Category of Banach spaces

Solution 1:

$\text{Ban}_1$ does have cokernels! The cokernel of a map $f : X \to Y$ of Banach spaces is the quotient of $Y$ by the closure of $\text{im}(f)$. It's true that the quotient $Y/\text{im}(f)$ isn't necessarily a Banach space, but that doesn't imply that cokernels don't exist, only that they aren't preserved by the forgetful functor to vector spaces.

$\text{Ban}_1$ is still not an abelian category because it isn't enriched over abelian groups, but it's very close: we can't arbitrarily add morphisms but we can take linear combinations $c_1 f_1 + \dots + c_n f_n$ such that $|c_1| + \dots + |c_n| \le 1$ (see totally convex space), and this is still good enough to e.g. reduce the computation of (co)equalizers to (co)kernels, etc.

Anyway, your questions:

Q1. Yes, and in fact, as Mark Kamsma says in the comments, $\text{Ban}_1$ is both complete and cocomplete. You can find a proof, for example, in this blog post.

Q2. Yes, although some care needs to be taken because some of the equivalences between different definitions of a short exact sequence that are valid in abelian categories fail here. I'm not sure which is standard in the Banach space literature.

I found one paper do basically the obvious thing and define an exact sequence using the usual kernel = image definition of exactness. The main subtlety here is that one could imagine defining the image categorically as either the cokernel of the kernel or the kernel of the cokernel, and these are not equivalent. The former is the regular coimage and the latter the regular image. The regular coimage of a map $f : X \to Y$ is the set-theoretic image with the quotient norm inherited from $X$, and the regular image is the closure of the set-theoretic image with the subspace norm inherited from $Y$. For the "kernel = image" definition of exactness we want the regular coimage.

There's a further question of whether we want "equals" to mean that the induced map from the image to the kernel is a set-theoretic bijection or an isomorphism in $\text{Ban}_1$ (hence an isometry). I don't have an abstract answer to this question; it depends on what you want to consider short exact sequences for.

Q3: I'm not sure how to answer this question without more clarity on what you want short exact sequences for. In this paper you can find, in the Remark after Definition 2.4, a counterexample showing that filtered colimits in $\text{Ban}_1$ aren't exact.