Finite choices

The axiom of choice says that an arbitrary product $\prod_{i\in I} A_i$ of non-empty sets $A_i$ indexed by a set $I$ is non-empty. It is well known that this axiom does not follow from the other axioms of Zermelo-Fraenkel theory. Even finite choices, that is, this statement restricted to the case where all sets are finite, is not a consequence. Even 2-choices, when one assumes that $A_i$ has two elements!

For each integer $n$, call  ${\rm AC}(n)$ the axiom of choice restricted to families $(A_i)$ where $A_i$ has $n$ elements. 

Tarski proved the funny following fact: ${\rm AC}(2) \Rightarrow {\rm AC}(4)$—if you know how to choose between 2 elements, you can choose between 4.

The proof is in fact quite easy. Consider a family $(A_i)$ of sets with 4 elements. I will use choice functions furnished by ${\rm AC}(2)$ to pick-up one preferred element from $A_i$. For simplicity, label the elements of $A_i$ as $\{a,b,c,d\}$ and remove the index $i$. Then, consider the set  $\{\{a,b\},\{a,c\},\{a,d\},\{b,c\},\{b,d\},\{c,d\}\}$ of all pairs of elements of $A_i$. The hypothesis ${\rm AC}(2)$ allows to choose, for each of those pairs, one preferred element. Call $n_a,n_b,n_c,n_d$ the number of times $a,b,c,d$ has been chosen; one thus has $n_a+n_b+n_c+n_d=6$ and consider those elements which have been chosen the most often, those for which $n_?$ is maximal.

• If there is only one, let's choose it. (This happens in repartitions like $(3,1,1,1)$, etc.)
• If there are three such elements (the repartition must be $(2,2,2,0)$), let's choose the unique one which has never been chosen.
• There can't be four such elements because 4 does not divides 6.
• If there are two (repartition $(2,2,1,1)$), then use your 2-choice function on this pair!

The other funny, but more difficult, thing, is that ${\rm AC}(2)$ does not imply ${\rm AC}(3)$! Why? because the group $\{\pm1\}$ can act without fixed points on a 2-elements set but cannot on a 3-elements set.  I hope to be able to say more on this another day.

Misconceptions about $K_X$

This is the title of a very short paper by Steven Kleiman published in L'enseignement mathématique, and which should be studied by every young student in scheme theory.

Here, $X$ is a scheme and $K_X$ is the sheaf of rational functions on $X$.

The misconceptions are the following, where we write $\mathop{Frac}(A)$ for the total ring of fractions of a ring $A$, namely the localized ring with respect to all element which are not zero divisors.

1. $K_X$ is not the sheaf associated to the presheaf $U\mapsto \mathop{Frac}(\Gamma(U,O_X))$; indeed, that map may not be a presheaf.
2. The germ $K_{X,x}$ of $K_X$ at a point $x$ may not be the total ring of fractions of the local ring $O_{X,x}$, it may be smaller.
3. If $U=\mathop{Spec}(A)$ is an affine open subset of $X$, then $\Gamma(U,K_X)$ is not necessarily equal to $\mathop{Frac}(A)$.

These mistakes can be found in the writings of very good authors, even Grothendieck's EGA IV... 

By chance, the first one is corrected in a straightforward way, and the other two work when the scheme $X$ is locally noetherian.

Thanks to Antoine D. for indicating to me this mistake, and to Google for leading me to Kleiman's paper.

The category of sets and its opposite

In the book Categories and sheaves by Kashiwara and Shapira, I found a nice argument for the fact that the category of sets is not equivalent to its opposite: they write « every morphism to the initial object is an isomorphism ». Of course!

In the category Sets, the initial object is the empty set, which means that for every set $A$, there is a unique map from $\emptyset$ to $A$. Now, if we reverse the process, namely, if we consider a set $A$ and a map $f\colon A\to \emptyset$, we see that $A$ must be empty and $f$ is a bijection, hence an isomorphism in the category of sets.

In the opposite category, all arrows are reversed, the initial object becomes the terminal object, etc. Isomorphisms are (reversed) maps which have an inverse, so isomorphisms are still given by bijections. A terminal object of Sets is one-element set, $\{x\}$ (you could take the set $1=\{\emptyset\}$ if, like von Neumann, you believe that numbers are sets). Indeed, there is a unique map from any set to a one-element set. Reverse the process again and consider a set $A$ and a map $f\colon \{x\} \to A$. This amounts to choosing an element of $A$, but such maps are not bijections in general, unless $A$ has itself only one element.

