Michael Shulman extensive 2-category

A coproduct $A+B$ in a 2-category is said to be disjoint if we have comma squares

$\array{ A^{\mathbf{2}} & \to & A & \quad & B^{\mathbf{2}} & \to & B & \quad & 0 & \to & A & \quad & 0 & \to & B \\ \downarrow & \Downarrow & \downarrow & \quad & \downarrow & \Downarrow & \downarrow & \quad & \downarrow & \Downarrow & \downarrow & \quad & \downarrow & \Downarrow & \downarrow & \quad\\ A & \to & A+B & \quad & B & \to & A+B & \quad & B & \to & A+B & \quad & A & \to & A+B & \quad}$

The first two say that $A\to A+B$ and $B\to A+B$ are ff and the second two say that they are disjoint subobjects of $A+B$.

A coproduct $A+B$ is said to be universal if for any morphism $Z\to A+B$, the pullbacks

$\array{X & \to & Z & \leftarrow & Y\\ \downarrow & & \downarrow & & \downarrow\\ A & \to & A+B & \leftarrow & B}$

exist and exhibit $Z$ as a coproduct $X+Y$.

Finally, we say a 2-category is extensive if it has finite coproducts which are disjoint and universal. If it also has finite limits we say it is lextensive, and if it is also coherent we call it positive. (Note that disjoint coproducts in a coherent 2-category are always universal.)

An extensive 2-category does satisfy 2-categorical versions of the additional characterizations of an extensive 1-category, but unlike in the 1-categorical case, these alternate conditions do not seem to suffice to characterize extensivity. In particular, though, a 1-category is extensive as a 1-category iff it is so as a homwise-discrete 2-category.

Preservation

If $K$ is extensive, so is $K^{co}$, obviously. Less obvious is:

Lemma

If $K$ is extensive, so are $gpd(K)$, $pos(K)$, and $disc(K)$. In other words, if $K$ is extensive, so is its $(n+1)$-category $trunc_n(K)$ of $n$-truncated objects for $0\le n$.

Proof

Since the three given categories are closed in $K$ under limits and strict initial objects, it suffices to show they are closed under coproducts. First suppose given two morphisms $f,g:Z\to A_1+A_2$. Then $f$ decomposes $Z=X_1+X_2$, and $g$ decomposes $Z=Y_1+Y_2$. Then the inclusions $X_i \to Z = Y_1+Y_2$ also decompose each $X_i=X_{i 1} + X_{i 2}$. Now if there exists a 2-cell $f\to g$, it induces a map from each $X_{i j}$ to the comma object of $A_1$ and $A_2$. Since coproducts are disjoint and initials are strict, this implies that $X_{12}=X_{21}=0$. Therefore, we have a decomposition $Z=X_{11}+X_{22}$ so that $f=f_1+f_2$ and $g=g_1+g_2$, where $f_i:X_{i i} \to A_i$ and $g_i:X_{i i} \to A_i$.

Now, by universality of the coproduct $X_{11}+X_{22}$, it follows that 2-cells $f\to g$ are determined uniquely by pairs of 2-cells $f_1\to g_1$ and $f_2\to g_2$. Therefore, if $A_1$ and $A_2$ are groupoidal, any 2-cells $f_1\to g_1$ and $f_2\to g_2$ are invertible, and thus so is any 2-cell $f\to g$; so $A_1+A_2$ is groupoidal. And if $A_1$ and $A_2$ are posetal, any parallel 2-cells $f_1 \;\rightrightarrows\; g_1$ and $f_2 \;\rightrightarrows\; g_2$ are equal, and thus so are any $f \;\rightrightarrows\; g$; so $A_1+A_2$ is posetal. And of course the discrete case follows by combining these.

However, the (0,1)-category (= poset) $Sub(1)$ of (-1)-truncated objects (= subterminal objects) does not inherit extensivity, and in fact posets are almost never extensive: the only disjoint coproduct is $0+0$.

We also have:

Lemma

If $K$ is extensive, so are the fibrational slices $Opf(X)$ and $Fib(X)$ for any $X\in K$.

Revised on February 17, 2009 17:56:29 by Mike Shulman (75.3.140.11)