@BD107 No, because the police bureau can place the officers wherever they wish. If the thief could escape no matter the officer placements from some point $p$, they can escape from any point by first heading to $p$.

For a K3 surface with no points: Does the Fermat quartic $w^4+x^4+y^4+z^4=0$ over $\mathbb{F}_5$ work for you?

@Iteraf A direct summand of a zero object must also be a zero object. (Indeed, if there were nontrivial maps to or from it, that would give nontrivial maps to or from the zero object.) So triviality of $X\oplus (-1)\cdot X$ implies triviality of $X$.

@Iteraf I think you will get better answers if you align the details in your question with the details in your comment.

@Iteraf The zero object is unique (up to isomorphism.) I'm not sure what category you are thinking of with both $0$ and $\mathbb{Z}$ (there are a few possible categories); in each of them only one of those two is a zero object. Re the Krull-Schmidt remark: Set $A=0\cdot X$. Then your axiom (b) gives $A=A\oplus A.$ In a Krull-Schmidt category this implies $A=0$, which in turn implies $X=0$ by the logic of my previous comment.

On the other hand, if what you are really interested in is the case where $R$ is a semi-ring, this is a natural notion (and recovers the case of group actions that you mention). Could you clarify which case you are interested in?

Can you give an example of such a category? In particular, the following seems strange to me: For any $X$ we have $(1+(-1))\cdot X=0\cdot X$, so if $X$ is nontrivial then $0\cdot X$ must also be nontrivial. This seems hard to reconcile with your axioms, though I don't see an immediate contradiction. (It certainly cannot hold in a Krull-Schmidt category.)

It may be helpful to note that the original setting in which all of these theorems arose was for p-adic groups. If you take $G$ a split group over $\mathbb{Q}_p$, then the maximal compact subgroup of $G(\mathbb{Q}_p)$ is $G(\mathbb{Z}_p)$. For loop groups, $G(\mathbb{Q}_p)$ is replaced by $LG$ and $G(\mathbb{Z}_p)$ is replaced by $L^+G$. The resulting decomposition is called "Cartan decomposition" by analogy to the case of real Lie groups, but I don't know of any uniform proof in all settings.

The Cartan decomposition is a decomposition of a group (here $LG$) as a product $KAK$, where $K$ is a maximal compact and $A$ is a "maximal torus". $L^+G$ is a maximal compact of $LG$, hence the relation. That being said, I personally find the fact you stated easier to think about in the context of Bruhat decomposition, which gives another proof.

To rephrase, it sounds like you are considering a moduli space of maps from $\mathbb{A}^1$ to $X$. There are lots of papers studying this for specific $X$. But it's quite hard to make any positive statements for general $X$. If $X$ is smooth then you should be able to write down an expected dimension for $X(d)$, which may or may not match with the actual dimension (and if it does, this will be at best a nontrivial theorem and at worst a conjecture beyond current technology.) Same for irreducibility.

(I know this is tautological.) I've never found it useful to learn subjects of math that I don't care for just because they were supposed to be "important." If they really are important, usually I eventually see those tools used in some very cool way, and then I feel motivated to learn them.

...There have been a lot of comments on the interconnectedness of mathematics, and how categorical notions appear in other subjects, etc. A more indirect (but IMO more important) argument: every field has some standard collection of proof strategies and philosophies. Learning other fields exposers you to a wider variety. E.g. I have a purely algebraic AG paper where the main argument is inspired by the use of Sobolev spaces to construct Donaldson invariants. On the other hand, my experience is that this sort of inspiration only happens when you find the field in question inspiring... (cont.)

Two conflicting comments: 1. I think that the "minimal required amount of knowledge" about any subject for a mathematician, even one working in that subject, is quite a bit lower than often claimed. E.g. Most algebraic geometers I know would only be able to read 10% or so of what gets published in the math.AG tag on arxiv. A significant portion of them know very little category theory, despite AG being more categorical than most subjects. If you really don't like any one subject, it's not that hard to avoid it. 2. You can draw inspiration from almost any part of math... (cont.)

Phrased like this the crux is to show that the loci in $\mathbb{A}^n$ corresponding to having eigenvalues in $D_1\cup D_2\cdots\cup D_n$ is disconnected and has the correct connected components. This can be seen by looking at the preimage of this loci under the map $(\mathbb{A}^1)^n\rightarrow\mathbb{A}^n.$

It is the quotient topology. Alternatively, there is an isomorphism $\operatorname{Sym}^n(\mathbb{A}^1)\cong\mathbb{A}^n$ (this is the statement that elementary symmetric polynomials generate all symmetric polynomials) and so the map $\mathbb{A}^1\rightarrow\mathbb{A}^n$ comes from taking coefficients of the characteristic polynomial and so is easily seen to be continuous.