The van Kampen Theorem

Let me recall the statement of this theorem.

Theorem. Let $X$ be a topological space, let $U,V$ be connected open subsets of $X$ such that $W=U\cap V$ is connected and let $x$ be a point of $U\cap V$. Then, the fundamental group $\pi_1(X,x)$ is the amalgamated product $\pi_1(U,x) *_{\pi_1(W,x)} \pi_1(V,x)$, that is, the quotient of the free product of the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ by the normal subgroup generated by the elements of the form $i_U(c)i_V(c)^{-1}$, where $i_U$ and $i_V$ are the natural injections from the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ respectively in their free product.

The classical proof of this result in topology books relies decomposes a loop at $x$ as a product of loops at $x$ which are either contained in $U$, or in $V$.

(In fact, van Kampen proves a theorem which is quite different at first sight.)

It has been long recognized that there is a completely different approach is possible, from which all loops are totally absent. For this proof we make a supplementary assumption, namely that our spaces are « semi-locally simply connected » : Any point $a$ of $X$ has a neighborhood $A$ such that the morphism $\pi_1(A,a)\to \pi_1(X,a)$ is trivial.

When $X$ is a connected slsc space together with a point $x$, the theory of the fundamental group is related to the theory of coverings,under the form of an equivalence of categories between coverings of $X$ and sets with an action of $\pi_1(X,x)$. The equivalence of categories is explicit; it maps a covering $p\colon Y\to X$ to the fiber $p^{-1}(x)$ on which $\pi_1(X,x)$ acts naturally via the path-lifting property of coverings (given $y\in p^{-1}(x)$, any loop $c$ at $x$ lifts uniquely to a path with origin $y$, the endpoint of which is $c\cdot y$).

Given this equivalence, one can prove the van Kampen Theorem very easily in two steps. First of all, one observes that it is equivalent to have a covering of $X$ as to have a covering of $U$ and a covering of $V$ together with an identification of these coverings above $W$. A covering of $U$ corresponds to a set $A$ with an action of $\pi_1(U,x)$; a covering of $V$ corresponds to a set $B$ with an action of $\pi_1(V,x)$; an identification of these coverings above $W$ corresponds to a bijection from $A$ to $B$ which is compatible with the two actions of $\pi_1(W,x)$ acting on $A$ via the morphism $\pi_1(W,x)\to \pi_1(U,x)$ and on $B$ via the morphism $\pi_1(W,x)\to \pi_1(V,x)$. It is harmless to assume that $A=B$ and that the bijection from $A$ to $B$ is the identity. Now, a covering of $X$ corresponds to a set $A$ together with two actions of the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ such that the two actions of $\pi_1(W,x)$ are equal. This is precisely the same as a set $A$ together with an action of the amalgamated product $\pi_1(U,x)*_{\pi_1(W,x)}\pi_1(V,x)$. CQFD.

The same proof applies and allows to describe the fundamental group of an union of spaces in more general contexts. For example, let us use the same method to understand the fundamental group of the circle $\mathbf S_1$. It is clear that a circle is nothing but an interval $ [0,1]$ of which the two endpoints are glued, and a covering of the circle corresponds to a covering $p\colon X\to [0,1]$ of the interval $[0,1]$ together with an identification of the fibers at $0$ and~$1$. Now, a covering of the interval can be written as a product $A\times [0,1]$ (where $A$ is the fiber at $0$, say). Consequently, identifying the fibers at $0$ and $1$ means giving yourself a bijection of $A$ to $A$. In other words, a covering of the circle « is »  a set $A$ together with a permutation of $A$, in other words, a set $A$ with an action of the additive group $\mathbf Z$. Moreover, the obvious loop is the image by the glueing map $[0,1]\to\mathbf S_1$ of the obvious path joining $0$ to $1$ so that this loop is the generator of $\pi_1(\mathbf S_1,p(0))$.

Observe that the latter example is not an instance of the van Kampen Theorem. One could get it via a groupoid-version of van Kampen.

All of this is more or less explained in the following texts:

• Adrien and Régine Douady, Algèbre et théories galoisiennes, Cassini 2005.
• Ronald Brown, Topology and Groupoids, Booksurge Publishing, 2006.
• I remember having read an old Bourbaki Tribu from the 50sby Cartan, Eilenberg and/or Weil explaining this, but I cannot find it anymore on the archive. :-(