Suppose 0 \rightarrow F \rightarrow G \rightarrow H \rightarrow 0 is a short exact sequence of quasicoherent sheaves.     If two out of three of \{ F, G, H \} are locally free, what can we say about the third?  

If F and G are locally free, then H need not be.    This is a fundamental example.

If F and H are locally free, then G is too.    This is important and not too hard.

If G and H are locally free and finite rank, then F is too.  This is also useful.

But if G and H are locally free, and infinite rank, then F need not be locally free.  The purpose of this post is to give Daniel Litt‘s explanation to me of a simple example of this.  I thought this example was not right to include in the Rising Sea (it would be distracting), but I wanted to preserve it for posterity (and for myself).

I will describe a surjective map of free modules whose kernel is not locally free. Viewed geometrically, this provides a surjective map of trivial (infinite-dimensional) vector bundles whose kernel is not a vector bundle. Of course, the kernel is projective.

Let M be a projective module over a ring R which is not locally free.  (Such things exist, see for example the Stacks Project tag 05WG.  To be explicit, there R = \prod_{i \in {\mathbb Z}^+} {\mathbb F}_2 and M is the ideal ( e_n : e_n = (1,1,   \cdots ,1,0,0, \cdots) ) (where there are n 1’s).)

Then M is a direct summand of a free module, say M \oplus  M' = R^{\oplus I}.  Writing

R^{ \oplus I'} = (M\oplus M') \oplus (M \oplus M') \oplus \cdots =M \oplus (M' \oplus M) \oplus (M' \oplus M) \oplus \cdots =M \oplus R^{\oplus J'}

gives M as a direct summand of a free module with free complement. Then there is a short exact sequence

0 \rightarrow M \rightarrow R^{\oplus I'} \rightarrow R^{\oplus J'} \rightarrow 0

exhibiting M as the kernel of a surjective map of free modules, where the map R^{\oplus I'} \rightarrow  R^{\oplus J'} is the projection.