two open covers that are equal by this equivalence relation will give the same cohomology.

  • 3
    $\begingroup$ Re: 2, the issue is that an indexed set is more than just its range (that is, the set of sets occurring in it): it also takes into account the indexing set. Precisely, an indexed set is a function with domain the indexing set, assigning $A_i$ to $i$. If you change the indexing set, you change the whole indexed set, in the same way that two functions with different domains are not equal even if their ranges are. $\endgroup$ – Noah Schweber Jan 16 at 20:28
  • $\begingroup$ For 2., essentially, the "set' of all sets is not a set. $\endgroup$ – Somos Jan 16 at 22:30
  • $\begingroup$ So Serre is saying, for some open cover we can always choose it to have a certain index set, I don't really get what he is saying that index set is - but he seems to be saying there is a canonical way to use a subset of the power set, or an element of the power set of the power set, thus our new collection forms a set? $\endgroup$ – Prince M Jan 16 at 23:17
  • $\begingroup$ @Somos something is missing from your response - the part where you tell me what the " 'set' of all sets" and "collection of open covers" have in common. $\endgroup$ – Prince M Jan 16 at 23:27

1) The concept of the direct limit of a direct system is defined in any category via a universal property (see the wikipedia article). So we must understand the concept of a direct system. You can define it for index sets which are directed preordered sets, or even more generally for small filtered categories. However, many authors restrict to directed partially ordered sets which suffices for most practical applications.

Let us work with directed preordered sets $(I,\le)$. We define a subset $I'' \subset I$ to be cofinal if each $i \in I$ admits $i'' \in I''$ such that $i \le i''$. Giving $I''$ the induced preorder, it is easy to verify that the direct limit of a direct system indexed by $I$ agrees with the direct limit of the cofinal subsystem indexed by $I''$.

Given a directed preordered set $(I,\le)$, we define $i \sim j$ iff $i \le j$ and $j \le i$. This is an equivalence relation and the set of equivalence classes $I'$ is a directed partially ordered set with $[i] \le [j]$ defined by $i \le j$. We may choose a subset $I'' \subset I$ containing precisely one representative from each equivelence class in $I'$. It inherits a preorder from $I$ and it is easy to see that $I''$ and $I'$ are isomorphic as preordered sets. Obviously $I''$ is a cofinal subset of $I$ which is a directed partially ordered set. This explains why many authors only consider the more special case of directed partially ordered sets.

2) The problem is that a cover is defined as an indexed family of sets, and the index set can be any set. But there is no set of all sets (it is a class), so there is no set of all covers of $X$.

For example, take any set $I$ and let $U_i = X$ for all $i$. This is not very interesting, but it is a cover.

3) The solution is that we may identify the family $\mathcal{U} = (U_i)_{i \in I}$ with the subset $s(\mathcal{U}) = \{ U_i \mid i \in I \} \subset \mathfrak{P}(X)$ which can be regarded again as a cover of $X$ (it is indexed by itself).

Doing so, we see that $s(\mathcal{U}) \prec \mathcal{U}$ and $\mathcal{U} \prec s(\mathcal{U})$, i.e. $s(\mathcal{U})$ and $\mathcal{U}$ are equivalent. This concept of equivalence resembles the definition in 1), the difference being that in 1) we started with a set, and here we start with a class. The effect is that we get a canonical set of representatives in the class of all covers.

4) In fact set theoretic issues are settled. But recall 1) - intuitively we may regard the set of coverings $s(\mathcal{U})$ as a "cofinal subsystem" of the class of all coverings. Note, however, that this set is not partially ordered which gives evidence that a more general concept of direct system is useful.

  • $\begingroup$ Ok - I think part of my problem was that in my head I was already identifying open covers with the range of the function indexing them, so I saw no need to refine them to form a set. However, the equivalence relation we put on the covers is actually more than just identifying open covers who differ by index set, correct? For example, in R^2 with usual topology the open cover of all open boxes and the open cover of all open disks should be equivalent? $\endgroup$ – Prince M Jan 16 at 23:31
  • $\begingroup$ That's correct! $\endgroup$ – Paul Frost Jan 16 at 23:32
  • $\begingroup$ And in examples such as the latter (different open covers getting identified), we want to identify those because as is shown in the next paragraph (not included above) they will produce isomorphic cohomology? So there is no harm in identifying them? If all we wanted to do was whittle the collection down to form a set, and only identify open covers who differ by choice of index set we could have required a bijection to exist between the index sets that yields a refinement in both directions between the covers. So what is our motivation for doing more, as Serre does, and only require $\endgroup$ – Prince M Jan 16 at 23:36
  • $\begingroup$ refinement in both directions? $\endgroup$ – Prince M Jan 16 at 23:41
  • $\begingroup$ In 1) we have seen that keeping only one representative from each equivalence class produces a cofinal subsystem which does not affect the direct limit. However, we do not really identify the groups associated to equivalent covers (I could not imagine how that should work), we only state that their contribution the direct limit is the same. Perhaps a naive analogy is that all subsequences of a convergent sequence have the same limit although you cannot identify the terms in two different subsequences. $\endgroup$ – Paul Frost Jan 17 at 0:03

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