If possible, how could one prove that every short exact sequence $0 \to A \xrightarrow f B \xrightarrow g C \to 0$ of vector spaces (here $A$, $B$ and $C$) splits without using any basis of $A$, $B$ or $C$. I was not able to exhibit a morphism $h \colon B \to A$ such that $h \circ f=Id_A$ without considering a basis.
[Math] Exact short sequences of vector spaces
linear algebrashort-exact-sequences
Related Solutions
The answer is no, I think. Here is a proof sketch. (An unclear point in a previous version has now been removed, by slightly modifying the construction of the sequence.)
Let $(S_n)_{n\in\omega}$ be a family of ``pairs of socks''; that is, each $S_n$ has 2 elements, the $S_n$ are disjoint, but there is no set which meets infinitely many $S_n$ in exactly one point. Let $S$ be the union of the $S_n$.
Let $V$ be a vector space with basis $S$ over the 3-element field. For each $v\in V$, each $s\in S$ let $c_s(v)$ be the $s$-coordinate of $v$. (In your notation: $v(s)$.)
Consider the subspace $W$ of all vectors $w$ with the following property: For all $n$, if $S_n = \{a,b\}$, then $c_a(w)+c_b(w)=0$. The set of all $n$ such that for both/any $a\in S_n$ we have $c_a(w) \neq0$ will be called the domain of $w$. Clearly, each domain is finite, and for each finite subset of $\omega$ of size $k$ there are $2^k$ vectors $w\in W$ with this domain.
[Revised version from here on.]
I will show
- From any basis $C$ of $W$ we can define a 1-1 sequence of elements of $W$.
- From any 1-1 sequence of elements of $W$ we can define a 1-1 sequence of elements of $S$. Together, this will show that there is no basis, as $S$ contains no countably infinite set.
For each set $D$ which appears as the domain of a basis vector, let $x_D$ be the sum of all basis vectors with this domain. So $x_D \neq 0$, and for $D\neq D'$ we get $x_D\neq x_{D'}$. From a well-order of the finite subsets of $\omega$ we thus obtain a well-ordered sequence of nonzero vectors. Since there must be infinitely many basis vectors, and only finitely many can share the same set $D$, we have obtained an infinite sequence of vectors in $W$.
We are now given an infinite sequence $(w_n)$ of distinct vectors of $W$. The union of their domains cannot be finite, so we may wlog assume that the sequence $k_n:= \max(dom(w_n))$ is strictly increasing. (Thin out, if necessary.)
Now let $a_n$ be the element of $S_{k_n}$ be such that $c_{a_n}(w_n)=1$. Then the set of those $a_n$ meets infinitely many of the $S_k$ in a singleton.
Hi John. I'd say there is no generalization of short exact sequence to the category of monoids, although I suppose it really depends on what you want to do with it. What you probably want is an internal equivalence relation. So you could say a diagram $A\rightrightarrows B \to C$ of monoid maps (where the two compositions $A\rightrightarrows C$ agree) is short exact if the map $B\to C$ is surjective and the induced map $A\to B\times_C B$ is an isomorphism. This is equivalent to requiring the induced map $A\to B\times B$ to be injective, its image to be an equivalence relation, and the induced map $B/A\to C$ is to be an isomorphism.
My general feeling is that this is the right concept in most categories of sets with algebraic structure (e.g. the category of sets itself, semi-rings). It's only in categories where the objects have some group structure that you can re-express it using kernels.
Best Answer
You can prove this using Zorn's lemma on the pairs (D,h) where D is a subspace of B and h is a partial section.
And yes, you do need Zorn's lemma: without it, there may exist vector spaces none of whose nontrivial subspaces has a complement (Herrlich, Axiom of Choice, LNM 1876, Disaster 4.43).