Michael Shulman

In general, the idea is that an n-congruence in an m-category K, where nm, is an “internal (n1)-category” in K. Of course, we only deal formally with the case m2, although we allow n and m to be of the form (r,s); see n-prefix.


Let D be a 2-congruence in a 2-category K.

  • D is a (2,1)-congruence if it is an internal groupoid, i.e. there is a map D 1D 1 providing “inverses”.
  • It is a (1,2)-congruence if D 1D 0×D 0 is ff.
  • It is a 1-congruence if it is both a (2,1)-congruence and a (1,2)-congruence.
  • it is a (0,1)-congruence if D 1D 0×D 0 is an equivalence.

Note that in a 1-category,

  • a 2-congruence is just an internal category (a 1-category),
  • a (2,1)-congruence is an internal groupoid (a (1,0)-category),
  • a (1,2)-congruence is an internal poset (a (0,1)-category), and
  • a 1-congruence is an internal equivalence relation (a 0-category).

Of course, a (0,1)-congruence in any 2-category is completely determined by any object D 0.


Let q:XY be a morphism in K. If Y is n-truncated for n1, then ker(q) is an (n+1)-congruence. This means that:

  1. If Y is groupoidal, then ker(q) is a (2,1)-congruence.
  2. If Y is posetal, then ker(q) is a (1,2)-congruence.
  3. If Y is discrete, then ker(q) is a 1-congruence.
  4. If Y is subterminal, then ker(q) is a (0,1)-congruence.

In all these cases the converse is true if K is regular and q is eso.


The forward directions are fairly obvious; it is the converses which take work. Suppose first that ker(q) is a (2,1)-congruence, and let α:fg:XY be any 2-cell. Pulling back the eso q along f and g gives P 1T and P 2T; let r:PT be the pullback P 1× XP 2. Since K is regular, r is eso. By definition of kernels, the 2-cell αr corresponds to a map P(q/q). But (q/q)C is a (2,1)-congruence, so composing this map with the “inverse” map (q/q)(q/q) gives another map P(q/q), and thereby another 2-cell frgr which is inverse to αr. Finally, since r is eso, precomposing with it reflects invertibility, so α must also be invertible. Thus Y is groupoidal.

Now suppose that ker(q) is a (1,2)-congruence, and let α,β:fg:TY be two parallel 2-cells. With notation as in the previous paragraph, the 2-cells αr and βr correspond to morphisms P(q/q) which become isomorphic in X. But since (q/q)X is a (1,2)-congruence, this implies that the two maps P(q/q) are isomorphic, and hence αr=βr. And since r is eso, precomposing with it is faithful, so α=β; thus Y is posetal.

The discrete case follows by combining the posetal and groupoidal cases, so it remains to show that if ker(q) is a (0,1)-congruence then Y is subterminal. We know it is discrete, so it suffices to show that given two f,g:TY we have a 2-cell fg. Continuing with the same notation, and letting h,k:PX be the induced maps with qhfr and qkgr, we have (h,k):PX×X=(q/q), and therefore the 2-cell defining the fork (q/q)XqY gives us a 2-cell qhqk and therefore frgr. Now r is the quotient of its kernel, so for this 2-cell to induce a 2-cell fg it suffices for it to be an action 2-cell for the actions of ker(r) on fr and gr; but this is automatic since we know Y to be posetal. Thus we have a 2-cell fg as desired, so Y is subterminal.

Revised on February 17, 2009 17:49:22 by Mike Shulman (