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(n,r)-congruence

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Idea

In general, the idea is that an kk-congruence in an n-category KK, where knk\le n, is an “internal (k1)(k-1)-category” in KK.

Here we consider the case m2m\le 2, although we allow nn and mm to be of the form (r,s)(r,s); see (n,r)-category.

Definition

Definition

Let DD be a 2-congruence in a 2-category KK. * DD is a (2,1)-congruence if it is an internal groupoid, i.e. there is a map D 1D 1D_1\to D_1 providing “inverses”. * It is a (1,2)-congruence if D 1D 0×D 0D_1\to D_0\times 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 0D_1\to D_0\times 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 0D_0.

Theorem

Let q:XYq:X\to Y be a morphism in KK. If YY is nn-truncated for n1n\ge -1, then ker(q)ker(q) is an (n+1)(n+1)-congruence. This means that:

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

In all these cases the converse is true if KK is regular and qq is eso.

Proof

The forward directions are fairly obvious; it is the converses which take work. Suppose first that ker(q)ker(q) is a (2,1)-congruence, and let α:fg:XY\alpha: f \to g: X \rightrightarrows Y be any 2-cell. Pulling back the eso qq along ff and gg gives P 1TP_1\to T and P 2TP_2\to T; let r:PTr:P \to T be the pullback P 1× XP 2P_1\times_X P_2. Since KK is regular, rr is eso. By definition of kernels, the 2-cell αr\alpha r corresponds to a map P(q/q)P\to (q/q). But (q/q)C(q/q)\rightrightarrows C is a (2,1)-congruence, so composing this map with the “inverse” map (q/q)(q/q)(q/q)\to(q/q) gives another map P(q/q)P\to (q/q), and thereby another 2-cell frgrf r \to g r which is inverse to αr\alpha r. Finally, since rr is eso, precomposing with it reflects invertibility, so α\alpha must also be invertible. Thus YY is groupoidal.

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

The discrete case follows by combining the posetal and groupoidal cases, so it remains to show that if ker(q)ker(q) is a (0,1)-congruence then YY is subterminal. We know it is discrete, so it suffices to show that given two f,g:TYf,g:T \rightrightarrows Y we have a 2-cell fgf\to g. Continuing with the same notation, and letting h,k:PXh,k:P\to X be the induced maps with qhfrq h \cong f r and qkgrq k \cong g r, we have (h,k):PX×X=(q/q)(h,k):P\to X\times X = (q/q), and therefore the 2-cell defining the fork (q/q)XqY(q/q) \;\rightrightarrows\;X \overset{q}{\to} Y gives us a 2-cell qhqkq h \to q k and therefore frgrf r \to g r. Now rr is the quotient of its kernel, so for this 2-cell to induce a 2-cell fgf\to g it suffices for it to be an action 2-cell for the actions of ker(r)ker(r) on frf r and grg r; but this is automatic since we know YY to be posetal. Thus we have a 2-cell fgf\to g as desired, so YY is subterminal.

References

Revised on August 24, 2012 20:25:45 by Urs Schreiber (82.113.106.24)