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.

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

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 $0\le n$.

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:

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

Last revised on February 17, 2009 at 17:56:29. See the history of this page for a list of all contributions to it.