Michael Shulman
extensive 2-category

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

A 2 A B 2 B 0 A 0 B A A+B B A+B B A+B A A+B \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 AA+BA\to A+B and BA+BB\to A+B are ff and the second two say that they are disjoint subobjects of A+BA+B.

A coproduct A+BA+B is said to be universal if for any morphism ZA+BZ\to A+B, the pullbacks

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

exist and exhibit ZZ as a coproduct X+YX+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 KK is extensive, so is K coK^{co}, obviously. Less obvious is:

Lemma

If KK is extensive, so are gpd(K)gpd(K), pos(K)pos(K), and disc(K)disc(K). In other words, if KK is extensive, so is its (n+1)(n+1)-category trunc n(K)trunc_n(K) of nn-truncated objects for 0n0\le n.

Proof

Since the three given categories are closed in KK under limits and strict initial objects, it suffices to show they are closed under coproducts. First suppose given two morphisms f,g:ZA 1+A 2f,g:Z\to A_1+A_2. Then ff decomposes Z=X 1+X 2Z=X_1+X_2, and gg decomposes Z=Y 1+Y 2Z=Y_1+Y_2. Then the inclusions X iZ=Y 1+Y 2X_i \to Z = Y_1+Y_2 also decompose each X i=X i1+X i2X_i=X_{i 1} + X_{i 2}. Now if there exists a 2-cell fgf\to g, it induces a map from each X ijX_{i j} to the comma object of A 1A_1 and A 2A_2. Since coproducts are disjoint and initials are strict, this implies that X 12=X 21=0X_{12}=X_{21}=0. Therefore, we have a decomposition Z=X 11+X 22Z=X_{11}+X_{22} so that f=f 1+f 2f=f_1+f_2 and g=g 1+g 2g=g_1+g_2, where f i:X iiA if_i:X_{i i} \to A_i and g i:X iiA ig_i:X_{i i} \to A_i.

Now, by universality of the coproduct X 11+X 22X_{11}+X_{22}, it follows that 2-cells fgf\to g are determined uniquely by pairs of 2-cells f 1g 1f_1\to g_1 and f 2g 2f_2\to g_2. Therefore, if A 1A_1 and A 2A_2 are groupoidal, any 2-cells f 1g 1f_1\to g_1 and f 2g 2f_2\to g_2 are invertible, and thus so is any 2-cell fgf\to g; so A 1+A 2A_1+A_2 is groupoidal. And if A 1A_1 and A 2A_2 are posetal, any parallel 2-cells f 1g 1f_1 \;\rightrightarrows\; g_1 and f 2g 2f_2 \;\rightrightarrows\; g_2 are equal, and thus so are any fgf \;\rightrightarrows\; g; so A 1+A 2A_1+A_2 is posetal. And of course the discrete case follows by combining these.

However, the (0,1)-category (= poset) Sub(1)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+00+0.

We also have:

Lemma

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

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